# 1 Opening items

## 1.1 Module introduction

Oscillations occur in many branches of physics, but in this module we will examine just two: mechanical systems and electrical circuits. At first sight a mass oscillating on a spring and the tuning circuit of a radio appear to have little in common; but the mathematics that models them is almost indistinguishable, and both can be described in terms of a second-order differential equation with constant coefficients. In Section 2 of this module we examine the way in which such equations arise and consider some of the oscillatory phenomena their solutions represent. In particular we look at simple, damped and driven oscillations, and we pay particular attention to the way in which solutions of the latter kind typically consist of a steady state term that oscillates and gradually becomes dominant, and a transient term that may be important initially but gradually dies away and eventually becomes insignificant. In some circumstances the amplitude of these dominant driven oscillations can become very large; this is the phenomenon of resonance which we consider in Subsection 2.7.

Very often it is only the steady state behaviour of the system that is of interest, and we may then assume that the transient term is zero. In such cases we are able to abandon the differential equation approach, and use a much simpler method based on complex numbers. This technique is particularly relevant to the analysis of alternating currents in electrical circuits, and in Section 3 we use it to develop a complex version of Ohm’s law. We will see that the current and the applied voltage oscillate at the same rate, but they do not necessarily do so in phase due to the complex impedance of the circuit concerned. In Subsection 3.3 we use complex methods to calculate the power dissipated in an electrical circuit, and in the final section we examine how complex numbers may be used to combine simple harmonic motions.

Study comment Having read the introduction you may feel that you are already familiar with the material covered by this module and that you do not need to study it. If so, try the following Fast track questions. If not, proceed directly to the Subsection 1.3Ready to study? Subsection.

## 1.2 Fast track questions

Study comment Can you answer the following Fast track questions? If you answer the questions successfully you need only glance through the module before looking at the Subsection 5.1Module summary and the Subsection 5.2Achievements. If you are sure that you can meet each of these achievements, try the Subsection 5.3Exit test. If you have difficulty with only one or two of the questions you should follow the guidance given in the answers and read the relevant parts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised to study the whole module.

Question F1

Write down the differential equation for the current I(t) in a circuit containing a resistance R, a capacitance C and an inductance L in series, that is driven by an applied voltage V0sin(Ωt). What is the general form of the steady state solution to this equation?

$L\dfrac{d^2I(t)}{dt^2} + R\dfrac{dI(t)}{dt} + \dfrac1CI(t) = \dfrac{dV(t)}{dt}$

This is derived as Equation 19 in the solution to a P2-questionSeed Question.

In the steady state, I(t) = I0sin(Ωtδ) where general expressions for I0 and δ are given by Equation 63 in the text:

$I(t) = \dfrac{V_0}{Z}\sin({\it\Omega}t - \delta)$(Eqn 63a)

$Z^2 = R^2 + \left(\dfrac{1}{{\it\Omega}C}-{\it\Omega}L\right)^2\;\text{and}\;\tan\delta = \dfrac1R\left({\it\Omega}L-\dfrac{1}{{\it\Omega}C}\right)$(Eqn 63b)

Question F2

If ${\mathcal Z} = R + i\omega L + 1/(i\omega C)$ (where R, ω, L and C are all real and positive) find expressions for ${\rm Re}({\mathcal Z}),\;{\rm Im}({\mathcal Z})\text{ and }\lvert{\mathcal Z}\rvert$. For what value of ω does $\lvert{\mathcal Z}\rvert$ take its least value?

Write down the principal values of the arguments of R, ${\mathcal Z}_L\text{ and }{\mathcal Z}_C$, where

${\mathcal Z}_L = i\omega L\quad\text{and}\quad{\mathcal Z}_C = 1/(i\omega C)$

and illustrate these complex numbers on an Argand diagram.

${\mathcal Z} = R + i\left(\omega L -\dfrac{1}{\omega C}\right)$ and therefore, since R, ω, L and C are real,

${\rm Re}({\mathcal Z}) = R\quad\text{and}\quad{\rm Im}({\mathcal Z}) = \omega L -\dfrac{1}{\omega C}$

Also $\lvert\,{\mathcal Z}\,\rvert = \sqrt{R^2 + \left(\omega L -\dfrac{1}{\omega C}\right)^2}$ and this takes its least value when

$\left(\omega L -\dfrac{1}{\omega C}\right)^2$, i.e. when $\omega = 1/\sqrt{LC\os}$.

The arguments of R, iωL and 1/(iωC) are 0, π/2 and −π/02, respectively, and they are illustrated in Figure 8a, except that ω has been replaced by Ω there.

Figure 8 Geometrical interpretations of Z and δ in terms of XL, XC and R.

Question F3

An inductance of 3.00 H and a capacitance 0.10 F are connected in parallel, and this combination is then connected in series with a resistance of 5.00 Ω. Find the current that passes through the resistor when a voltage V(t) = acos(Ωt), where a = 4.00 V and Ω = 3.00 Hz, is applied to the circuit.

If $\mathcal Z$ is the complex impedance of the capacitor and inductance in parallel, then

$\dfrac{1}{\mathcal Z} = \dfrac{1}{{\mathcal Z}_C} + \dfrac{1}{{\mathcal Z}_L} = i{\it\Omega}C + \dfrac{1}{i{\it\Omega}L} = i{\it\Omega}C - \dfrac{i}{{\it\Omega}L} = 0.3i-\dfrac i9 \approx 0.189i\,{\Omega}^{-1}$

and therefore ${\mathcal Z} = -5.29i\,\Omega$. The impedance of the circuit is therefore $R + {\mathcal Z} \approx (5 - 5.29i)\,\Omega$. The current is the real part of

$\dfrac{4\exp[i(3.00\,{\rm s^{-1}})t]}{5-5.29i}\,{\rm A} \approx [0.55\exp(0.81i)]\times\exp[i(3.00\,{\rm s^{-1}})t]\,{\rm A} = 0.55\exp\left\{\left[(3.00\,{\rm s^{-1}})t+0.81\right]i\right\}\,{\rm A}$

so thatI(t) ≈ 0.55cos[(3.00 s−1)t + 0.81] A

Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route through the module and to proceed directly to the following Ready to study? Subsection.

Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the Section 5Closing items.

Study comment This module is intended to form a link between the maths and physics strands of FLAP. It therefore makes much heavier mathematical demands than most modules, though it assumes less knowledge of physics. To begin the study of this module you will need to be familiar with the solution of second–order differential equations with constant coefficients (though we provide a brief review of this topic in Subsection 2.3), and you should also know how such equations arise from Newton’s second law of motion. You should be able to manipulate trigonometric identities. You should also be familiar with the Cartesian coordinate system, complex numbers, including their exponential_representation_of_a_complex_numberexponential representation (z = reiθ), the Argand diagram and the real part, imaginary part, modulus_of_a_complex_numbermodulus argument and complex conjugate of a complex number (although we provide you with a short summary of the subject in Subsection 3.1). You should also be able to differentiate and integrate standard functions such as sin(x), cos(x) and exp(x). A familiarity with geometric progressions would also be useful, although not essential. It would be helpful if you have seen how oscillatoryoscillatory systems arise in physics, but we assume no prior knowledge in this area. If you are unfamiliar with any of these topics you can review them by referring to the Glossary, which will indicate where in FLAP they are developed. The following Ready to study questions will help you to establish whether you need to review some of the above topics before embarking on this module.

Question R1

Sketch the graph of y = 3sin(ωt + δ) for:

(a) ω = 2 s−1   and   δ = 0,   (b) ω = 2 s−1   and   δ = −π/2.

How would the first graph you drew change if:

(c) ω = 2 s−1   and   δ = −4,   (d) ω = 2 s−1   and   δ = 4.

(e) Describe (without drawing a diagram) the graph of y = sin(ωt + π/2).

(a) A sketch of y = 3sin[(2 s−1)t] is shown as the solid line in Figure 16. The period of this curve is π seconds.

(b) When δ = −π/2 we have y = 3sin[(2 s−1)tπ/2], and the curve moves a quarter of a period (π/4 seconds) to the right, as shown by the dashed line. It may then be described by the expression

$y = 3\sin\left[\omega\left(t-\dfrac{\pi}{2\omega}\right)\right]$

(c) When δ = −4, y = 3sin(ωt − 4) = 3sin[ω(t − 4/ω)] and the graph moves 4/ω units to the right along the t–axis.

(d) When δ = 4, y = 3sin(ωt + 4) = 3sin[ω(t + 4/ω)] and the graph moves 4/ω units to the left along the t–axis.

(e) The graph of y = sin(ωt + π/2) is the same shape as y = sin(ωt) but shifted π/(2ω) units to the left along the t–axis. However sin(ωt + π/2) = cos(ωt) so that the graph is actually a cosine curve.

Consult amplitude, angular frequency and phase in the Glossary for further information.

Question R2

Calculate the value of ϕ given that

$\cos\phi = \dfrac{2}{\sqrt{2^2+3^2}}\quad\text{and}\quad\sin\phi = \dfrac{-3}{\sqrt{2^2+3^2}}$ i

Use the trigonometric identity cos(A + B) = cosAcosB − sinAsinB to express 2cos(ωt) + 3sin(ωt) in the form Rcos(ωt + ϕ).

From Figure 17 we have ϕ ≈ −0.9828.

It is easy to check that:

$\cos(-0.9828) \approx 0.5547 \approx \dfrac{2}{\sqrt{2^2+3^2}}$

and$\sin(-0.9828) \approx -0.8321 \approx \dfrac{-3}{\sqrt{2^2+3^2}}$

\begin{align}2\cos(\omega t)+3\sin(\omega t) & = \sqrt{13\os}\left(\dfrac{2}{\sqrt{13\os}}\cos(\omega t) - \dfrac{-3}{\sqrt{13\os}}\sin(\omega t\right) \\ & = \sqrt{13\os}\left[\cos\phi\cos(\omega t) - \sin\phi\sin(\omega t)\right] \\ & = \sqrt{13\os}\cos(\omega t+\phi) \\ & = \sqrt{13\os}\cos[\omega t+(-0.9828)]\end{align}

Consult trigonometric identities in the Glossary for further information.

Question R3

Solve the quadratic equation h2 + 5h + 6 = 0.

The quadratic equation can be factorized as follows:

h2 + 5h + 6 = (h + 2)(h + 3) = 0

so the equation has solutions h1 = −2 and h2 = −3.

These solutions may also be obtained by means of the standard expression

$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Consult quadratic equation in the Glossary for further information.

Question R4

If z is defined by z = 4e5/4, what are the principal values of arg(z), |z|, Re(z) and Im(z)? If you are unsure about any of these terms consult complex numbers in the Glossary.

To find the principal value of arg(z) we need z in the form 4e where −π < θπ. Since e = 1, we can express z as follows:

z = 4exp(5/4) × exp(−8/4) = 4exp(−3/4)

and the principal value of arg(z) is −3π/4 and |z| = 4.

Using Euler’s formula (e = cosθ + isinθ), we also have:

$z = 4\left[\cos\left(\dfrac{3\pi}{4}\right) - i\sin\left(\dfrac{3\pi}{4}\right)\right]$

with${\rm Re}(z) = 4\cos\left(\dfrac{3\pi}{4}\right)$

and${\rm Im}(z) = -4\sin\left(\dfrac{3\pi}{4}\right)$

Consult the relevant terms in the Glossary for further information.

Question R5

(Optional) What is the sum of the geometric series 1 + r + r2 + r3 + ... + rn−1

What is the sum for the particular case of r = i (where i2 = −1) and n = 9?

$\displaystyle 1 + r + r^2 + r^3 + \ldots + r^{n-1} = \sum_{m=0}^{n-1}r^m = \dfrac{1-r^n}{1-r}$

For r = i and n = 9 we find

$\dfrac{1-i^9}{1-i} = \dfrac{1-i}{1-i} = 1$

Consult geometric series and summation in the Glossary for further information.

# 2 Oscillations and differential equations

## 2.1 Introducing mechanical and electrical oscillations

Figure 1 shows a small body of mass m held between two stretched springs on a smooth horizontal table. Under the influence of the springs, the body is able to move to and fro along a line that we will take to be the x–axis of a system of Cartesian coordinates.

The equilibrium position of the body will be taken to be the point x = 0, so the position coordinate of the body at any time t determines its displacement from equilibrium at that time.

If the body is released from rest at a point slightly to the right of its equilibrium position at some initial time t = 0 (diagram A), it will subsequently oscillate back and forth about its equilibrium position, as indicated in diagrams B, C, D and E.

As a result, the instantaneous position of the body will be a function of time and can be denoted by x(t). The dashed line in Figure 1 is a graphical representation of this function, although it would be more natural for us to draw the graph of x(t) with the t–axis horizontal and the x–axis vertical, as in Figure 2a (with a suitably scaled t–axis to make the figure a manageable size). In the absence of dissipationdissipative effects, such as friction or air resistance, the total energy (kinetic_energykinetic plus potentialpotential_energy) of the oscillator will be constant, and the displacement–time graph of Figure 2a will be sinusoidal (i.e. of the same general shape as the graph of a sine or cosine function). A displacement–time graph of this kind is characteristic of the particular kind of motion known as simple harmonic motion (SHM) which occurs in many branches of physics and engineering. The function whose graph is shown in Figure 2a may be represented algebraically by an expression of the form

x(t) = A0sin(ω0t + π/2)

and is a special case of the class of functions that provide the most general mathematical description of simple harmonic motion

x(t) = A0sin(ω0t + ϕ)(1) i

where A0, ω0 and ϕ are constants that characterize the motion and are, respectively, referred to as the amplitude, the angular frequency and the phase constant (or initial phase) of the oscillation. The amplitude is equal to the magnitude of the maximum displacement from equilibrium that occurs during each cycle of oscillation.

The angular frequency is related to the period T (i.e. the time required for one complete oscillation such as that from A to E in Figure 1) and to the frequency f by the relation

ω0 = 2π/T = 2πf

Thus the frequency (f = 1/T) is the number of oscillations per second, and the angular frequency is just 2π times that value. The phase constant determines the value of x at t = 0, since x(0) = A0sin(ϕ). Note that the alternative name for ϕ, the initial phase, arises because the quantity (ω0t + ϕ) which determines the stage that the oscillator has reached in its cycle at any time t is called the phase, and ϕ is simply the value of the phase at t = 0.

Figure 2b shows the effect of changing the initial phase of an oscillator. The equation describing the dashed line was given above as x(t) = A0sin(ω0t + π/2), so it corresponds to a phase constant ϕ = π/2. This may be contrasted with the equation describing the solid curve, which may be written

x(t) = Asin(ωt + π/2 − ω0t0)

and which corresponds to a phase constant ϕ = π/2 − ω0t0. The quantity t0 indicates the extent to which the behaviour of the oscillator represented by the solid curve lags behind that represented by the dashed curve. We can therefore say that there is a phase difference between the two oscillators, and that the former (represented by the solid curve) lags the latter (represented by the dashed curve) by ω0t0.

In practice, an oscillating system of the kind shown in Figure 1 would be subject to friction and other dissipative effects and would lose energy to its environment. As a result of this energy transfer, the maximum displacement attained during each oscillation generally tends to decrease with time, resulting in the sort of damped oscillations indicated in Figure 2c. Provided the damping is sufficiently light it is possible to describe this kind of oscillation in a similar way to the simple harmonic oscillation described above. Of course, the description is not exactly the same; the damping generally tends to reduce the angular frequency from ω0 to some lower value ω, and causes the amplitude to become a (decreasing) function of time A(t), but apart from these changes we can often describe the damped oscillations by a function of the form

x(t) = A(t)sin(ωt + ϕ)(2)

Now consider the system shown in Figure 3a, in which a body of mass m is attached to one end of a horizontal spring, the other end of which is attached to a fixed point P. The body can slide back and forth along a straight line, which we will again take to be the x–axis, but this time it is subject to an externally imposed force (acting along the x–axis) in addition to the force due to the spring and any dissipative force that may act. In this situation the externally imposed force is called a driving force and the oscillations that it helps to produce and sustain in the oscillator are called forced or driven oscillations. If the driving force varies sinusoidally with time, at angular frequency Ω, i so that it may be described by an expression of the form F0sin(Ωt), we will eventually find that the motion of the oscillating body under the influence of the driving force is described by

x(t) = Asin(Ωtδ)(3) i

where A and δ are constants whose values depend on the angular frequency of the driving force, Ω, and the characteristics of the oscillator, but are independent of time. The steady nature of the eventual motion shows that in this case work done by the driving force is somehow compensating the oscillator for the energy it loses due to dissipative effects.

