vibrations and waves – problem sheet 1€¦ · vibrations and waves (jan-feb 2011) problem sheet...

Vibrations and Waves (Jan-Feb 2011) Problem Sheet 1

Vibrations and Waves – Problem Sheet 1(Covers material in Lectures 1 & 2)

1. Some simple harmonic oscillators were found to follow:

(i) x(t) = 2 exp( i6t)

(ii) x(t) = i3 exp( i5t)

(iii) x(t) = (2 + i3) exp( i6t)

(iv) x(t) = (1 − i5) exp( i6t)

Find the real part of these solutions.

2. Consider two springs of stiffness k1 and k2 connected in series, as shown in the figure below.Let x1 be the displacement of spring 1 and x2 the displacement of spring 2.

(i) Apply Hooke’s law to each spring separately to show that x1 = k2x2/k1. Hint: the force isthe same on each spring.

(ii) We want to find the effective spring constant, keff , for the two springs in series, i.e. F =

−keff(x1 + x2). Use this together with the result from (i) to show that keff = (1/k1 + 1/k2)−1.

(iii) What would be the effective spring constant for the two springs in parallel?

3. A spring is hung vertically from a support. A mass is attached to the end of the spring. Thevertical displacement of the mass is observed to follow the equation:

x(t) = 0.05 cos(7.51t)

where everything is in SI units. What, in SI units, is (i) the amplitude A, (ii) the natural frequencyω0, (iii) the frequency f and (iv) the period T . If the mass is 0.1 kg, what is the spring constantk of the spring?

How far would the spring have stretched when the mass was initially attached to its end?

Take g = 9.8 m/s2.

4. The motion of the horizontal mass on a spring has the general solution:

x(t) = A cos(4t + φ)

Work out the variation of the velocity v(t) with time. By considering the initial conditions, workout the value of A and φ for the following cases:

(i) t = 0, x = 0.3 m, v = 0

(ii) t = 0, x = −0.5 m, v = 0

(iii) t = 0, x = 0, v = 1.2 m/s

5. A U shaped tube of cross-sectional area A is filled with a liquid of density ρ. A total length L isfilled, as illustrated below:

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L

x

A

ρ

The liquid is initially displaced by +h on one side and h on the other side and is then released.

(i) Show that the Equation of Motion is given by:

LAρd2xdt2 = −2Aρgx

(ii) By solving the Equation of Motion, find the variation of the vertical displacement x(t) of theliquid with time t.

(iii) At what natural frequency ω0 does the liquid oscillate?

(iv) What is the velocity v(t) of the oscillating liquid?

(v) What is the acceleration a(t)?

(vi) Derive the variation of the potential energy PE of the liquid with x and t.

(vii) Derive the variation of the kinetic energy KE of the oscillating liquid with t.

(viii) Find the total energy TE of the oscillating liquid.

(ix) Find the variation of KE with x.

6. Show that the restoring force about any stable equilibrium point is linear (i.e. Hooke’s Law isvalid) for sufficiently small displacements. Since a linear restoring force leads to SHM, thisexplains why SHM is so widespread and important.

[Hint: Consider the potential U(x) which is related to the force by F(x) = − dU(x)/ dx. Make aTaylor series of U(x) about the equilibrium point x0 and then think what is meant by x0 being astable equilibrium point.]

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Vibrations and Waves – Problem Sheet 2(Covers material in Lectures 3–6)

1. The equation of motion for a damped system is mx + bx + kx = 0. Show that for a criti-cally damped system ω0 = γ/2, where the symbols have their usual meanings. For criticallydamped SHM, the displacement is given by x(t) = (A + Bt) exp [−(γ/2)t], where A and B areconstants. Show that this is a solution of the equation of motion. [Hint: First write the equationof motion and x(t) in term of A, B, ω0 and t only.]

2. For lightly damped SHM, the displacement can be written as x(t) = A e−γt/2 cos(ωdt + φ). Findthe constants A, φ, for the following initial conditions with a mechanical system: mass pulledto x = x0 and released with zero velocity.

3. The electrical equivalent of a damped mass on a spring driven by an external force F(t) =

F0 cosωt is an LCR circuit with an A/C voltage source V(t) = V0 cosωt.

The equation of motion of this system is

Ld2qdt2 + R

dqdt

+qC

= V0 cosωt

(i) Identify each of the parameters in this equation of motion with its equivalent parameterin the mass-on-a-spring system.

(ii) Rearrange the equation of motion into the standard form and hence write down the quan-tities for γ and ω0 in terms of the electrical parameters.

(iii) Show that in the steady state solution, the complex amplitude is

A =(V0/L)

ω20 − ω2 + iωγ

[Hint: use a trial solution A e iωt.] Calculate the magnitude A and the phase φ of A.

(iv) The time-average power dissipated by the mechanical system when driven by an exter-nal force F(t) = F0 cosωt is Pav = 1

2 bω2A2 where A is the amplitude of the oscillation.Without detailed calculation write down the equivalent formula for the time-average powerdissipated by the LCR system.

(v) Show that the maximum power dissipation Pav occurs at ω0 and show that this maxi-mum value is V2

0/(2R). Explain in physical terms why the average power increases as Rdecreases in the electrical system and as b decreases in the mechanical system.

(vi) Find the driving frequencies ω1, ω2 (ω1 < ω2) for which Pav drops to half its maximumvalue. [Hint: you should be able to derive an equation that gives four values for ω, two ofwhich you can discard.] Hence show that ∆ω = γ and ω0/∆ω = Q, where ∆ω = ω2 − ω1is the full-width-at-half-maximum. and Q is the Q-factor. Note that ω0/∆ω is a measureof the “sharpness” of the peak in the plot of Pav versus ω.

(vii) Show that the average power dissipated by the resistance is exactly the average powersupplied by the driving voltage V(t), [Hint: You may find the result from (iii) useful tocalculate the current in the circuit.]

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4. In an experiment, the driving frequency, ω, of a damped mass on a spring system is graduallyincreased and Pav is measured as a function of ω to give the following curve:

(i) From the curve above, what is the Q of the system?

(ii) If the mass is 1.0 kg, what are the spring and damping constants?

5. In Problem Sheet 1, we showed that the equation of motion for liquid in a U-shaped tube is

LAρx = −2Aρgx

where L is the total length of liquid in the tube, A the cross-sectional area of the tube, ρ thedensity of the liquid and g = 9.8 m/s2 is the acceleration due to gravity.

A more careful analysis reveals that liquid moving in a tube is subject to a damping forceFdamp = −8πηLv where η is the viscosity of the liquid and v is its velocity.

(i) Write down the new equation of motion including the damping force.

(ii) Write down expressions for the damped frequency of oscillation ωd =

√ω2

0 − γ2/4, the

resonant frequency ωres =

√ω2

0 − γ2/2 and the Q-factor in terms of the parameters forthe U-tube.

Let A = 0.5 × 10−4 m2 and L = 5 m. For the two cases (iii) and (iv) below, determine whetherthe damping is light, critical or heavy and if lightly damped, calculate the damped frequencyωd and Q:

(iii) U-tube filled with water. Take ρwater = 103 kg/m3 and ηwater = 10−3 Ns/m2.

(iv) U-tube filled with oil. Take ρoil = 0.9 × 103 kg/m3 and ηoil = 200 × 10−3 Ns/m2.

(v) An air pump is attached to one end of the tube that provides an oscillatory pressure ofamplitude P0 = F0/A = 5 × 103 N/m2 at a driving frequency of ω. What is the resonantfrequency ωres of water in the U-tube [use 3 decimal places]. Calculate the steady-stateamplitude of the motion and the average power supplied by the pump both at resonanceωres and at ω0. Explain in physical terms why the peak power is not at the peak amplitude.