✦ What physical interpretation can you give to the parameters A and δ that appear in Equation 3?

A represents the amplitude of the forced oscillations. δ represents the difference in phase between the steady state displacement oscillation, and the driving force. Because of the minus sign in Equation 3,

x(t) = Asin(Ωtδ)(Eqn 3)

δ actually measures the extent by which the displacement oscillation lags behind the force, as we might expect it to do on physical grounds.

(If, despite our expectations, the displacement oscillations led the force we would have to assign a negative value to δ.)

Oscillations, whether simple, damped or driven are not confined to mechanical systems. All sorts of physical systems exhibit oscillations. Temperatures may oscillate from day to day or season to season; concentrations of different chemicals may rise and fall in oscillating chemical reactions; electric charges may oscillate back and forth in appropriately constructed electrical circuits; and so on. The properties of electrical oscillations are particularly important and provide interesting analogies with mechanical oscillations. We will now briefly describe some of the situations in which electrical oscillations arise, and then investigate the reasons why such apparently different systems should exhibit such closely similar behaviour.

An electric circuit is a closed path around which electric charge may flow. A typical circuit, such as that of Figure 3b, contains a number of electrical components that assist or retard the flow of charge and thereby give the circuit its particular characteristics. In order for the flow of charge to occur at all there must generally be a potential difference between one part of the circuit and another; this is measured in volts (V, where 1 V = 1 J C−1) and is often referred to as a voltage. It might be supplied by a battery, but in the case of Figure 3b there is a voltage generator, shown by the symbol at the top of the diagram, which produces a time dependent potential difference V(t) between its terminals. The instantaneous rate of flow of charge at any point in the circuit constitutes the instantaneous current I(t) at that point, and may be measured in amperes (A, where 1 A = 1 C s−1). The conventional current direction is taken to be that of positive charge flow. The rest of the circuit shown in Figure 3b consists of a resistor (shown as the rectangle), a capacitor (shown as parallel bars) and an inductor (shown as the coil), connected in series, so that the same current flows through each component. Such circuits are called series LCR circuits.

A resistor is a component that dissipates energy. When there is a potential difference VR across a resistor, a current will flow through it. The current and the voltage will be related by Ohm’s law

VR(t) = I(t)R(4)

where R is a constant, called the resistance of the resistor, which may be measured in ohms (Ω, where 1 Ω = 1 V A−1). i The energy dissipated per second (i.e. the power dissipated) when a current I flows through a resistor of resistance R is

P = I2R = VR2/R = IVR(5)

A capacitor is a device for storing electrical charge (and thereby storing energy in the associated electric field). The simplest such device consists of a pair of parallel metal plates placed a short distance from one another and the symbol used to represent the capacitor reflects this. When charged, these plates carry charges of equal magnitude but opposite sign, +q and −q. The charge q is measured in coulomb (C, where 1 C = 1 A s), and is related to the voltage VC across a capacitor by the relation

$V_C(t) = \dfrac{q(t)}{C}$(6)

where the constant C that characterizes the capacitor is called its capacitance and is measured in farad (F, where 1 F = 1 A s V−1).

An inductor is another device that can be used to store energy, in the magnetic field produced when a current flows through it. In the case of an inductor, the instantaneous voltage across the inductor is proportional to the rate of change of the instantaneous current through the inductor, hence

$V_L (t) = L\dfrac{dI(t)}{dt}$(7)

where the constant L that characterizes the inductor is called its inductance and is measured in henry (H, where 1 H = 1 V s A−1). The direction or polarity of the induced voltage is always such as to oppose the change which causes it – this is known as Lenz’s law.

The quantities VR, VC, VL that have been introduced above have polarities and so may each be positive or negative. We must ensure that we understand the significance of the signs of these quantities. Figure 3b shows the situation at the instant where the capacitor charge is increasing on the upper plate and VC is growing with the polarity shown. We are also illustrating the case where the current is increasing in the direction shown, so Lenz’s law gives the polarity of VL as being opposite to the increasing current direction.

The voltage VR has a polarity opposite to the direction of the current. Each of these instantaneous voltages and that of the generator are conveniently shown by arrows, with the arrows pointing in the direction of increasing positive voltage as shown. A positive value for the voltage across the capacitor implies that the upper plate is at a higher voltage than the lower plate and a positive value for VL and VR implies that in VR(t) each case the end nearer to the generator is positive.

If we let q represent the instantaneous charge on the upper plate of the capacitor, then q will be positive when the capacitor voltage is positive. Moreover, if the instantaneous current is positive, then positive charge will be flowing onto the upper plate of the capacitor and q will be increasing, this tells us nothing about the sign of q but it does ensure that

$I = \dfrac{dq}{dt}$(8)

As you can see from the above discussion, the circuit shown in Figure 3b will be characterized by the relevant values of R, C and L. Given these three values and the specific form of the externally supplied voltage V(t), it is possible to determine the current I(t) that flows through the circuit, and the associated charge q(t) on the upper plate of the capacitor. Interestingly, the circuit turns out to be an electrical analogue of the driven mechanical oscillator shown in Figure 3a. In particular, if the external voltage is of the form V(t) = V0sin(Ωt), then eventually, after any transient currents have died away:

q(t) = Asin(Ωtδ)(9)

Consequently,

$I(t) = \dfrac{dq}{dt} = A{\it\Omega}\cos({\it\Omega}t - \delta)$(10)

If you compare Equation 9 with Equation 3,

x(t) = Asin(Ωtδ)(Eqn 3)

you will see that they are both of the same form. It is in this sense that the charge oscillations in the series LCR circuit driven by an externally supplied sinusoidal voltage may be said to be analogous to the displacement oscillations of the mechanical oscillator driven by an externally supplied sinusoidal force.

In the next subsection you will see why these two very different physical systems give rise to essentially identical oscillatory phenomena. The essential point is that the underlying physics of both systems is described by very similar equations.

## 2.2 Mathematical models of mechanical and electrical oscillators

### A driven mechanical oscillator

The oscillating body in Figure 3a is subject to three forces:

1

A restoring force F1x due to the spring that tends to return the body to its equilibrium position. This will be taken to be:

F1x(t) = −kx(t)(11a)

where k is the positive spring constant that characterizes the spring.

2

A damping force F2x, due to friction and air resistance, that opposes the motion of the body. We will assume that the magnitude of this force is proportional to the instantaneous velocity of the sliding body, so that:

$F_{2x}(t) = -b\dfrac{dx(t)}{dt}$(11b)

where b is a positive constant that characterizes the dissipative forces.

3

A driving force F3x provided by an external agency. We will assume that this force varies with time in a periodic way, and has the relatively simple form:

F3x(t) = F0sin(Ωt)(11c)

where F0 is the maximum magnitude that the driving force attains, and Ω is the angular frequency of the driving force. Note that the angular frequency Ω is externally imposed and is not necessarily related in any way to the natural frequency of the system in Figure 3a.

Using Newton’s second law of motion we can therefore say that the sliding body must obey an equation of motion of the form:

$m\dfrac{d^2x(t)}{dt^2} = F_{1x}(t) + F_{2x}(t) + F_{3x}(t)$

so, in this case:

$m\dfrac{d^2x(t)}{dt^2} = -kx(t) - b\dfrac{dx(t)}{dt} + F_0\sin({\it\Omega}t)$(12)

which can be rearranged to isolate the time–dependent driving term as follows:

$m\dfrac{d^2x(t)}{dt^2} + kx(t) + b\dfrac{dx(t)}{dt} = F_0\sin({\it\Omega}t)$(13)

This is known as the equation of motion of a harmonically driven linearly damped oscillator. The function x(t) = Asin(Ωtδ) that we introduced in Subsection 2.1 (Equation 3) to describe the steady state behaviour of the driven oscillator is a solution of this equation, provided we choose A and δ appropriately, as we will demonstrate in the next subsection.

### A series LCR circuit

In order to determine the differential equation that describes the behaviour of the series LCR circuit of Figure 3b we need to introduce two basic principles of circuit analysis (based on Kirchhoff’s laws):

1

In a series circuit the instantaneous current I(t) through each component is the same. The physical basis for this is the principle of the conservation of electric charge.

2

In a series circuit the sum of the instantaneous voltages across each passive component i is equal to the externally supplied voltage V(t). The physical basis for this is the principle of the conservation of energy.

Now, we already know from Equations 4, 6 and 7 that

VR(t) = I(t)R(Eqn 4)

$V_C(t) = \dfrac{q(t)}{C}$(Eqn 6)

$V_L(t) = L\dfrac{dI(t)}{dt}$(Eqn 7)

So we can use the second of the two principles given above (conservation of energy) to write:

V(t) = VR(t) + VC(t) + VL(t)

Using the first principle (conservation of electric charge), together with Equations 4, 6 and 7, this gives us

$V(t) = RI(t) + \dfrac{q(t)}{C} + L\dfrac{dI(t)}{dt}$(14)

However, we also know that the current in the circuit is given by the rate of change of the charge q on the upper plate of the capacitor, so we can write

$I(t) = \dfrac{dq(t)}{dt}$(Eqn 8)

and hence   $dI(t) = \dfrac{d^2q(t)}{dt^2}$(15)

Substituting Equations 8 and 15 into Equation 14 we see that

$V(t) = R\dfrac{dq(t)}{dt} + \dfrac1C q(t) + L\dfrac{d^2q}{dt^2}$

which may be rearranged to give

$L\dfrac{d^2q}{dt^2} + R\dfrac{dq(t)}{dt} + \dfrac1C q(t) = V(t)$(16)

Finally, substituting the relevant expression for V(t) we obtain

$L\dfrac{d^2q}{dt^2} + R\dfrac{dq(t)}{dt} + \dfrac1C q(t) = V_0\sin({\it\Omega}t)$(17)

Now, if you compare Equation 13,

$m\dfrac{d^2x(t)}{dt^2} + kx(t) + b\dfrac{dx(t)}{dt} = F_0\sin({\it\Omega}t)$(Eqn 13)

and Equation 17 you will see that they have the same form. One may be obtained from the other by making the following substitutions:

qx   Lm i   Rb   1/Ck   V0F0

Note that we are not claiming some sort of mystical link between charge and displacement, or inductance and mass, but simply drawing attention to the fact that the two very different physical systems can both be described by similar equations. It is the mathematical model that is the same in both cases, not the system it is representing.

Equations 13 and 17 are examples (essentially the same example) of second–order differential equations coefficientwith constant coefficients. (From a mathematical point of view they are also linear and inhomogeneous_differential_equationinhomogeneous.) Solving such equations is an inherently mathematical process, but it is of great interest to physicists since the possible solutions include the various forms of harmonic motion that were described in Subsection 2.1.

✦ By differentiating Equation 16,

$\dfrac{d^2q}{dt^2} + R\dfrac{dq(t)}{dt} + \dfrac1C q(t) = V(t)$(Eqn 16)

with respect to time, show that the instantaneous current in a series LCR circuit obeys a differential equation similar to that satisfied by the charge q(t).

✧ Differentiating Equation 16 gives

$L\dfrac{d^3q(t)}{dt^3} + R\dfrac{d^2q(t)}{dt^2} + \dfrac1C \dfrac{dq(t)}{dt} = \dfrac{dV(t)}{dt}$(18)

However dq/dt = I, so we can rewrite this as

$L\dfrac{d^2I(t)}{dt^2} + R\dfrac{dI(t)}{dt} + \dfrac1C I(t) = \dfrac{dV(t)}{dt}$(19)

Although this equation has dV/dt on the right–hand side rather than V(t), it is still the case that the right–hand side represents a given function of t. It is therefore correct to say that the current in the series LCR circuit obeys a similar (though not necessarily identical) equation to the charge.

## 2.3 Second–order differential equations – a brief review

This subsection describes the mathematical principles involved in solving second–order differential equations coefficientwith constant coefficients. This topic is discussed from the same point of view, but in greater detail, in the maths strand of FLAP.

The general linear second–order differential equation with constant coefficients is of the form:

$a\dfrac{d^2x(t)}{dt^2} + b\dfrac{dx(t)}{dt} + cx(t) = f(t)$(20) i

where a, b and c are constants and f(t) is independent of x.

If f(t) = 0 for all values of t the equation is said to be homogeneous_differential_equationhomogeneous, otherwise the equation is said to be inhomogeneous_differential_equationinhomogeneous. The equation is said to be linear because the dependent variable x only appears once, and only to the first power, in each of the terms that involves it at all – there are no terms involving x2 or (dx/dt)2 or x(dx/dt) or anything else of that kind. The equation is second_order_differential_equationsecond order because it involves no derivative of x higher than the second derivative.

As you can see, Equations 13 and 17,

$m\dfrac{d^2x(t)}{dt^2} + kx(t) + b\dfrac{dx(t)}{dt} = F_0\sin({\it\Omega}t)$(Eqn 13)

$\dfrac{d^2q}{dt^2} + R\dfrac{dq(t)}{dt} + \dfrac1C q(t) = V_0\sin({\it\Omega}t)$(Eqn 17)

are both of this general form.

In the cases that are of interest to us the constants a, b, and c are all positive, and the function f(t) corresponds to the external driving term.

When confronted with an equation such as Equation 20 our usual aim is to find its general solution. For such a second–order differential equation this general solution expresses x in terms of t, the given constants that appear in the equation and two additional arbitrary constants. The values of the arbitrary constants cannot be determined from Equation 20 itself but must be found from supplementary conditions such as the initial values of x and its derivative, x(0) and $\dfrac{dx(0)}{dt}$. These supplementary conditions are generally referred to as boundary conditions or initial conditions as appropriate.

In the case of Equation 20 the general solution is the sum of two parts, a particular solution xp(t), which may be any solution of Equation 20 that does not contain arbitrary constants, and a complementary function xc(t) which does contain two arbitrary constants, and which satisfies the corresponding homogeneous equation:

$a\dfrac{d^2x(t)}{dt^2} + b\dfrac{dx(t)}{dt} + cx(t) = 0$(21)

Question T1

Show that x(t) = xc(t) + xp(t) will satisfy Equation 20 and will contain two arbitrary constants, as the general solution should.

We know that x(t) = xc(t) + xp(t) contains just two arbitrary constants, because there are two such constants in the complementary function xc(t) and none in the particular solution xp(t). Moreover, we know that

$a\dfrac{d^2x_{\rm c}(t)}{dt^2} + b\dfrac{dx_{\rm c}(t)}{dt} + cx_{\rm c}(t) = 0$

and$a\dfrac{d^2x_{\rm p}(t)}{dt^2} + b\dfrac{dx_{\rm p}(t)}{dt} + cx_{\rm p}(t) = f(t)$

Adding corresponding terms in these two equations and using the sum rule of differentiation we can write:

$a\dfrac{d^2[x_{\rm p}(t)+x_{\rm p}(t)]}{dt^2} + b\dfrac{[x_{\rm p}(t)+x_{\rm p}(t)]}{dt} + c[x_{\rm p}(t)+x_{\rm p}(t)] = f(t)$

Thus we can also see that x(t) = xc(t) + xp(t) is a solution of Equation 20,

$a\dfrac{d^2x(t)}{dt^2} + b\dfrac{dx(t)}{dt} + cx(t) = f(t)$(Eqn 20)

We will now outline the procedure for determining the complementary function in any give case, and then comment on the determination of a particular solution.

### Finding the complementary function

1

Given the values of a, b and c in Equation 21 write down the so–called auxiliary equation:

ap2 + bp + c = 0

2

Find the solutions p1 and p2 (of this quadratic equation, see Question R3).

3

(a) If b2 > 4ac the solutions will be two different real numbers and the complementary function will be:

xc(t) = Bexp(p1t) + Dexp(p2t)(22) i

(b) If b2 < 4ac the solutions will be two different complex numbers which may be written:

p1 = −γ +   and   p2 = −γ

whereγ = b/a   and   $\omega = \sqrt{\dfrac ca - \dfrac{\gamma^2}{4}}$(23)

and the complementary function will be:

xc(t) = eγt/2[Ecos(ωt) + Gsin(ωt)](24)

(c) If b2 = 4ac the solutions will be two equal real numbers p1 = p2 = −b/2a = −γ/2, and the complementary function will be:

xc(t) = (H + Jt)eγt/2(25)

### Finding a particular solution

Determining a particular solution is generally much more difficult, and usually comes down to educated guesswork. However, in the cases that will be of interest to us in this module the driving term f(t) will usually have the general form

f(t) = f0sin(Ωt)(26)

and the particular solution will have the corresponding form

xp(t) = Asin(Ωtδ)(27)

where$A = \dfrac{f_0}{\sqrt{\smash[b]{(c/a-{\it\Omega}^2)+(\gamma{\it\Omega})^2}}}\quad\text{and}\quad\delta = \arctan\left(\dfrac{\gamma{\it\Omega}}{c/a-{\it\Omega}^2}\right)$(28) i

Note that the constants A and δ appearing in the particular solution are not arbitrary constants; their values are determined by the given values of a, b, c, f0 and Ω and not by any initial or boundary conditions.

Note also that for the homogeneous equation the particular solution can always be set equal to zero since f0 can then be set equal to zero, so A = 0.

Using Equations 22 to 28 it is now possible to solve a wide a range of oscillatory problems.

### Example 1

Write down the complementary function, a particular solution, and finally the general solution of the differential equation

$L\dfrac{d^2q(t)}{dt^2} + R\dfrac{dq(t)}{dt} + \dfrac{q(t)}{C} = V_0\sin({\it\Omega}t)$(29)

when L = 1.0 H, R = 5.0 Ω, C = 1/6 F, V0 = 0.6 V and Ω = 5.0 s−1.

#### Solution

Comparing Equation 29 with Equation 20,

$a\dfrac{d^2x(t)}{dt^2} + b\dfrac{dx(t)}{dt} + cx(t) = f(t)$(Eqn 20)

and identifying a with L, b with R and c with 1/C, we see that in this case b2 > 4ac. Solving the auxiliary equation leads to p1 = −2 s−1 and p2 = −3 s−1 so the complementary function takes the form of Equation 22,

xc(t) = Bexp(p1t) + Dexp(p2t)(Eqn 22)

and is given by

qc(t) = Bexp[−(2s − 1)t]+ Dexp[−(3s − 1)t]

where B and D are arbitrary constants.

A particular solution is qp(t) = Asin[(5 s−1)tδ] where, from Equation 28,

$A = \dfrac{f_0}{\sqrt{\smash[b]{(c/a-{\it\Omega}^2)+(\gamma{\it\Omega})^2}}}\quad\text{and}\quad\delta = \left(\dfrac{\gamma{\it\Omega}}{c/a-{\it\Omega}^2}\right)$(Eqn 28)

A = 1.9 × 10−2 C   and   δ = −0.92 i

The general solution of the differential equation is therefore:

q(t) = qc(t) + qp(t) = Bexp[−(2s−1)t] + Dexp[−(3s−1)t] + (1.9×10−2 C)sin[(5s−1)t + 0.92]

Note that in this case the charge q(t) and hence Vc(t) lead the applied voltage V0 = Asin(Ωt).

The constants B and D in the above example are determined by the initial state of the system, i.e. the initial charge on the capacitor and the initial current. In practice however their values are usually unimportant, as the following question invites you to show.

✦ Suppose that in Example 1 we have B = 0.6 C and D = 0.6 C. Calculate the values of q(t), qc(t) and qp(t) when t = 0 s, t = 1 s, t = 3 s, t = 5 s and t = 8 s. What do you notice about the values of q(t), qc(t) and qp(t) as t increases? What part do B and D play in determining the eventual behaviour of q(t)?

Table 1 The contribution to the charge
on the capacitor from the complementary
function and the particular solution for
various values of t.
t/s q(t)/C qc(t)/C qp(t)/C
0 1.185 1.200 −0.015
1 0.096 0.111 −0.015
3 0.021 0.002 0.019
5 −0.017 0.000 −0.017
8 0.019 0.000 0.019

✧ The value qc(t) rapidly tends to zero as t increases (see Table 1), so that the values of q(t) and qp(t) become indistinguishable for large values of t. Hence B and D play no part in determining the eventual behaviour of q(t).

Figure 4 shows the graphs of qp(t) (the dashed sinusoidal curve) and q(t) = qc(t) + qp(t) (the solid curve). Initially they are quite different, but for t > 4 s they are indistinguishable. Because of this behaviour qp(t) is said to represent the steady state behaviour, whereas qc(t) is said to represent the transient behaviour.

The important points to notice from Example 1 are:

• the transient part of the general solution is insignificant for large values of t, so that eventually the steady state term Asin(Ωtδ) will dominate the solution;
• the constants B, D, E, G, H and J determine the initial state of the system, but if we are only interested in what happens for large values of t their values are irrelevant;
• the steady state term Asin(Ωtδ) is a sinusoidal function with the same angular frequency as the driving term V(t).

### An important consequence

Figure 3 (a) A mass subject to restoring, damping and driving forces.

If we start to drive the mechanical oscillator shown in Figure 3a, or close a switch so as to complete the circuit shown in Figure 3b, there will be a short period of time when the transient behaviour is significant, but in most practical situations the behaviour rapidly moves to a steady state in which the oscillation is sinusoidal. If we are only interested in the steady state, which is usually the case, then we can totally ignore the transient term in the solution of the differential equation. Even more significantly, if we are only interested in the steady state behaviour, we can often abandon this approach entirely and avoid the difficult business of solving differential equations altogether. In Section 3 we will introduce a much simpler method of analysing harmonically driven oscillators based on the assumption that the oscillation is sinusoidal.

Question T2

A mechanical oscillator satisfies the differential equation:

$a\dfrac{d^2x(t)}{dt^2} + b\dfrac{dx(t)}{dt} + cx(t) = 0$

where a = 1 s2, b = 4 s and c = 3.

Write down the auxiliary equation and solve it. Hence find the general solution of this (homogeneous) differential equation.

The auxiliary equation is (1 s2)h2 + (4 s)h + 3 = 0 which factorizes to give (h + 1 s)(h + 3 s) = 0. Thus h1 = −1 s and h2 = −3 s. The general solution of the differential equation is therefore in the form of Equation 22,

xc(t) = Bexp(p1t) + Dexp(p2t)(Eqn 22)

x(t) = Be(−1 s)t + De(−3 s)t

Since the equation is homogeneous, a particular solution is x(t) = 0.

## 2.4 Harmonic oscillations: simple, damped and driven

In this subsection we consider some special cases of the second–order linear differential equation:

$a\dfrac{d^2x(t)}{dt^2} + b\dfrac{dx(t)}{dt} + cx(t) = f_0\sin({\it\Omega}t)$(30)

Each of the cases we consider will correspond to a particular kind of oscillatory behaviour, and may be applied to mechanical or electrical systems (or any other system that may be similarly modelled).

### b = 0, f0 = 0; the case of simple harmonic oscillation

In this case Equation 30 may be written in the form

$a\dfrac{d^2x(t)}{dt^2} + cx(t) = 0$   where a ≠ 0

It is conventional to rewrite this by dividing both sides by a and introducing

the natural angular frequency   $\omega_0 = \sqrt{c/a\os}$(31)

so that$\dfrac{d^2x(t)}{dt^2} + \omega_0^2x(t) = 0$(32)

In this (homogeneous) case a particular solution is xp(t) = 0. Moreover, b2 < 4ac, so the complementary function, and hence the general solution, takes the form of Equation 24 with γ = b/c = 0, and $\omega = \sqrt{c/a\os} = \omega_0$

i.e.x(t) = Ecos(ω0t) + Gsin(ω0t)(33)

✦ Show that this solution can be written in the equivalent form

x(t) = A0sin(ω0t + ϕ)(34)

and hence confirm that it can be used to represent simple harmonic motion with amplitude A0, angular frequency $\omega_0 = \sqrt{c/a\os}$ and phase constant ϕ.

✧ Using the trigonometric identity sin(θ + ϕ) = sinθcosϕ + cosθsinϕ, Equation 34 can be written in the form:

x(t) = A0sin (ω0t)cosϕ + A0cos(ω0t)sinϕ

i.e.x(t) = Ecos(ω0t) + Gsin (ω0t)

as required, with E = A0sinϕ and G = A0cosϕ.

Question T3

A simple series circuit consists of a capacitor connected in series with an inductor. If the charge on the capacitor at time t = 0 is q0, and there is no current in the circuit at that time, determine the differential equation that describes the variation of q with time, write down its general solution, and show that the charge exhibits simple harmonic oscillations with angular frequency $\omega_0 = \sqrt{1/(LC)\os}$.

From Equation 17,

$L\dfrac{d^2q}{dt^2} + R\dfrac{dq(t)}{dt} + \dfrac1C q(t) = V_0\sin({\it\Omega}t)$(Eqn 17)

with R = 0 and V0 = 0,

$\dfrac{d^2q}{dt^2} + \dfrac{q(t)}{C} = 0$   i.e. $\dfrac{d^2q(t)}{dt^2} + \omega_0^2q(t) = 0$

This is the equation of simple harmonic oscillations with angular frequency $\omega_0 = \sqrt{1/LC\os}$, which has the general solution:

q(t) = A0sin(ω0t + ϕ)

In this case we know that q(0) = q0 and $\dfrac{dq(0)}{dt} = 0$, so

A0sinϕ = q0   and   A0ω0cosϕ = 0

It follows that   ϕ = π/2 and A0 = q0.

### f0 = 0; the case of linearly damped oscillation

In this case Equation 30,

$a\dfrac{d^2x(t)}{dt^2} + b\dfrac{dx(t)}{dt} + cx(t) = f_0\sin({\it\Omega}t)$(Eqn 30)

may be written in the form

$a\dfrac{d^2x(t)}{dt^2} + b\dfrac{dx(t)}{dt} + cx(t) = 0$(35)

It is conventional to rewrite this by dividing both sides by a and introducing

the damping constant   γ = b/a(36a)

and the natural angular frequency $\omega_0 = \sqrt{c/a\os}$(36b)

so that$\dfrac{d^2x(t)}{dt^2} + \gamma\dfrac{dx(t)}{dt} + \omega_0^2 x(t) = 0$(36c)

In this case a particular solution is xp(t) = 0, and the complementary function may take any of the forms described by Equations 22, 24 or 25,

xc(t) = Bexp(p1t) + Dexp(p2t)(Eqn 22)

xc(t) = eγt/2[Ecos(ωt) + Gsin(ωt)](Eqn 24)

xc(t) = (H + Jt)eγt/2(Eqn 25)

depending on the values of γ and ω0:

(a) If $\gamma^2 \gt 4\omega_0^2$ the oscillator is said to be overdamped and the general solution has the form

x(t) = Bexp(p1t) + Cexp(p2t)(37)

where$p_1 = \dfrac{-\gamma}{2} + 4\sqrt{\dfrac{\gamma^2}{4}-\omega_0^2}\quad\text{and}\quad p_2 = \dfrac{-\gamma}{2} + 4\sqrt{\dfrac{\gamma^2}{4}-\omega_0^2}$(38)

(b) If $\gamma^2 \lt 4\omega_0^2$ the oscillator is said to be underdamped and the general solution has the form

x(t) = eγt/2[Ecos(ωt) + Gsin(ωt)](39)

where$\omega = \sqrt{\omega_0^2-\dfrac{\gamma^2}{4}}$(40)

(c) If $\gamma^2 \lt 4\omega_0^2$ the oscillator is said to be critically damped and the general solution has the form

x(t) = (H + Jt)eγt/2(41) i

Some typical examples of damped oscillatory behaviour are shown in Figure 5.

Notice that only in the underdamped case is it mathematically justified to claim that there is a well defined angular frequency associated with the oscillation since only then is at least one full cycle completed. Also notice that the vibrations decrease more rapidly as the value of b (and hence γ) is increased.

In practical situations there is always resistance in a circuit, or friction in a mechanical system, so we are generally justified in assuming that b > 0.

Question T4

Show that in the case of underdamped oscillations, the general solution may be written in the form:

x(t) = eγt/2[A0sin(ωt + ϕ)](42)

This only involves showing that A0sin(ωt + ϕ) may be written Ecos(ωt) + Gsin(ωt), which is certainly true since

A0sin(ωt + ϕ) = A0sin(ωt)cos(ϕ) + A0cos(ωt)sin(ϕ) = Ecos(ωt) + Gsin(ωt)

## The general case of harmonically driven linearly damped oscillation

In this case we take Equation 30,

$a\dfrac{d^2x(t)}{dt^2} + b\dfrac{dx(t)}{dt} + cx(t) = f_0\sin({\it\Omega}t)$(Eqn 30)

and, as with the undriven case, it is conventional to rewrite this by dividing both sides by a and introducing

the damping constant   γ = b/a(43)

the natural angular frequency $\omega_0 = \sqrt{c/a\os}$(44)

anda0 = f0/a(45)

so that$\dfrac{d^2x(t)}{dt^2} + \gamma\dfrac{dx(t)}{dt} + \omega_0^2x(t) = a_0\sin({\it\Omega}t)$(46)

In this case the general solution is the sum of a transient term (a complementary function) given by the appropriate solution to the linearly damped oscillator equation, and a steady state term (particular solution) as given by Equations 27 and 28,

xp(t) = Asin(Ωtδ)(Eqn 27)

$A = \dfrac{f_0}{\sqrt{\smash[b]{(c/a-{\it\Omega}^2)+(\gamma{\it\Omega})^2}}}\;\;\text{and}\;\;\delta = \arctan\left(\dfrac{\gamma{\it\Omega}}{c/a-{\it\Omega}^2}\right)$(Eqn 28)

with f0 replaced by a0.

Thus, in the physically important case of underdamping

x(t) = eγt/2[A0sin(ωt + ϕ)] + Asin(Ωtδ)(47)

where A0 and ϕ are arbitrary constants, $\omega = \sqrt{\omega_0^2-\dfrac{\gamma^2}{4}}$(48)

$A = \dfrac{a_0}{\sqrt{\smash[b]{(\omega_0^2-{\it\Omega}^2)+(\gamma{\it\Omega})^2}}}\;\;\text{and}\;\;\delta = \arctan\left(\dfrac{\gamma{\it\Omega}}{\omega_0^2-{\it\Omega}^2}\right)$(49)

Note that as t increases the relative importance of the first term on the right in Equation 47 decreases, so in the steady state the solution is effectively

x(t) = Asin(Ωtδ)(50)

where Ω is the driving angular frequency, and δ is the extent to which the phase of the oscillator lags behind that of its driver. Once again, note that A and δ are not arbitrary constants but are determined by the given values of γ, ω0, Ω and a0.

✦ A series circuit of negligible total resistance consists of a switch, a 2.4 V battery, a capacitor of capacitance 0.01 F and an inductor of inductance of 5.0 H. Write down the differential equation that determines the current I(t) at a time t after the circuit is completed. Write down an expression for I(t) assuming that the capacitor is initially uncharged.

✧ The general case for a driven LCR circuit is given by Equation 19,

$L\dfrac{d^2I(t)}{dt^2} + R\dfrac{dI(t)}{dt} + \dfrac1CI(t) = \dfrac{dV(t)}{dt}$(Eqn 19)

In our case R = 0 and V(t) = 2.4 V = constant, so this differential equation becomes that for undamped SHM:

$L\dfrac{d^2I(t)}{dt^2} + \dfrac1CI(t) = 0$

or$L\dfrac{d^2I(t)}{dt^2} = -\dfrac{1}{LC}I(t) = -\omega_0^2I(t)$(51)

where $\omega_0 = \sqrt{1/LC\os}$ is the natural SHM frequency.