Numerical Answers:

4(vi) γ ≈ πs−1, Q ≈ 6 4(vii) k = 36π2 N/m = 355 N/m, b = πNm−1s = 3.14 Nm−1s 5(iii)water is lightly damped, ωd(water) = 1.964 s−1, Qwater = 3.94, 5(iv) oil is heavily damped, 5(v)ωres = 1.948 s−1, Aav(ωres) = 1.013 m, Aav(ω0) = 1.005 m, Pav(ωres) = 0.245 W, Pav(ω0) = 0.249 W,

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Vibrations and Waves – Problem Sheet 3(Covers material in Lectures 7–10)

1. It is observed that a pulse takes 0.1 sec to travel from one end to the other of a long stretchedstring. The tension in the string is provided by passing one end over a pulley and attaching itto a hanging weight of mass 153 times that of the string. What is the length of the string? Takeg = 9.8 m/s2.

2. A transverse wave travels down a stretched string in the positive z direction. The resultingmotion of two points on the string at z1 = 0 and z2 = 2 m is given by

ψ1(z1, t) = 0.2 cos(3πt)

ψ2(z2, t) = 0.2 cos(3πt − π/8)

i) What is the frequency of the wave in Hz?

ii) What is the wavelength? Hint: is your answer the only one consistent with the giveninformation?

iii) What is the velocity of the wave?

3. Prove the principle of superposition for the non-dispersive wave equation.

4. If f is an arbitrary function, show that ψ(z, t) = f (z− vt) is a solution to the wave equation

∂2ψ

∂t2 = v2 ∂2ψ

∂z2

Hint: write f (z− vt) as f (y) and use the chain rule to differentiate w.r.t t and z.

5. To convince yourself that f (z − vt) is indeed the function f (z) moving at velocity v along the zaxis, consider a “top-hat” function given by

f (z) =

{1 m −0.5 m ≤ z ≤ 0.5 m0 z < −0.5 m, z > 0.5 m

(The reason for the name “top-hat” will become evident in a moment).

i) Sketch f (z) as a function of z

ii) Assume v = 1 m/s. Sketch on the same axes f (z − vt) for t = 1 s and t = 5 s. Byconsidering the displacement of the top-hat in ∆t = 5 s − 1 s = 4 s, show that its velocityis indeed v.

iii) Repeat (b) for v = −1 m/s.

6. Make the following functions solutions to the wave equation:

i) f (z) = A exp(−z2/w2) moving in the positive z direction at speed 5 m/s. f (z) here is aGaussian function of width w.

ii) f (z) = (1 + z2/w2)−1 moving in the negative z direction at speed 2 m/s. f (z) here is aLorentzian function of width w

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7. The diagram below shows a pair of equal pendulums coupled together by a light rigid rod.Construct such a coupled oscillator system for yourself. Find two equal masses, use stringor cotton thread and (for example) a drinking straw for the light rod. (Make sure the physicaldimensions of your masses are not too large compared with the rest of the system). Calculatethe two normal modes for motion in the plane of the system, either using the general approachwe used in the lectures or by using symmetry. Check to see if your calculated motions andfrequencies agree with what you measure. Try to observe beating and check the beat periodagrees with your predictions.

8. (More challenging) Most real travelling waves lose energy as they propagate through a medium—a process known as attenuation. We can model attenuation on our stretched string by imagin-ing the string being immersed in a viscous oil. Then each segment of the string is subject to avelocity-dependent damping force, just like we considered for damped SHM.

i) Assume the damping force on a small segment of string of length ∆z is Fdamp = −βv∆zwhere v is the transverse velocity of the segment and β is the damping constant per unitlength. By considering the forces on the segment as we did in the derivation of the waveequation for an ideal, undamped string, show that the new damped wave equation is

∂2ψ

∂t2 = v2 ∂2ψ

∂z2 − Γ∂ψ

∂t

where ψ is the transverse string displacement, v =√

T/µ is the wave speed, T is thetension, µ is the mass per unit length of the string, and Γ = β/µ.

ii) Show that ψ(z, t) = A exp[ i(ωt − kz)] is a solution and derive the dispersion relation forthe damped string.

iii) Hence show that the wavenumber is complex for the damped case. You are not requiredto evaluate k and you may assume ω is real.

iv) Assume the wavenumber can be written as k = K − iκ where K and κ are real. Showthat this leads to a travelling wave with an exponentially decaying amplitude. What is thephysical interpretation?