The particular solution in this case is Ip(t) = 0 when the capacitor becomes fully charged. The complementary function in this case is given by Equation 34,

x(t) = A0sin (ω0t + ϕ)(Eqn 34)

soI(t) = I0sin(ω0t + ϕ)(52)

In this case we have taken the resistance to be zero so there is no damping and the complementary function does not disappear in the limit as t → ∞.

To find I0 and ϕ we differentiate Equation 52 to give:

$\dfrac{dI(t)}{dt} = I_0\omega_0\cos(\omega t+\phi)$(53)

Using Kirchhoff’s voltage law (Equation 14),

$V(t) = RI(t) + \dfrac{q(t)}{C} + L\dfrac{dI(t)}{dt}$(Eqn 14)

with R = 0 gives:

$V(t) = V_C(t) + V_L(t) = \dfrac{q(t)}{C} + L\dfrac{dI(t)}{dt}$

so$V(t) = \dfrac{q(t)}{C} + LI_0\omega_0\cos(\omega t+\phi)$(54)

The initial conditions fix V(0) = 2.4 V, q(0) = 0 and I(0) = 0, so from Equation 52, since I0 ≠ 0, sinϕ = 0

soϕ = 0(55)

and from Equation 54,

$I_0 = \dfrac{V(0)}{\omega_0L\cos\phi} = \dfrac{V(0)}{\omega_0L} =V(0)\sqrt{\dfrac CL} = 2.4\times10^{-2}\sqrt{20\os}\,{\rm A}$(56)

Finally we have:

$I(t) = 2.4\times10^{-2}\sqrt{20\os}\sin\sqrt{20\os}t$

## 2.5 Electrical impedance

We will now determine the steady state current that flows in three simple electrical circuits, where each contains a single component, and is driven by an externally applied voltage V(t) = V0sin(Ωt).

(a) For the circuit containing the resistor (Figure 6a), Equation 14,

$V(t) = RI(t) + \dfrac{q(t)}{C} + L\dfrac{dI(t)}{dt}$(Eqn 14)

gives

$\underbrace{~~~~~~V(t)~~~~~~}_{\color{purple}{\large\substack{\text{potential difference}\\[2pt]\text{supplied by the}\\[2pt]\text{voltage generator}}}} = \underbrace{~~~~~RI(t)~~~~~}_{\color{purple}{\large\substack{\text{potential difference}\\[2pt]\text{across the resistor}}}}$

so that V0sin(Ωt) = RI(t) and therefore

$I(t) = \dfrac{V_0}{R}\sin({\it\Omega}t)$(57)

Notice that the current I(t) and the applied voltage are in phase in this case.

(b) For the circuit containing the inductor (Figure 6b), Equation 14,

$V(t) = RI(t) + \dfrac{q(t)}{C} + L\dfrac{dI(t)}{dt}$(Eqn 14)

gives

$\underbrace{~~~~~~V(t)~~~~~~}_{\color{purple}{\large\substack{\text{potential difference}\\[2pt]\text{supplied by the}\\[2pt]\text{voltage generator}}}} = \underbrace{~~~~~L\dfrac{dI(t)}{dt}~~~~~}_{\color{purple}{\large\substack{\text{potential difference}\\[2pt]\text{across the inductor}}}}$

so that$V_0\sin({\it\Omega}t) = L\dfrac{dI(t)}{dt}\quad\text{implying that}\quad\dfrac{dI(t)}{dt} = \dfrac{V_0}{L}\sin({\it\Omega}t)$

Hence, in the steady state integration gives us

$I(t) = -\dfrac{V_0}{{\it\Omega}L}\cos({\it\Omega}t) = \dfrac{V_0}{{\it\Omega}L}\sin({\it\Omega}t-\pi/2)$(58) i

Comparing Equations 58 and 57 we see that an inductor behaves rather like a resistor with effective resistance ΩL, but the phase of the current lags that of the voltage by π/2.

(c) For the circuit containing the capacitor (Figure 6c), Equation 19,

$L\dfrac{d^2I(t)}{dt^2} + R\dfrac{dI(t)}{dt} + \dfrac1CI(t) = \dfrac{dV(t)}{dt}$(Eqn 19)

gives

$\dfrac{I(t)}{C} = \dfrac{dV(t)}{dt}$

so that

$I(t) = C\dfrac{dV(t)}{dt} = {\it\Omega}CV_0\cos({\it\Omega}t) = \dfrac{V_0}{1/({\it\Omega}C)}\sin({\it\Omega}t + \pi/2)$(59)

Comparing Equations 59 and 57 we see that a capacitor behaves rather like a resistor with effective resistance 1/(ΩC), but the phase of the current leads that of the voltage by π/2.

The frequency dependent ‘effective resistance’ of an inductor, XL = ΩL is known as its inductive reactance, while the corresponding quantity for a capacitor, XC = 1/(ΩC), is known as its capacitive reactance. These quantities, together with the resistance R of a resistor, play an important part in determining the current I(t) that will flow through a component when a voltage V(t) is applied across it. We can summarize these relationships contained in Equations 57, 58 and 59 as follows:

If V(t) = V0sin(Ωt) and I(t) = I0sin(Ωtδ), then:

for an inductor V0/I0 = XL = ΩL and δ = π/2

for a resistor V0/I0 = R and δ = 0

for a capacitor V0/I0 = XC = 1/(ΩC) and δ = −π/2 i

The ability of inductors and capacitors to ‘react’ to an applied voltage by altering their ‘effective resistance’ according to its frequency is part of the reason for their significance in electronics. In particular, it allows them to play an important role in filter circuits designed to pass signals (varying voltages) in certain frequency ranges while inhibiting the passage of others. The differences in behaviour between resistors, capacitors and inductors mean that appropriately designed combinations of these components can be used to manipulate signals in a variety of ways.

The analysis of circuits in terms of the differential equations that represent them is a sophisticated study in its own right, but provided we are only concerned with the steady state behaviour of networks of passive components (resistors, capacitors and inductors), driven by applied voltages that vary sinusoidally with time, the subject can be greatly simplified. As an example we will state without proof four more results for simple series circuits. (These can be established by finding particular solutions for the appropriate versions of Equation 19).

$L\dfrac{d^2I(t)}{dt^2} + R\dfrac{dI(t)}{dt} + \dfrac1CI(t) = \dfrac{dV(t)}{dt}$(Eqn 19)

In each case the result consists of a description of the current I(t) = I0sin(Ωtδ) that flows in response to an applied voltage V(t) = V0sin(Ωt), and in each case this requires that we provide an explicit expression relating I0 to the known quantities V0, R, C and L. In order to do this each result provides an explicit expression for the quantity Z = V0/I0, which is known as the impedance of the circuit. The impedance is measured in ohm (Ω), and represents a generalization of resistance and reactance.

##### Resistor and capacitor in series

For the circuit shown in Figure 7a:

IfV(t) = V0sin(Ωt)   then   $I(t) = \dfrac{V_0}{Z}\sin({\it\Omega}t - \delta)$

where$Z^2 = R^2 + \dfrac{1}{({\it\Omega}C)^2}\quad\text{and}\quad\tan\delta = \dfrac{-1}{{\it\Omega}CR}$(60)

✦ Given that V0 = 3 V, Ω = 5 s−1, R = 2 Ω and C = 0.2 F for the circuit shown in Figure 7a, calculate the current I(t).

✧ $Z^2 = R^2 + \dfrac{1}{({\it\Omega}C)^2} = \left[(2)^2 + \dfrac{1}{(5\times0.2)^2}\right]\,\Omega^2 = 5\,\Omega^2$

so that$Z = \sqrt{5\os}\,\Omega$

Also:

$\tan\delta = \dfrac{-1}{{\it\Omega}CR} = \dfrac{-1}{5\times0.2\times0.2} = -0.5$

so thatδ ≈ −0.46.

Thus$I(t) = \dfrac{V_0}{Z}\sin({\it\Omega}t - \delta) \approx \dfrac{3}{\sqrt{5\os}}\sin\left[(5\,{\rm s^{-1}})t + 0.46\right]\,{\rm A}$

##### Resistor and inductor in series

For the circuit shown in Figure 7b:

IfV(t) = V0sin(Ωt)   then   $I(t) = \dfrac{V_0}{Z}\sin({\it\Omega}t - \delta)$

whereZ2 = R2 + (ΩL)2   and   tanδ = ΩL/R(61)

##### Inductor and capacitor in series

For the circuit shown in Figure 7c:

IfV(t) = V0sin(Ωt)   then   $I(t) = \dfrac{V_0}{Z}\sin({\it\Omega}t - \delta)$

where$Z = \left\lvert\dfrac{1}{{\it\Omega}C}-{\it\Omega}L\,\right\rvert\quad\text{and}\;\;\delta = \begin{cases}\pi/2 & \text{if}\;\;\dfrac{1}{{\it\Omega}C} \lt {\it\Omega}L\\-\pi/2 & \text{if}\;\;\dfrac{1}{{\it\Omega}C} \gt {\it\Omega}L \end{cases}$(62)

##### Resistor, inductor and capacitor in series

For the circuit shown in Figure 3b:

IfV(t) = V0sin(Ωt)   then   $I(t) = \dfrac{V_0}{Z}\sin({\it\Omega}t - \delta)$(63a)

where$Z^2 = R^2 + \left(\dfrac{1}{{\it\Omega}C}-{\it\Omega}L\right)^2\;\;\text{and}\;\;\tan\delta = \dfrac1R\left({\it\Omega}L-\dfrac{1}{{\it\Omega}C}\right)$(63b)

The above results can be neatly summarized and generalized if we introduce a total reactance X = XLXC, for we can then say:

When a voltage V(t) = V0sin(Ωt) is applied across any series circuit of components of total resistance R and total reactance X, the resulting steady state current will have the form I(t) = I0sin(Ωtδ)

where V0 = I0 Z  and  δ = arctan(X/R)

and the impedance Z is given by $Z = \sqrt{R^2 + X^2}$.

Figure 8 Geometrical interpretations of Z and δ in terms of XL, XC and R.

It is possible to give a simple geometric interpretation to the relationship between impedance, resistance and reactance. This is indicated in Figure 8a, where XL is treated as a ‘vector quantity’ directed vertically upwards, XC is treated as a vector directed vertically downwards, and R is treated as a vector directed to the right. (The length of each ‘vector’ represents the relevant value of resistance or reactance.)

As Figures 8b to 8e indicate, the value of Z in each of the cases discussed above will be represented by the length of the ‘vector sum’ of XL, XC and R, and the value of δ will be given by the angle (measured in the anticlockwise direction) from the horizontal axis to the ‘vector’ representing Z. i We will return to this geometric interpretation of impedance in Section 3.

### Example 2

Calculate the impedance of a circuit which consists of a resistor of 10 Ω, a capacitor of 0.05 F and an inductor of 2.0 H in series. If a voltage V(t) = 3sin[(2 s−1)t] V is applied to the circuit, then by how much do the current and voltage differ in phase in the steady state? Write down an expression for the steady state current.

#### Solution

From Equation 63b, $Z^2 = R^2 + \left(\dfrac{1}{{\it\Omega}C}-{\it\Omega}L\right)^2$

so in this case, with Ω = 2 Hz,

$Z = \sqrt{10^2+\left(\dfrac{1}{0.05\times2}-2\times2\right)^2}\,\Omega \approx 11.66\,\Omega$

and$\tan\delta = \dfrac1R\left({\it\Omega}L-\dfrac{1}{{\it\Omega}C}\right) = \dfrac{1}{10}\left(2\times2-\dfrac{1}{2\times0.05}\right) \approx -0.60$

soδ ≈ −0.54.

Thus$I(t) = I_0\sin({\it\Omega}t - \delta) = \rm \dfrac{3}{11.66}\sin[(2\,s^{-1})t+0.54]\,A$

i.e.$I(t) \approx (0.26)\sin[(2\,s^{-1})t+0.54]\,A$

Question T5

When the series LCR circuit of Figure 3b is driven by an applied voltage V(t) = V0sin(Ωt), the current I(t) satisfies Equation 19,

$L\dfrac{d^2I(t)}{dt^2} + R\dfrac{dI(t)}{dt} + \dfrac1CI(t) = \dfrac{dV(t)}{dt}$(Eqn 19)

Show that Equation 63,

$I(t) = \dfrac{V_0}{Z}\sin({\it\Omega}t - \delta)$(Eqn 63a)

$Z^2 = R^2 + \left(\dfrac{1}{{\it\Omega}C}-{\it\Omega}L\right)^2\;\;\text{and}\;\;\tan\delta = \dfrac1R\left({\it\Omega}L-\dfrac{1}{{\it\Omega}C}\right)$(Eqn 63b)

really does provide a particular solution of this equation, as claimed above. i

If$I(t) = \dfrac{V_0}{Z}\sin({\it\Omega}t - \delta)$

then$\dfrac{dI(t)}{dt} = \dfrac{{\it\Omega}V_0}{Z}\cos({\it\Omega}t-\delta)$

and$\dfrac{d^2I(t)}{dt^2} = \dfrac{-{\it\Omega}^2V_0}{Z}\sin({\it\Omega}t-\delta)$

so that the left–hand side of Equation 19,

$L\dfrac{d^2I(t)}{dt^2} + R\dfrac{dI(t)}{dt} + \dfrac1CI(t) = \dfrac{dV(t)}{dt}$(Eqn 19)

becomes:

$L\left[\dfrac{-{\it\Omega}^2V_0}{Z}\sin({\it\Omega}t-\delta)\right] + R\left[\dfrac{{\it\Omega}V_0}{Z}\cos({\it\Omega}t-\delta)\right] + \dfrac1C\left[\dfrac{V_0}{Z}\sin({\it\Omega}t-\delta)\right]$

which, after expanding the sine and cosine terms, becomes:

$\dfrac{{\it\Omega}V_0}{Z}\left\{\left[R\sin\delta+\left(\dfrac{1}{{\it\Omega}C} - {\it\Omega}L\right)\cos\delta\right]\sin({\it\Omega}t) + \left[R\cos\delta - \left(\dfrac{1}{{\it\Omega}C} - {\it\Omega}L\right)\sin\delta\right]\cos({\it\Omega}t)\right\}$

but the coefficient of sin(Ωt) is zero by virtue of Equation 63, so the left–hand side of Equation 19 gives us:

$\dfrac{{\it\Omega}V_0}{Z}\left[R\cos\delta - \left(\dfrac{1}{{\it\Omega}C} - {\it\Omega}L\right)\sin\delta\right]\cos({\it\Omega}t)$

As can be seen from the geometrical representation of Z in Figure 8e,

$\dfrac RZ = \cos\delta\quad\text{and}\quad\dfrac1Z\left({\it\Omega}L- \dfrac{1}{{\it\Omega}C}\right) = \sin\delta$

and we know that cos2δ + sin2δ = 1, so that the above expression simplifies to give ΩV0cos(Ωt), which is equal to the right–hand side of Equation 19. Hence, Equation 63,

$I(t) = \dfrac{V_0}{Z}\sin({\it\Omega}t - \delta)$(63a)

$Z^2 = R^2 + \left(\dfrac{1}{{\it\Omega}C}-{\it\Omega}L\right)^2\;\;\text{and}\;\;\tan\delta = \dfrac1R\left({\it\Omega}L-\dfrac{1}{{\it\Omega}C}\right)$(63b)

does indeed describe a particular solution of Equation 19.

Question T6

Calculate the impedance of a circuit which consists of a resistor of 5.0 Ω, a capacitor of 1/6 F and an inductor of 1.0 H in series.

A voltage $V(t) = \frac35\sin[(5\,{\rm s} - 1)t]\,{\rm V}$ is applied to the circuit and the current is allowed to reach its steady state. By how much do the steady state current and voltage differ in phase? Write down an expression for the steady state current. Compare your answer with that of Example 1.

From Equation 63,

$I(t) = \dfrac{V_0}{Z}\sin({\it\Omega}t - \delta)$(63a)

$Z^2 = R^2 + \left(\dfrac{1}{{\it\Omega}C}-{\it\Omega}L\right)^2\;\;\text{and}\;\;\tan\delta = \dfrac1R\left({\it\Omega}L-\dfrac{1}{{\it\Omega}C}\right)$(63b)

we have

$Z^2 = R^2 + \left(\dfrac{1}{{\it\Omega}C}-{\it\Omega}L\right)^2$

with R = 5 Ω, L = 1 H, C = 1/6 F,  Ω = 5 s−1 and V0 = 3/5 V, so that in this case

$Z^2 = \left\{5^2 + \left(\dfrac{1}{5\times(1/6)} - 5\times1\right)^2\right\}\,\Omega^2$

and therefore Z ≈ 6.28013 Ω

Also, $\tan\delta = \dfrac1R\left({\it\Omega}L-\dfrac{1}{{\it\Omega}C}\right)$, so that in this case

$\tan\delta = \dfrac15\left(5\times1 - \dfrac{1}{5\times(1/6)}\right) = 0.76$

and therefore δ ≈ 0.65.

Since I0 = V0/Z it follows that

$I(t) = I_0\sin({\it\Omega}t-\delta) = \dfrac{3/5}{6.28}\sin[(5\,{\rm s^{-1}})t-0.65]\,{\rm A} \approx (0.096)\sin[(5\,{\rm s^{-1}})t-0.65]\,{\rm A}$

This problem is identical with that of Example 1 except that here we are not interested in the transient term and also Example 1 was dealing with charge.

## 2.6 Mechanical impedance

We saw earlier that the mathematical description of a harmonically driven series LCR circuit is essentially identical to that of a harmonically driven, linearly damped mechanical oscillator. In particular we saw that charge oscillations in the circuit are directly analogous to the displacement oscillations of the mechanical system. However, we have just seen that the circuit also displays current oscillations, the amplitude of which can be expressed in terms of an impedance that depends on the angular frequency of the driving voltage. Does the mechanical oscillator exhibit oscillations analogous to the current oscillations? If so, what are they, how do they behave and what is the mechanical analogue of the impedance?

The current in the series LCR circuit is related to the charge on the capacitor by

$I = \dfrac{dq}{dt}$(Eqn 8)

Since the mechanical analogue of the charge q is the displacement x, we should expect the mechanical analogue of the current to be the velocity

$\upsilon_x = \dfrac{dx}{dt}$

✦ According to Equation 19,

$L\dfrac{d^2I(t)}{dt^2} + R\dfrac{dI(t)}{dt} + \dfrac1CI(t) = \dfrac{dV(t)}{dt}$(Eqn 19)

sinusoidally driven current oscillations satisfy a differential equation of the form

$L\dfrac{d^2I(t)}{dt^2} + R\dfrac{dI(t)}{dt} + \dfrac1C I(t) ={\it\Omega}V_0\cos{\it\Omega}(t)$

Write down the analogous differential equation that you might expect mechanical velocity oscillations to satisfy, and show that the mechanical oscillator of Figure 3a does in fact obey such an equation.

✧ Since L corresponds to m, R to b, 1/C to k and V0 to F0, we expect the velocity oscillations to satisfy

$m\dfrac{d^2\upsilon_x(t)}{dt^2} + b\dfrac{d\upsilon_x(t)}{dt} + k\upsilon(t) = {\it\Omega}F_0\cos({\it\Omega}t)$(64)

The fact that the mechanical oscillator obeys such an equation can be seen by differentiating Equation 13:

$m\dfrac{d^2x(t)}{dt^2} + b\dfrac{dx(t)}{dt} + kx(t) = F_0\sin({\it\Omega}t)$(Eqn 13)

✦ Using the description of the current oscillations given in the last subsection, write down the corresponding description of the velocity oscillations in the driven mechanical oscillator, and hence identify the mechanical impedance Zm.

✧ If the driving force is Fx(t) = F0sin (Ωt), the velocity oscillations will be described by

υx = υ0sin(Ωtδ)(65)

where

$\upsilon_0 = \dfrac{F_0}{Z_{\rm m}} = \dfrac{F_0}{\sqrt{b^2+\left(\dfrac{k}{{\it\Omega}}-{\it\Omega}m\right)^2}}$

and

$\tan\delta = \dfrac1b\left({\it\Omega}m-\dfrac{k}{\it\Omega}\right)$

Hence$Z_{\rm m} = \sqrt{b^2+\left(\dfrac{k}{{\it\Omega}}-{\it\Omega}m\right)^2}$

By analogy with the electrical case, it is possible to identify the mass m and the spring constant k as ‘reactive’ parts of the mechanical oscillator, since their contribution to the mechanical impedance depends on the angular frequency of the driving force.

## 2.7 Resonance and driven oscillations

The displacement oscillations described by Equations 49 and 50,

$A = \dfrac{a_0}{\sqrt{\smash[b]{(\omega_0^2 - {\it\Omega}^2)^2+(\gamma{\it\Omega})^2}}}\quad\text{and}\quad\arctan\left(\dfrac{\gamma{\it\Omega}}{\omega_0^2 - {\it\Omega}^2}\right)$(Eqn 49)

x(t) = Asin(Ωtδ)(Eqn 50)

and the velocity oscillations described by Equations 65 and 66,

υx = υ0sin(Ωtδ)(Eqn 65)

$\tan\delta = \dfrac1b\left({\it\Omega}m - \dfrac{k}{\it\Omega}\right)$(Eqn 66)

both have an amplitude that depends sensitively on the angular frequency Ω of the driving force. The charge and current oscillations in the driven LCR circuit show a similar sensitivity to the angular frequency of the driving voltage.

As an example of this behaviour, Figure 9 shows the way in which the amplitude I0 of the steady state current varies with driving frequency for fixed values of C, L and V0 at a variety of values of R. It is clear from Equation 63,

$I(t) = \dfrac{V_0}{Z}\sin({\it\Omega}t - \delta)$(Eqn 63a)

$Z^2 = R^2 + \left(\dfrac{1}{{\it\Omega}C}-{\it\Omega}L\right)^2\;\text{and}\;\tan\delta = \dfrac1R\left({\it\Omega}L-\dfrac{1}{{\it\Omega}C}\right)$(Eqn 63b)

that in this case, for any fixed value of R, the impedance is a minimum and the current amplitude a maximum when ${\it\Omega} = 1/\sqrt{(LC)\os}$.

This is an example of the phenomenon of resonance, the production of a large response in a driven oscillator by driving it at a frequency close to the natural frequency it would have in the absence of any driving or damping. As you can see, the smaller the value of R, the taller and narrower the peak, i.e. the sharper the resonance. In this particular case the resonant frequency at which the response is a maximum is identical to the natural frequency $\omega_0 = 1/\sqrt{(LC)\os}$, but that is not always the case.

✦ The current and velocity oscillations have a resonant frequency that is independent of the resistance or damping (R and b respectively). Is the same true of the resonant frequency of the charge and displacement oscillations? If not, what is the relationship between the resonant frequency, the natural frequency and the resistance or damping in this case?

✧ According to Equation 49,

$A = \dfrac{a_0}{\sqrt{\smash[b]{(\omega_0^2-{\it\Omega}^2)+(\gamma{\it\Omega})^2}}}\;\;\text{and}\;\;\delta = \arctan\left(\dfrac{\gamma{\it\Omega}}{\omega_0^2-{\it\Omega}^2}\right)$(Eqn 49)

the amplitude of the steady state displacement oscillations is given by

$A = \dfrac{a_0}{\sqrt{\smash[b]{(\omega_0^2-{\it\Omega}^2)+(\gamma{\it\Omega})^2}}}$

The value of Ω at which this amplitude is a maximum can be found by differentiating with respect to Ω, and setting the resulting derivative equal to 0. Using the chain rule to perform the differentiation:

$\dfrac{dA}{d{\it\Omega}} = \dfrac{-1}{2}a_0\left[(\omega_0^2-{\it\Omega}^2)^2 + (\gamma{\it\Omega})^2\right]^{-3/2}\times\left[2(\omega_0^2-{\it\Omega}^2)(-2{\it\Omega})+2(\gamma^2{\it\Omega})\right]$

Hence the resonant frequency is given by

$\gamma^2{\it\Omega}-2{\it\Omega}(\omega_0^2-{\it\Omega}^2) = 0$

i.e.${\it\Omega} = \sqrt{\smash[b]{\omega_0^2-\gamma^2/2}}$

Remembering that γ = b/m we see that in this case the resonant frequency is not independent of the damping, and by implication we should not expect it to be independent of the resistance in the electrical case either.

Question T7

A radio aerial for BBC Radio 4 contains an inductor L = 0.001 H (i.e. 1.0 mH) and a variable capacitor C in series. The transmitter induces a voltage in the aerial, and produces a potential difference V(t) = V0sin(ωt) across the open circuit, where ω = 2π × 198 kHz. To what value should the capacitor be set in order to maximize the amplitude of the current?

We need to ensure that ω = 2π 198 × 103 s−1 is the resonant angular frequency of the circuit so that

$\pi\times1.98\times10^5\,{\rm s^{-1}} = \dfrac{1}{\sqrt{0.001\,{\rm H}\times C\os}}$

and therefore   C ≈ 6.5 × 10−10 F = 650 pF

# 3 Oscillations and complex numbers

Complex numbers are often used in the analysis of sinusoidal oscillations. They can greatly simplify many problems, so much so that they constitute the standard approach in most advanced work. As you pursue your studies of physics it is inevitable that you will frequently encounter discussions of oscillatory phenomena based on complex methods. This is particularly true in quantum physics, where complex numbers are not just useful, but essentially unavoidable.

## 3.1 Complex numbers – a brief review

1

Any complex number, z, may be written as

z = x + iy

where x and y are real numbers and i satisfies i2 = −1.

2

If z = x + iy (with x and y real ) then x is known as the real part of z, written as Re(z), and y is known as the imaginary part of z, written as Im(z). i Thus,

z = Re(z) + iIm(z)

3

Complex numbers obey the rules of normal algebra except that i2 can be replaced by −1 whenever it appears.

4

The complex conjugate of z (written z*) is defined by

z* = xiy = Re(z) − iIm(z)

5

The modulus_of_a_complex_numbermodulus of z = x + iy (written as |z|) is defined by

$\lvert\,z\,\rvert = \sqrt{x^2 +y^2} = \sqrt{[{\rm Re}(z)]^2 + [{\rm Im}(z)]^2}$

6

For arbitrary complex numbers z and w

${\rm Re}(z) = \dfrac12 (z+z*)\quad {\rm Im}(z) = \dfrac{1}{2i}(z-z*)$   (zw)* = z*w* (z*)* = z  and  |z|2 = zz*

7

A complex number, z = x + iy, is said to be in cartesian_form_of_a_complex_numberCartesian form or a Cartesian representation. Such a complex number may also be written in the form:

z = r(cosθ + isinθ)

where r and θ are real numbers, in which case it is said to be in polar_form_of_a_complex_numberpolar form or a polar representation. When written in polar form, the modulus of z is then given by |z| = r, and θ is referred to as the argument of z (written as arg(z)). Adding 2π to the argument of a complex number does not change that complex number. The principal value of the argument of a complex number is the value θ which lies in the range −π < θπ.

8

We can convert from Cartesian to polar form using

$r = \sqrt{\smash[b]{x^2 + y^2}},\;\;\cos\theta = \dfrac{x}{\sqrt{\smash[b]{x^2 + y^2}}}\;\;\text{and}\;\;\sin\theta = \dfrac{y}{\sqrt{\smash[b]{x^2 + y^2}}}$

and from polar to Cartesian form by means of

x = rcosθ  and  y = rsinθ

9

A complex number can be represented by a point on an Argand diagram (complex plane) by using (x, y) as the Cartesian coordinates or (r, θ) as the polar coordinates of the point. (By convention, θ is measured anticlockwise from the positive x–axis.)

10

A complex number may also be written in exponential_form_of_a_complex_numberexponential form or exponential representation by using Euler’s formula

e = cosθ + isinθ

If z = re, then |z| = r, arg(z) = θ and z* = re.

11

Addition and subtraction of complex numbers is simplest in Cartesian form:

z + w = (x + iy) + (u + ) = (x + u) + i(y + υ)

zw = (x + iy) − (u + ) = (xu) + i(yυ)

Multiplication and division of complex numbers is simplest in exponential form:

zw = (re) (se) = (rs)ei(θ + ϕ)

z/w = (re)/ (se) = (r/s)ei(θϕ)

The following result, known as Demoivre’s theorem is valid for any real value of n

eniθ = (cosθ + isinθ)n = cos() + isin()

## 3.2 Complex impedance

Study comment It is important to appreciate that in the following discussion of electrical circuits we are only interested in the steady state, and we are therefore assuming that sufficient time has elapsed for the transient part of the current to be negligible. In this subsection when we refer to ‘the current’ we always mean the steady state current. Since we are only concerned with the steady state, the phase of the applied voltage is unimportant – it is the phase difference between the applied voltage and the current that is critical. Previously it was convenient to assume that the applied voltage was of the form V0sin(Ωt), but here it is more conventional to choose V0cos(ωt), and to describe the steady state current by I0cos(ωtδ). The effect is minimal and serves only to make the mathematics a little easier, and more standard.

The impedance Z and phase lag δ determine the relationship between the voltage that drives the LCR circuit of Figure 3b, and the steady state current it eventually produces. There is however a very simple method of determining these quantities in terms of the values of R, C and L, and, as we will see, this new method may be used to analyse far more complicated circuits.

Figure 8 Geometrical interpretations of Z and δ in terms of XL, XC and R.

We begin by referring once again to Figure 8, the geometric interpretation of impedance. If we interpret the direction assigned to the ‘vector’ representing the resistance R as the real axis of an Argand diagram, and if we allow the heads of the various ‘vectors’ in Figure 8 to denote complex numbers, then the values of both Z and δ can be found very easily as the modulus and argument of the sum of the various complex numbers involved.

Adopting this approach, and noting that in this case we are dealing with an applied voltage with angular frequency ω, we can identify the following complex quantities from Figure 8a:

• the complex inductive reactance iωL
• the complex capacitive reactancei/ωC
• the resistance (a real quantity) R

We can then define a new quantity, the complex impedance $\mathcal Z$ of a series LCR circuit by the relation

${\mathcal Z} = R + i\omega L + \left(\dfrac{-i}{\omega C}\right) = R + i\left(\omega L -\dfrac{1}{\omega C}\right)$(67)

From Figures 8b to 8e it can be seen that this complex impedance has the following properties:

• its modulus is equal to the impedance, so $\lvert{\mathcal Z}\rvert = Z$ i
• its argument is equal to the phase lag, so $\arg({\mathcal Z}) = \delta$

Thus, the complex impedance can be written in the form ${\mathcal Z} = Z\,{\rm e}^{i\delta}$. We can take this representation further by expressing the applied voltage V(t) as the real part of a complex

voltage defined by

${\mathscr V}(t) = V\,{\rm e}^{i\omega t}$(68)

so that$V(t) = {\rm Re}({\mathscr V}(t)) = {\rm Re}(V_0\,{\rm e}^{iωt}) = V_0\cos(\omega t)$

It then follows that the instantaneous current I(t) in the series LCR circuit is given by the real part of the complex current ${\mathcal V}(t)$ defined by

${\mathscr I}(t) = \dfrac{{\mathscr V}(t)}{\mathcal Z}$(69)

since${\rm Re}[{\mathscr I}(t)] = {\rm Re}\left(\dfrac{{\mathscr V}(t)}{\mathcal Z}\right) = {\rm Re}\left(\dfrac{V_0\,{\rm e}^{i\omega t}}{Z\,{\rm e}^{i\delta}}\right) = {\rm Re}\left(\dfrac{V_0}{Z}\,{\rm e}^{i(\omega t-\delta)}\right) = I_0\cos(\omega t-\delta)$

i.e.${\rm Re}[{\mathscr I}(t)] = I(t)$

Equation 69 plays an important role in the analysis of circuits carrying sinusoidally varying currents. Such currents are generally referred to as alternating currents, or a.c. currents, and Equation 69 is sometimes described as the complex or a.c. form of Ohm’s law.

### Example 3

Use Equation 67,

${\mathcal Z} = R + i\omega L + \left(\dfrac{-i}{\omega C}\right) = R + i\left(\omega L -\dfrac{1}{\omega C}\right)$(Eqn 67)

to calculate the impedance of a circuit which consists of a resistor of 10 Ω, a capacitor of 0.05 F and an inductor of 2 H in series. If a voltage V(t) = 3cos[(2 s−1)t] V is applied to the circuit, what is the phase difference between the steady state current and the applied voltage? Use the complex form of Ohm’s law to write down an expression for the steady state current. Plot the complex numbers representing ${\mathscr V}(t)$ and ${\mathscr I}(t)$ at some arbitrary time t on an Argand diagram. Indicate the quantities corresponding to V(t) and I(t) on your diagram.

#### Solution

From Equation 67 we have:

${\mathcal Z} = R + i\omega L - \left(\dfrac{-i}{\omega C}\right) = \left(10 + (2\times2)i - \dfrac{i}{(2\times0.05}\right)\,\Omega = (10-6i)\,\Omega$

then$Z = \lvert{\mathcal Z}\rvert = \lvert10-6i\rvert = \sqrt{10^2+6^2} \approx 11.66\,\Omega$

and$\delta = \arg({\mathcal Z}) = \arctan(-6/10) \approx -0.54$

and it follows that   ${\mathcal Z} \approx 11.66\,{\rm e}^{-0.54i}\,\Omega$

In this case   ${\mathscr V}(t) = 3\exp[(2\,{\rm s^{-1}})it]\,{\rm V}$

so, from Equation 69,   ${\mathscr I}(t) = \dfrac{{\mathscr V}(t)}{\mathcal Z} = \dfrac{3\exp[(2\,{\rm s^{-1}})it]}{10-6i}\,{\rm A} = \dfrac{3\exp[(\,{\rm s^{-1}})it]\,{\rm V}}{\,{\rm e}^{-0.54i}}\,{\rm A}$

It follows that

\begin{align}I(t) & \approx {\rm Re}\left\{\left(3\exp[(2\,{\rm s^{-1}})it]\right)\left(\dfrac{{\rm e}^{0.54i}}{11.66}\right)\right\}\,{\rm A} \\& \approx {\rm Re}\left\{\dfrac{3}{11.66}\exp[(2\,{\rm s^{-1}})it + 0.54i]\right\}\,{\rm A}\end{align}

i.e.I(t) = Re{0.26exp[(2 s−1)it + 0.54i]} A

I(t) = (0.26)cos[(2 s−1)t + 0.54] A

If we plot the complex numbers ${\mathscr V}(t)/{\rm V}$ and ${\mathscr V}(t)/{\rm A}$ i on the same Argand diagram, as in Figure 10 (not drawn to scale), then they will lie on two circles, of radius 3 units and 0.26 units, respectively. The real parts of these quantities, indicated on the horizontal axis, represent the instantaneous ωt voltage and current. The essential point to notice is that, although the two points move around the circles as t increases, they are fixed in relation to each other. The magnitude of the angle ϕ in Figure 10 is 0.54 radians, which is the magnitude of the argument of Z. In this particular case we can see immediately from the diagram that the current leads the voltage by this amount.

Aside It is worth noting that the conversion of complex numbers from Cartesian to exponential (or polar) form that was at the heart of this example is available as a standard function on many modern calculators. My calculator stores a complex number 3 + 5i as (3 5) so to check this example I keyed in:

(10, 0) for the 10 Ω resistance and then stored it as R,
(0, 2 × 2) for the iωL term and stored it as Z1,
then $\left(0, -\dfrac{1}{2\times0.05}\right)$ for the 1/(iωC) term and stored it as Z2.

Then I calculated 3/(R + Z1 + Z2) and finally converted it into polar form using a function labelled c → p on my calculator. You may find that you can perform similar calculations on your own calculator.

If you compare the solution to Example 3 with the solution to Example 2, you will find that the answers are the same (apart from a change of sin to cos), but this method is simpler.

Question T8

Use Equation 67,

${\mathcal Z} = R + i\omega L + \left(\dfrac{-i}{\omega C}\right) = R + i\left(\omega L -\dfrac{1}{\omega C}\right)$(Eqn 67)

to calculate the impedance of a resistor R = 15 Ω, a capacitor of C = 5 μF and an inductor L = 4 mH in series. Find the complex impedance, $\mathcal Z$, and hence find $1/\mathcal Z$, when a voltage V(t) = 10cos[(104 s−1)t] V is applied to the circuit. By how much do the steady state current and the applied voltage differ in phase? Write down an expression for the steady state current.

From Equation 67,

${\mathcal Z} = R + i\omega L + \left(\dfrac{-i}{\omega C}\right) = R + i\left(\omega L -\dfrac{1}{\omega C}\right)$(Eqn 67)

we have

${\mathcal Z} = R + i\omega L + \dfrac{1}{i\omega C} = \left(15+[10^4\times(4\times10^{-3})]i + \dfrac{1}{10^4\times(5\times10^{-6})i}\right)\,\Omega = (15+20i)\,\Omega$

then   $Z = \lvert{\mathcal Z}\rvert = \lvert15+20i\,\rvert\,\Omega = \sqrt{15^2+20^2}\,\Omega = 25\,\Omega$

We also see that

$\dfrac{1}{\mathcal Z} = \left(\dfrac{1}{15+20i}\right)\,\Omega^{-1} = \dfrac{1}{125}(3-4i)\,\Omega^{-1} \approx 0.04\,{\rm e}^{-0.93i}\,\Omega^{-1}$

so the argument of Z (and hence the phase difference) is 0.93.

In this case V(t) = 10eiωt V, with ω = 104 s−1, and from Equation 69,

${\mathscr I}(t) = \dfrac{{\mathscr V}(t)}{\mathcal Z}$(Eqn 69)

${\mathscr I}(t) \approx \left[(0.04)\,{\rm e}^{-0.93i}\right](10\,{\rm e}^{\omega it})\,{\rm A} \approx 0.4\,{\rm e}^{(\omega t-0.93)i}\,{\rm A}$

It follows that $I(t) = {\rm Re}[{\mathscr I}(t)] \approx 0.4\cos(\omega t - 0.93)\,{\rm A}$

### Generalizing the complex method

We will now illustrate the power of the complex method by applying it to the circuit shown in Figure 11 in which the three components are connected in parallel_connectionparallel.

This is the first time we have considered a parallel circuit in this module, and we will need to use the principles (obtained from Kirchhoff’s laws) that

1

The current drawn from the voltage generator is equal to the sum of the currents through the separate components.

2

The voltage across each component is the same.

The first of these principles implies that I(t) = IR(t) + IL(t) + IC(t), which may be regarded as the real part of the following complex equation

${\mathscr I}(t) = {\mathscr I}_R(t) + {\mathscr I}_L(t) + {\mathscr I}_C(t)$(70)

and the common voltage across each component may be regarded as the real part of a common complex voltage V(t), which will be related to the complex current in each component and its impedance by Equation 69,

${\mathscr I}(t) = \dfrac{{\mathscr V}(t)}{\mathcal Z}$(69)

as follows:

${\mathscr V}(t) = R{\mathscr I}_R(t)$

${\mathscr V}(t) = (i\omega L){\mathscr I}_R(t)$

${\mathscr V}(t) = \dfrac{-1}{(\omega C)}{\mathscr I}_R(t)$ i

Using these relations and recognizing that −i/(ωC) can be more neatly written as 1/(iωC), Equation 70 becomes

${\mathscr I}(t) = \dfrac{{\mathscr V}(t)}{R}+\dfrac{{\mathscr V}(t)}{i\omega L}+\dfrac{{\mathscr V}(t)}{1/(i\omega C)} = {\mathscr V}(t)\left(\dfrac1R+\dfrac{1}{i\omega L}+\dfrac{1}{1/(i\omega C)}\right)$(71)

In other words, if we denote the complex impedance of a resistor, an inductor and a capacitor in parallel by Z, then

$\dfrac{1}{\mathcal Z} = \dfrac1R + \dfrac{1}{i\omega L} + \dfrac{1}{1/(i\omega C)}$(72)

We can use this expression for ${\mathcal Z}$, together with the above relations between ${\mathscr I}_R, {\mathscr I}_L, {\mathscr I}_C, {\mathscr I}\;\text{and}\; {\mathscr V}$ to solve a wide variety of problems involving parallel LCR circuits.

✦ Use Equation 72 to find an expression for the complex impedance Z of the components shown in Figure 11, in Cartesian form. Hence determine the (real) impedance of a 10 Ω resistance and a 0.2 H inductance, when connected in parallel and used in a context where the driving voltage has an angular frequency of 50 Hz.

✧ Rearranging the formula we obtain (after some algebra)

${\mathcal Z} = \dfrac{R-R^2[\omega C-1/(\omega L)]i}{1 + R^2[\omega C - 1/(\omega L)]^2}$

i.e.${\mathcal Z} = \dfrac{(10 - 10i)}{1+100[-1/(50\times0.2)^2]^2}\,\Omega = (5-5i)\,\Omega$

so$Z = \lvert{\mathcal Z}\rvert = \sqrt{5^2+5^2}\,\Omega \approx 7.07\,\Omega$

### Combining impedances

The expressions we have introduced for the complex impedance of components in series and parallel (Equations 67 and 72)

${\mathcal Z} = R + i\omega L + \left(\dfrac{-i}{\omega C}\right) = R + i\left(\omega L -\dfrac{1}{\omega C}\right)$(Eqn 67)

$\dfrac{1}{\mathcal Z} = \dfrac1R + \dfrac{1}{i\omega L} + \dfrac{1}{1/(i\omega C)}$(Eqn 72)

are in fact particular cases of two general rules for combining complex impedances. With these general rules we can analyse the behaviour of an enormous range of a.c. circuits, though we will not do so in this module.

Given a number of complex impedances ${\mathcal Z}_1,\,{\mathcal Z}_2, \dots {\mathcal Z}_n$, then in series their combined complex impedance ${\mathcal Z}$ is given by:

${\mathcal Z} = {\mathcal Z}_1 + {\mathcal Z}_2 + \dots + {\mathcal Z}_n$(73)

and in parallel their combined complex impedance Z is given by

$\dfrac{1}{\mathcal Z} = \dfrac{1}{\mathcal Z}_1 + \dfrac{1}{\mathcal Z}_2 + \dots + \dfrac{1}{\mathcal Z}_n$(74) i

Question T9

Find the complex impedance of the circuit shown in Figure 12 in terms of ω, R, C and L. The circuit consists of a resistor in parallel with a capacitor, in series with an inductor.

The complex impedances of a resistor, capacitor and inductor are, respectively,

${\mathcal Z}_R = R,\;{\mathcal Z}_C = \dfrac{1}{i\omega C}\text{ and }{\mathcal Z}_L = i\omega L$

If the complex impedance of the resistor in parallel with the capacitor is ${\mathcal Z}_1$ then from Equation 74,

$\dfrac{1}{\mathcal Z} = \dfrac{1}{\mathcal Z}_1 + \dfrac{1}{\mathcal Z}_2 + \dots + \dfrac{1}{\mathcal Z}_n$(Eqn 74)

$\dfrac{1}{\mathcal Z}_1 = \dfrac{1}{\mathcal Z}_R + \dfrac{1}{\mathcal Z}_C = \dfrac1R +i\omega C$

so that

${\mathcal Z}_1 = \dfrac{1}{1/R +i\omega C} = \dfrac{R}{1+i\omega RC} = \dfrac{R(1-i\omega C)}{1+\omega^2R^2C^2}$

Using Equation 73,

${\mathcal Z} = {\mathcal Z}_1 + {\mathcal Z}_2 + \dots + {\mathcal Z}_n$(Eqn 73)

the total complex impedance $\mathcal Z$ is given by

${\mathcal Z} = {\mathcal Z}_1 + {\mathcal Z}_L = \dfrac{R(1-i\omega C)}{1+\omega^2R^2C^2} + i\omega L = \dfrac{R-i\omega\left[R^2C-L(1+\omega^2R^2C^2)\right]}{1+\omega^2R^2C^2}$

## 3.3 The power dissipated

The power dissipated, or rate of energy transfer from devices is often of physical interest. The instantaneous power dissipated by a device in an electrical circuit is the product of the current flowing through it and the potential difference across it, i.e. P(t) = I(t)V(t). We may regard this as the product of the real parts of a complex current and a complex voltage, so we can write $P(t) = {\rm Re}[{\mathscr I}(t)]\times{\rm Re}[{\mathscr V}(t)]$. Such products must be treated with care, as the following question shows.

✦ Given that z = 2 + 3i and w = 1 − 2i, calculate Re(z) × Re(w) and Re(zw). i

✧ Re(z) × Re(w) = 2 × 1 = 2   and   Re(zw) = Re(8 − i) = 8.

P(t) varies from moment to moment, but for devices in a.c. circuits it is usually the average power dissipated over a full cycle of oscillation, $\langle P\rangle$, which is of interest. This is given by

$\displaystyle \langle P\rangle = \dfrac1T\int_0^T{\rm Re}[{\mathscr I}(t)]\,{\rm Re}[{\mathscr V}(t)]\,dt$(75)

where ${\mathscr V}(t)$ is the potential difference across the device and {\mathscr I}(t) is the current flowing through it. T = 2π/ω is the period of an oscillation.

We can simplify this result by introducing two new constants which may be complex, ${\mathscr V}_0\;\text{and}\;{\mathscr I}_0$. Any sinusoidally varying voltage may then be written as the real part of

${\mathscr V}(t) = {\mathscr V}_0\,{\rm e}^{i\omega t}$(76) i

and any sinusoidally varying current as the real part of

${\mathscr I}(t) = {\mathscr I}_0\,{\rm e}^{i\omega t}$(77)

With the aid of these generalized complex expressions, it is possible to show that Equation 75 leads to the following useful result:

$\langle P\rangle = \dfrac12 {\rm Re}\left({\mathscr I}_0^*\,{\mathscr V}_0\right)$(78) i

If you are interested in knowing how this formula is obtained you can answer the following (fairly difficult) question. If not, you can simply use Equation 78 to answer Question T11.

Question T10

Prove this claim.

i.e. that   $\langle P\rangle = \dfrac12 {\rm Re}\left({\mathscr I}_0^*\,{\mathscr V}_0\right)$(Eqn 78)

[Hint: First show that $\int_0^T\,{\rm e}^{ni\omega t}\,dt = 0$ for any non–zero integer, n, and then use that result to derive Equation 78 from Equation 75,

$\displaystyle \langle P\rangle = \dfrac1T\int_0^T{\rm Re}[{\mathscr I}(t)]\,{\rm Re}[{\mathscr V}(t)]\,dt$(Eqn 75)]

Using Euler’s formula we have

$\displaystyle \int_0^T{\rm e}^{ni\omega t}\,dt = \int_0^T\cos(n\omega t)\,dt + i\int_0^T\sin(n\omega t)\,dt = \int_0^T\cos\left(\dfrac{2\pi nt}{T}\right)\,dt + i\int_0^T\sin\left(\dfrac{2\pi nt}{T}\right)\,dt$

which means that we are integrating the sine and cosine functions over an integer (but non-zero) multiple of a whole period, and this gives zero. To be explicit, for n ≠ 0 we have

$\displaystyle \int_0^T\cos\left(\dfrac{2\pi nt}{T}\right)\,dt + i\int_0^T\sin\left(\dfrac{2\pi nt}{T}\right)\,dt = \dfrac{T}{2\pi n}\left\{\left[\sin\left(\dfrac{2\pi nt}{T}\right)\right]_0^T - i\left[\cos\left(\dfrac{2\pi nt}{T}\right)\right]_0^T\right\} = 0$

(You could also see this by looking at graphs of the functions and noticing that for one complete cycle the areas above and below the axis are equal.)

To derive Equation 78,

$\langle P\rangle = \dfrac12 {\rm Re}\left({\mathscr I}_0^*\,{\mathscr V}_0\right)$(Eqn 78)

we start from Equation 75,

$\displaystyle \langle P\rangle = \dfrac1T\int_0^T{\rm Re}[{\mathscr I}(t)]\,{\rm Re}[{\mathscr V}(t)]\,dt$(Eqn 75)

$\displaystyle \langle P\rangle = \dfrac1T\int_0^T\left(\dfrac{{\mathscr I}(t)+{\mathscr I}^*(t)}{2}\right)\left(\dfrac{{\mathscr V}(t)+{\mathscr V}^*(t)}{2}\right)\,dt$

$\displaystyle \langle P\rangle = \dfrac1T\int_0^T\dfrac12 \left({\mathscr I}_0\,{\rm e}^{i\omega t} + {\mathscr I}_0^*{\rm e}^{i\omega t}\right)\dfrac12 \left({\mathscr V}_0\,{\rm e}^{i\omega t} + {\mathscr V}_0^*{\rm e}^{i\omega t}\right)\,dt$

$\displaystyle \langle P\rangle = \dfrac{1}{4T} \left[{\mathscr I}_0{\mathscr V}_0\int_0^T{\rm e}^{2i\omega t}\,dt + {\mathscr I}_0^*{\mathscr V}_0^*\int_0^T{\rm e}^{-2i\omega t}\,dt + \left({\mathscr I}_0^*{\mathscr V}_0 + {\mathscr I}_0{\mathscr V}_0^* \right)\int_0^T\,dt\right]$

$\displaystyle \langle P\rangle = \dfrac14\left({\mathscr I}_0^*{\mathscr V}_0 + {\mathscr I}_0{\mathscr V}_0^* \right)$   (the first two integrals are zero from our first result)

$\langle P\rangle = \dfrac14\left[\left({\mathscr I}_0^*{\mathscr V}_0\right) + \left({\mathscr I}_0^*{\mathscr V}_0\right)^*\right] = \dfrac12 {\rm Re}\left({\mathscr I}_0^*{\mathscr V}_0\right)$

Question T11

Suppose that for the series LCR circuit shown in Figure 3b, we are told that the current is I(t) = I0cos(ωt) (where I0 is real). Find the average power,

$\langle P\rangle = \dfrac12 {\rm Re}\left({\mathscr I}_0^*\,{\mathscr V}_0\right)$(Eqn 78)

dissipated by the circuit in terms of L, C, R and I0. What effect would varying the values of L and C have on the value of $\langle P\rangle$?

$\langle P\rangle$ is given by

$\langle P\rangle = \dfrac12 {\rm Re}\left({\mathscr I}_0^*{\mathscr V}_0\right)$

In this case we are told the current I(t) = I0cos(ωt), so we put ${\mathscr I}(t) = I_0\,{\rm e}^{i\omega t}$. However, from Equation 69,

${\mathscr I}(t) = \dfrac{{\mathscr V}(t)}{\mathcal Z}$(Eqn 69)

we have

${\mathscr V}(t) = {\mathcal Z}I_0\,{\rm e}^{i\omega t}$

and therefore

${\mathscr I}_0^*{\mathscr V}_0 = I_0\times({\mathcal Z}I_0) = I_0^2{\mathcal Z}$

However we know that ${\mathcal Z} = R + i\omega L + \dfrac{1}{i\omega C}$

and therefore

$\langle P\rangle = \dfrac12 {\rm Re}\left({\mathscr I}_0^*{\mathscr V}_0\right) = \dfrac12 {\rm Re}\left[I_0^2\left(R+i\omega L+\dfrac{1}{i\omega C}\right)\right]$

Notice that $\langle P\rangle$ is independent of L and C, and this means that it is the resistance which is solely responsible for the transfer of energy from the system.

# 4 Superposed oscillations and complex algebra

Simple harmonic oscillation is common in nature, and situations often arise in which an oscillating system is subject to several independent influences each of which tends to promote SHM. For example, the steady state current in a circuit might be the response of two independent sinusoidal voltage supplies, each of which drives the circuit with a characteristic amplitude, angular frequency and phase constant. In such cases, provided the circuit concerned has the property of linearity, i.e. provided its behaviour can be modelled by linear differential equations, the response of the circuit at any time will be the sum of the responses it would have shown to each of the applied voltages independently. This general feature of the behaviour of linear systems is enshrined in the superposition principle:

When several oscillations are added, the resulting disturbance at any time is the sum of the disturbances due to each oscillation at that time.

In view of the wide applicability of the superposition principle it is hardly surprising that the problem of adding or rather superposing simple harmonic oscillations is frequently encountered in physics. From a mathematical point of view these additions can be carried out in a number of ways (the use of phasors and of trigonometric identities are both explored elsewhere in FLAP); however when large numbers of oscillations must be added together it is often advantageous to make use of complex methods. It is this process that we will discuss in this section.

In what follows we will make use of the following general result from the arithmetic of complex numbers. If α1 and α2 are arbitrary real numbers, then:

exp(1) + exp(2) = exp[i(α1 + α2)/2]exp[i(α1α2)/2] + exp[i(α1 + α2)/2]exp[−i(α1α2)/2]

exp(1) + exp(2) = exp[i(α1 + α2)/2]{exp[i(α1α2)/2] + exp[−i(α1α2)/2]}

exp(1) + exp(2) = exp[i(α1 + α2)/2]{cos[(α1α2)/2] + isin[(α1α2)/2]

exp(1) + exp(2)  + isin[(α1α2)/2] + cos[(α1α2)/2] − isin[(α1α2)/2]}

i.e. exp(1) + exp(2) = 2exp[i(α1 + α2)/2]cos[(α1α2)/2](79)

## 4.1 Superposition of two SHMs differing only in phase constant

Suppose we want to superpose the (real) oscillations Acos(ωt + ϕ1) and Acos(ωt + ϕ2), which might represent two simultaneous electric currents combining to produce a total current. We already know that each oscillation may be written as the real part of a complex expression of the form Aexp[i(ωt + ϕ)], so we can write the sum of the two oscillations as the real part of the quantity:

z(t) = Aexp[i(ωt + ϕ1)] + Aexp[i(ωt + ϕ2)]

z(t) = Aexp(iωt)[exp(1) + exp(2)]

Using Equation 79,

exp(1) + exp(2) = 2exp[i(α1 + α2)/2]cos[(α1α2)/2](Eqn 79)

with α1 = ϕ1 and α2 = ϕ2 this may be written:

$z(t) = 2A\cos\left(\dfrac{\phi_1-\phi_2}{2}\right)\exp(i\omega t)\exp\left[\dfrac{i(\phi_1+\phi_2)}{2}\right]$

and it follows that:

${\rm Re}[z(t)] = {\rm Re}\left\{2A\cos\left(\dfrac{\phi_1-\phi_2}{2}\right)\exp(i\omega t)\exp\left[\dfrac{i(\phi_1+\phi_2)}{2}\right]\right\}$

i.e.${\rm Re}[z(t)] = 2A\cos\left(\dfrac{\phi_1-\phi_2}{2}\right)\cos\left(\omega t + \dfrac{i(\phi_1+\phi_2)}{2}\right)$

which implies:

$A\cos(\omega t+\phi_1) + A\cos(\omega t+\phi_2) = \underbrace{2A\cos\left(\dfrac{\phi_1-\phi_2}{2}\right)}_{\color{purple}{\large\text{amplitude }}}\cos\left(\omega t + \dfrac{i(\phi_1+\phi_2)}{2}\right)$(80) i

This result shows that adding two SHMs with the same amplitude A and angular frequency ω, but different phase constants ϕ1 and ϕ2 gives a new SHM with:

• the same angular frequency, ω
• a new phase constant (ϕ1 + ϕ2)/2
• a new amplitude, 2Acos[(ϕ1ϕ2)/2].

The particular case A = 2.0 m, ω = 3.0 Hz, ϕ1 = −1.07 and ϕ2 = 2.57 is illustrated in Figure 13. The dotted and dashed curves combine to give the solid lined curve.

Question T12

Use the complex method to find the sum of two SHMs which have the form Asin(ωt) and Acos(ωt), where Α and ω are (real) constants.

Asin(ωt) = Re(−iAeiωt) and Acos(ωt) = Re(Aeiωt) so that:

\begin{align}A\sin(\omega t) + A\cos(\omega t) & = {\rm Re}\left(A\,{\rm e}^{i\omega t}-iA\,{\rm e}^{i\omega t}\right) \\ & = A\,{\rm Re}\left[(1-i)\,{\rm e}^{i\omega t}\right] = A\,{\rm Re}\left(\sqrt{2\os}\,{\rm e}^{-i\pi/4}\,{\rm e}^{i\omega t}\right) \\ & = {\rm Re}\left[\sqrt{2\os}\,{\rm e}^{i(\omega t-\pi/4)}\right] = \sqrt{2\os}A\cos(\omega t-\pi/4) \\ & = \sqrt{2\os}\sin(\omega t+\pi/4)\quad\text{because }1-i = \sqrt{2\os}\,{\rm e}^{-i\pi/4}\end{align}

## 4.2 Superposition of two SHMs differing in angular frequency and phase constant

Suppose now that we want to sum Acos(ω1t + ϕ1) and Acos(ω2t + ϕ2). We can do this by determining the real part of

z(t) = Aexp[i(ω1t + ϕ1)] + Aexp[i(ω2t + ϕ2)]

and using Equation 79,

exp(1) + exp(2) = 2exp[i(α1 + α2)/2]cos[(α1α2)/2](Eqn 79)

with α1 = ω1t + ϕ1 and α2 = ω2t + ϕ2, to obtain

z(t) = Aexp[i(ω1t + ϕ1)] + Aexp[i(ω2t + ϕ2)]

z(t) $= 2A\exp\left\{i[(\omega_1 +\omega_2)t + (\phi_1+\phi_2)]/2\right\}\cos\left(\dfrac{\omega_1-\omega_2}{2}t + \dfrac{\phi_1-\phi_2}{2}\right)$

Taking the real part we obtain

Acos(ω1t + ϕ1) + Acos(ω2t + ϕ2) = Re[z(t)]

Acos(ω1t + ϕ1) $=2A\cos\left[\dfrac{(\omega_1-\omega_2)t + (\phi_1-\phi_2)}{2}\right]\cos\left[\dfrac{(\omega_1+\omega_2)t + (\phi_1+\phi_2)}{2}\right]$(81)

This general expression is fairly complicated and so it is useful to consider a particular case as in the following exercise.

Question T13

Use Equation 79,

exp(1) + exp(2) = 2exp[i(α1 + α2)/2]cos[(α1α2)/2](Eqn 79)

to find the sum of two SHMs which have the form Acos(21ωt + π/4) and Acos(23ωtπ/4) where ω and A are real constants. Sketch the resulting function of t and comment on the form of your result.

$\exp(i\alpha_1) + \exp(i\alpha_2) = 2\exp\left[\dfrac{i(\alpha_1+\alpha_2)}{2}\right]\cos\left(\frac{\alpha_1\alpha_1}{2}\right)$

If we put α1 = 23ωtπ/4 and α2 = 21ωt + π/4 in Equation 79,

exp(1) + exp(2) = 2exp[i(α1 + α2)/2]cos[(α1α2)/2](Eqn 79)

we have

Acos(21ωt + π/4) + Acos(23ωtπ/4)

= 2Acos(ωtπ/4)cos(22ωt)

= 2Asin(ωt + π/4)cos(22ωt)

The cos(22ωt) term gives a rapidly varying oscillation whose amplitude is modulated by the slowly varying sin(ωt + π/4) term.

Oscillations having this form are said to display beats (as illustrated in Figure 18).

## 4.3 Superposition of many SHMs – the diffraction grating

Finally, we consider a problem that is of considerable importance in the study of optics. This concerns the pattern of illumination created on a distant screen when a uniform beam of light of a single colour encounters a diffraction grating. For our present purposes a diffraction grating may be thought of as consisting of a large number of narrow parallel slits, each of which acts as a source of light. Two of these slits, S1 and S2, together with the screen are indicated in Figure 14, though the figure has not been drawn to scale (D should be so much larger than d that the lines S1P and S2P are effectively parallel).

As explained elsewhere in FLAP, light may often be treated as a wave phenomenon; so at a given point P on the screen, the effect of the light spreading out from each of the slits is to create an oscillation that may be represented by acos(ωt + ϕ), where the angular frequency of the oscillation is determined by the colour of the light, and the phase constant ϕ is determined by the distance between the relevant slit and the point P.

Because each slit is at a different distance from the screen, the oscillation that each slit causes at P will be characterized by a particular value of ϕ, and it may be shown that for oscillations due to any pair of neighbouring slits these phase constants will differ by an amount

$\mu = \dfrac{2\pi d}{\lambda}\sin\theta$(82)

where d is the separation of adjacent slits on the diffraction grating, λ is another characteristic of the light (its wavelength), and θ is the angle between the points P and O, measured from the diffraction grating.

In order to determine the total superposition oscillation occurring at P due to the light arriving from the n slits that make up the grating, we need to determine the sum of n SHMs. In other words we need to evaluate a sum of the form

acos(ωt) + acos(ωt + μ) + acos(ωt + 2μ) + ... + acos[ωt + (n − 1)μ]

We can do this by finding the real part of the complex quantity

z(t) = aeiωt + aei(ωt + μ) + aei(ωt + 2μ) + ... + aei[ωt + (n − 1)μ] = aeiωt [1 + e + ei2μ + ... + ei(n − 1)μ]

The sum in the square brackets is a geometric series with first term 1, and common ratio e and (from Question R5) this can be written as:

$1 + {\rm e}^{i\mu} + {\rm e}^{i2\mu} + \dots + {\rm e}^{i(n-1)\mu} = \dfrac{1-{\rm e}^{in\mu}}{1-{\rm e}^{i/mu}}$(83)

and the resulting expression for z(t) is

$z(t) = a\,{\rm e}^{i\omega t}\left(\dfrac{1-{\rm e}^{in\mu}}{1-{\rm e}^{i/mu}}\right)$

The terms involving μ can be rearranged as follows

$\dfrac{1-{\rm e}^{in\mu}}{1-{\rm e}^{i/mu}} = \dfrac{1-{\rm e}^{in\mu/2}}{1-{\rm e}^{i/mu/2}}\left(\dfrac{{\rm e}^{i/mu/2}-{\rm e}^{-i/mu/2}}{{\rm e}^{i/mu/2}-{\rm e}^{-i/mu/2}}\right) = \dfrac{{\rm e}^{in\mu/2}\sin(n\mu/2)}{{\rm e}^{i\mu/2}\sin(\mu/2)}$(84)

Substituting this in the expression for z(t) and taking the real part gives:

${\rm Re}[z(t)] = a\dfrac{\sin(n\mu/2)}{\sin(\mu/2)}\cos\left(\omega t+\dfrac{(n-1)\mu}{2}\right)$

We can see from this result that:

• the angular frequency is ω;
• the phase constant is $\dfrac{(n-1)\mu}{2}$;
• the amplitude, A say, is given by:

$A = a\dfrac{\sin(n\mu/2)}{\sin(\mu/2)}$(85)

Substituting the value of μ from Equation 82,

$\mu = \dfrac{2\pi d}{\lambda}\sin\theta$(Eqn 82)

we can write this as:

$A = a\dfrac{\sin[n\pi d\sin(\theta)/\lambda]}{\sin[\pi d\sin(\theta)/\lambda]}$

The square of this amplitude will be proportional to the intensity of illumination at any point on the screen, provided that the slits are sufficiently narrow. Thus we can expect to observe an intensity distribution that varies with θ in proportion to:

$I(\theta) = a^2\dfrac{\sin^2[n\pi d\sin(\theta)/\lambda]}{\sin^2[\pi d\sin(\theta)/\lambda]}$

This pattern is shown in Figure 15 for the case of 2, 3, 4 and 5 slits. As you can see, increasing the number of slits makes the intensity peaks taller and narrower. In practice, diffraction gratings have a great many slits (~10 000), with the result that the observed pattern of illumination consists of well separated bright lines. This pattern is discussed in more detail in the block of FLAP modules devoted to light and optics.

Note The slits must be sufficiently narrow so that light from each slit diffracts to the point of superposition. If the slits are too wide, an additional effect, the diffraction pattern due to a single slit, will modify the intensity pattern, causing the various intensity maxima to reduce in brightness away from the θ = 0 maximum.

Question T14

Find the sum of asin(ωt) + asin(ωt + ϕ) + asin(ωt + 2ϕ) + ... + asin[ωt + (n − 1)ϕ]

Consider:

z(t) = aeiωt + aei(ωt+ϕ) + aei(ωt+2ϕ) + ... + aei[ωt+(n−1)ϕ] = aeiωt[1 + e + ei2ϕ + ... + ei(n−1)ϕ]

then:

Im[z(t)] = asin(ωt) + asin(ωt + ϕ) + asin(ωt + 2ϕ) + ... + asin(ωt + (n − 1)ϕ)

$= {\rm Im}\left[a\,{\rm e}^{i\omega t}\left(\dfrac{1-{\rm e}^{in\phi}}{1-{\rm e}^{i\phi}}\right)\right] = {\rm Im}\left[a\,{\rm e}^{i\omega t}\dfrac{{\rm e}^{in\phi/2}}{{\rm e}^{i\phi/2}}\left(\dfrac{{\rm e}^{in\phi/2} - {\rm e}^{-in\phi/2}}{{\rm e}^{i\phi/2} - {\rm e}^{-i\phi/2}}\right)\right]$

$= {\rm Im}\left[a\,{\rm e}^{(i\omega t+in\phi/2-i\phi/2)}\left(\dfrac{\sin(n\phi/2)}{\sin(\phi/2)}\right)\right] = a\sin[\omega t+(n-1)\phi/2]\left(\dfrac{\sin(n\phi/2)}{\sin(\phi/2)}\right)$

# 5 Closing items

## 5.1 Module summary

1

Differential equations of second order with constant coefficients arise from many physical situations, in particular: mechanical and electrical systems.

A mass m subject to a sinusoidal driving force F(t) = F0sin(Ωt), a damping force proportional to velocity and a restoring force proportional to displacement from the origin, will satisfy a differential equation of the form

$m\dfrac{d^2x(t)}{dt^2} + kx(t) + b\dfrac{dx(t)}{dt} = F_0\sin({\it\Omega}t)$(Eqn 13)

The charge q(t) on a capacitor in a series LCR circuit containing a resistance R, a capacitance C and an inductance L, subject to an externally applied voltage V(t) = V0sin(Ωt), will satisfy a similar differential equation

$\dfrac{d^2q}{dt^2} + R\dfrac{dq(t)}{dt} + \dfrac1C q(t) = V_0\sin({\it\Omega}t)$(Eqn 17)

Both equations give rise to harmonically driven, linearly damped harmonic oscillations.

2

In the absence of damping and driving, Equation 13,

$m\dfrac{d^2x(t)}{dt^2} + kx(t) + b\dfrac{dx(t)}{dt} = F_0\sin({\it\Omega}t)$(Eqn 13)

reduces to the equation of simple harmonic motion i (SHM):

$\dfrac{d^2x(t)}{dt^2} + \omega_0^2 x(t) = 0$, where $\omega_0 = \sqrt{k/m\os}$

This has the general solution,

x(t) = A0sin(ω0t + ϕ)(Eqn 34)

where A0 and ϕ are arbitrary constants that are determined by the initial conditions. A0 is the amplitude, ϕ the phase constant and ω0 is the (natural) angular frequency while the constants T = 2π/ω0 and f = 1/T = ω0/2π are known, respectively, as the period and the frequency of the motion.

3

In the absence of any driving term, Equation 13 reduces to the equation of damped harmonic motion $\dfrac{d^2x(t)}{dt^2} + \gamma\dfrac{dx(t)}{dt} + \omega_0^2 x(t) = 0$, where $\omega_0 = \sqrt{k/m\os}$ and γ = b/m. The general solution depends on the relative values of γ and ω0, but in the physically important case of underdamping it takes the form:

x(t) = eγt/2[A0sin(ωt + ϕ)](Eqn 42)

where A0 and ϕ are arbitrary constants and $\omega = \sqrt{\omega_0^2-\dfrac{\gamma^2}{4}}$.

4

The equation of harmonically driven linearly damped oscillation may be written in the form

$\dfrac{d^2x(t)}{dt^2} + \gamma\dfrac{dx(t)}{dt} + \omega_0^2 x(t) = a_0\sin({\it\Omega}t)$

and has a general solution that is the sum of a transient term and a steady state term. In the case of underdamping, this general solution takes the form:

x(t) = eγt/2[A0sin(ωt + ϕ)] + Asin(Ωtδ)(Eqn 47)

A0 and ϕ are arbitrary constants, $\omega = \sqrt{\omega_0^2-\dfrac{\gamma^2}{4}}$(Eqn 48)

$A = \dfrac{a_0}{\sqrt{\smash[b]{(\omega_0^2 - {\it\Omega}^2)^2+(\gamma{\it\Omega})^2}}}\quad\text{and}\quad\arctan\left(\dfrac{\gamma{\it\Omega}}{\omega_0^2 - {\it\Omega}^2}\right)$(Eqn 49)

In the steady state this reduces to x(t) = Asin(Ωtδ).

5

When a voltage V(t) = V0sin(Ωt) is applied across a series LCR circuit, the resulting steady state current has the form

I(t) = I0sin(Ωtδ)

where V0 = I0Z  and  δ = arctan(X/R)

Z being the impedance given by:

$Z = \sqrt{R^2 + X^2}$ and X = XLXC = ΩL − 1/(ΩC) the total reactance.

6

In the case of the mechanical system described by Equation 13 the Subsection 2.6mechanical impedance is defined by Zm = F0/υ0, where υ0 is the amplitude of the Subsection 2.7velocity oscillation and $Z_m = \sqrt{b^2 + \left(\dfrac{k}{\it\Omega} - {\it\Omega}m\right)}$.

7

Resonance is the phenomenon whereby a driven oscillator exhibits large amplitude oscillations when driven at a frequency close to the natural frequency it would have in the absence of any driving or damping.

8

When analysing a.c. circuits in their steady state, it is convenient to represent the applied voltage V(t) as the real part of a complex quantity ${\mathscr V}(t) = {\mathscr V}_0\,{\rm e}^{i\omega t}$ and the current I(t) as the real part of a complex quantity ${\mathscr I}(t) = {\mathscr I}_0\,{\rm e}^{i\omega t}$, where ${\mathscr V}_0\;\text{and}\;{\mathscr I}_0$ are complex constants. The complex form of Ohm’s law is then the equation ${\mathscr V}(t) = {\mathscr I}(t){\mathcal Z}$ where ${\mathcal Z}$ is the complex impedance. The value $Z = \lvert{\mathcal Z}\rvert$ is the impedance, and $\delta = \arg({\mathcal Z})$ is the extent by which the phase of the current lags behind that of the driving voltage.

9

For complex impedances combined in series

${\mathcal Z} = {\mathcal Z}_1 + {\mathcal Z}_2 + \dots + {\mathcal Z}_n$(Eqn 73)

For complex impedances combined in parallel

$\dfrac{1}{\mathcal Z} = \dfrac{1}{\mathcal Z}_1 + \dfrac{1}{\mathcal Z}_2 + \dots + \dfrac{1}{\mathcal Z}_n$(Eqn 74)

For a single resistor, ${\mathcal Z}_R = R$, for a single capacitor ${\mathcal Z}_C = 1/(i\omega C)$, and for a single inductor ${\mathcal Z}_L = i\omega L$.

10

The average power dissipated by a circuit carrying alternating current is

$\langle P\rangle = \dfrac12 {\rm Re}\left({\mathscr I}_0^*\,{\mathscr V}_0\right)$(Eqn 78)

11

The effect of superposing harmonic oscillations may be found by adding together appropriate complex quantities and then taking the real part of the result (or the imaginary part if appropriate).

## 5.2 Achievements

Having completed this module, you should be able to:

A1

Define the terms that are emboldened and flagged in the margins of the module.

A2

Use second–order differential equations to model a variety of oscillatory problems and to highlight the analogy between mechanical and electrical systems in situations where both are modelled by similar differential equations.

A3

To explain the significance of impedance and resonance in relation to the amplitude of the steady state solution to the sinusoidally driven, linearly damped harmonic oscillator.

A4

Use complex numbers to solve simple problems involving LCR circuits and to display relative magnitudes and phases of currents and voltages by means of an Argand diagram.

A5

To calculate the average power dissipated by a suitable component in a simple a.c. circuit.

A6

Use complex numbers to solve problems involving SHM, including the superposition of two or more SHMs.

Study comment You may now wish to take the following Exit test for this module which tests these Achievements. If you prefer to study the module further before taking this test then return to the topModule contents to review some of the topics.

## 5.3 Exit test

Study comment Having completed this module, you should be able to answer the following questions each of which tests one or more of the Achievements.

Question E1 (A2)

The general solution of the differential equation

$\dfrac{d^2x(t)}{dt^2} + 2q\omega\dfrac{dx(t)}{dt}+(n^2 +q^2)\omega^2x(t) = 0$

isx(t) = eqωt[Acos(nωt) + Bsin(nωt)]

and a particular solution of the differential equation

$\dfrac{d^2x(t)}{dt^2}+2q\omega\dfrac{dx(t)}{dt^2}+(n^2 +q^2)\omega^2x(t) = h\sin({\it\Omega}t)$

isxp(t) = ccos(Ωt) + dsin(Ωt)

where c = −2qωΩh/H  and  d = [(q2 + n2)ω2Ω2]h/H

withH = q4ω2 + 2q2ω2(n2 + 1) + (n2ω2Ω2)2   and   nΩ/ω.

Use this solution to write down the transient current and the steady state current in a series LCR circuit containing a resistor 6 Ω, a capacitor 1/13 F and an inductor 1 H, when it is driven by an applied voltage V(t) = V0cos(Ωt + π), where V0 = 1.00 V and Ω = 1.00 s−1. Describe an analogous mechanical system.

The differential equation that determines the current is (from Equation 19)

$L\dfrac{d^2I(t)}{dt^2} + R\dfrac{dI(t)}{dt} + \dfrac1CI(t) = \dfrac{dV(t)}{dt}$(Eqn 19)

which in this case becomes:

$(1\,{\rm H})\dfrac{d^2I(t)}{dt^2} + (6\,\Omega)\dfrac{dI(t)}{dt} + (13\,{\rm F}^{-1})I(t) = (1\,{\rm V}^{-1})\sin({\it\Omega}t)$

sinceif V(t) = V0cos(Ωt + π) = −V0cos(Ωt)   then   $\dfrac{dV(t)}{dt} = V_0{\it\Omega}\sin({\it\Omega}t)$

If we choose q = 3, n = 2 and ω = 1 s−1 in the given solution we obtain the transient term

Itr(t) = exp[−(3s−1)t]{Acos[(2s−1)t] + Bsin[(2s−1)t]}

where A and B are constants determined by the initial conditions.

For q = 3, n = 2  and  ω = 1 s−1 we also have   H = 180 s−2 so that c = −1/30 A and d = (1/15) A.

The steady state current is given by

Iss(t) = [−(1/30)cos(t s−1) + (1/15)sin(t s−1)] A

From Equation 13,

$m\dfrac{d^2x(t)}{dt^2} + kx(t) + b\dfrac{dx(t)}{dt} = F_0\sin({\it\Omega}t)$(Eqn 13)

a driving force of F(t) = cos(ωt) N applied to a mass 1 kg on a spring with spring constant (13 N m−1) and damping constant (6 N s m−1) will give the same differential equation.

Question E2 (A2)

What is the resonant frequency of the electrical circuit described in Question E1? What would you expect to happen when the angular frequency of the applied voltage is close to this value?

The resonant angular frequency ω0 of the circuit is

$\omega_0 = \dfrac{1}{\sqrt{LC\os}} = \dfrac{1}{\sqrt{1/13\os}}\,{\rm s^{-1}} \approx 3.61\,{\rm s^{-1}}$

If all the components are kept fixed and only the angular frequency of the current is changed, then when the angular frequency of the applied voltage is close to the value 3.61 s−1 the amplitude of the current will reach a maximum value. This maximum amplitude is not necessarily large; however, if the resistance is small we would expect the amplitude of the current to be very large when the circuit resonates.

Question E3 (A4 and A5)

Calculate the complex impedance of a resistance 6 Ω, a capacitor 1/13 F and an inductor 1 H in series. What is the current through the circuit if a voltage V(t) = V0cos(ωt + π), where ω = 1 s−1, is applied to the circuit? Sketch the complex current and voltage on an Argand diagram, and calculate the average power dissipated.

The complex impedance Z of a series LCR circuit is given by

${\mathcal Z} = R + i\omega L + \dfrac{1}{i\omega C} = \left(6 + i\omega + \dfrac{13}{i\omega}\right)\,Ω = \left(6 + i\omega - \dfrac{13i}{\omega}\right)\,Ω$

When ω = 1 s−1 we have ${\mathcal Z} = (6 - 12i)\,\Omega = 6(1 - 2i)\,\Omega \approx 13.4\,{\rm e}^{-1.1i}\,\Omega$

We have V(t) = V0cos(ωt + π) and we write

${\mathscr V}(t) = V_0\,{\rm e}^{(\omega t+\pi)i}\,{\rm V}$

which gives ${\mathscr I}(t) = \dfrac{\mathscr V}{\mathcal Z} \approx \dfrac{V_0\,{\rm e}^{(\omega t+\pi)i}}{13.4\,{\rm e}^{-1.1i}}\,{\rm A} \approx 0.07V_0\,{\rm e}^{(\omega t+\pi)i}\,{\rm A}$

and therefore

$I(t) = {\rm Re}[{\mathscr I}(t)] \approx 0.07V_0\cos(\omega t + 4.2)\,{\rm A}$

where ω = 1 s−1.

[You should notice that the answers to Questions E1 and Question E3E3 are in fact identical except for the factor V0 because, provided ω = 1 s−1, −(1/30)cos(t s−1) + (1/15)sin(t s−1) can be rewritten in the form 0.07cos(ωt + 4.2) using the trigonometric identity   cos(α + β) = cosαcosβ − sinαsinβ]

Question E4 (A6)

Use a complex representation to find the result of adding two SHMs of the form asin(ωtkx) and asin(2ωt + 2kx).

If we put α1 = 2ωt + 2kx and α2 = ωtkx in Equation 80,

$A\cos(\omega t+\phi_1) + A\cos(\omega t+\phi_2) = \underbrace{2A\cos\left(\dfrac{\phi_1-\phi_2}{2}\right)}_{\color{purple}{\large\text{amplitude }}}\cos\left(\omega t + \dfrac{i(\phi_1+\phi_2)}{2}\right)$(Eqn 80)

we obtain:

${\rm e}^{i(\omega t-kx)} + {\rm e}^{i(2\omega t+2kx)} = 2\,{\rm e}^{i(3\omega t+kx)/2} \cos\left(\dfrac{\omega t+3kx}{2}\right)$

and taking the imaginary part of this equation gives:

$a\sin(\omega t - kx) + a\sin(2\omega t - 2kx) = 2a\sin\left(\dfrac{3\omega t + kx}{2}\right)\cos\left(\dfrac{\omega t + 3kx}{2}\right)$