Numerical Answers:

1) 15 m; 2(i) 1.5 Hz, (ii)32/(16n + 1) m with n = 0, 1, 2, 3, . . . (iii) 48/(16n + 1) m/s;

Carl Paterson 2 of 2


Vibrations and Waves – Problem Sheet 4

1. The following two waves in a medium are superposed:

ψ1(z, t) = A cos(10t − 5z)

ψ2(z, t) = A cos(9t − 4z)

where z is in metres and t in seconds.

i) Write an equation for the combined disturbance.ii) What is the beat frequency in Hz?iii) What is its group velocity?

2. This question tests your understanding of phase and group velocities.

i) Using the chain rule for differentiation, show that dω2/ d(k2) = vgvp where ω, k are thefrequency and wavevector for a wave, and vg and vp are the group and phase velocities,respectively.

ii) Electromagnetic waves propagate through plasmas with a refractive index given by n =√1 − ω2

p/ω2 where ωp is a constant for given plasma conditions known as the plasma

frequency. Write down an expression for the phase velocity for waves in a plasma andhence show that ω and k are related by the plasma dispersion relation ω2 = c2k2 + ω2

p,where c is the speed of light.

iii) Using (i) show that this dispersion relation leads to the result that vg = c√

1 − ω2p/ω

2.

iv) Can vp exceed c? What about vg? Comment on your answers.

3. Short wavelength ripples (λ . 1 cm) moving on the surface of water are controlled by surfacetension. The phase velocity of such ripples is given by

vp =

√2πSρλ

where S is the surface tension and ρ the density of water.

i) Use dimensional analysis to show that S has units of force per unit length.ii) Determine the dispersion relation and sketch it as a function of wavenumber k. Is the

dispersion anomalous or normal. Explain your answer.iii) Show that the group velocity of a wavegroup of ripples is 1.5 times the phase velocity.iv) A pebble is tossed into a pond of 2 metres diameter at one side. This creates a ripple

wavegroup consisting of waves with λ in the range 0.1–1 cm. How long does it take forthe wavegroup to cross the pond? For water S = 0.073 N/m (at 20◦) and ρ = 103 kg/m3.

4. This question will test you ability to apply Snell’s law of refraction. Light is incident at an angleθ0 from a medium with refractive index n0 onto one with refractive index n1 and then onto onewith refractive index n2 (left diagram). The interfaces between the media are parallel.

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i) Use Snell’s law to obtain an expression for the angle θ2.

ii) Hence show that the angle in the bottom layer of a stack of parallel layers of differentrefractive indices does not depend on the intermediate layers (right diagram).

iii) The refractive index of air at sea level is n = 1.0003. Obtain an expression for theapparent angle θ of the sun from zenith (i.e., from the vertical) at sea level when its trueangle is θ0 .

iv) Plot the deviation θ − θ0 as a function of θ0.

v) Use your expression to obtain an estimate of the deviation of the sun from its true anglewhen the sun is on the horizon. Why is you result likely to be an overestimate?

5. The following is probably the most complicated situation you can get for the Doppler effect—moving source and echo from moving target!

A bat flying at speed uB is out hunting. Using echo-location it detects a moth flying at speeduM in a straight line away from it towards a tree. The bat emits a sound wave of frequency fBtowards the moth. The speed of sound in air is v. You may use f ′ = (v+uR)/(v−uS) f for source(S) and receiver (R) moving with speeds uR and uS respectively towards each other relative tothe medium. (The double dash in (c),(d) indicates that two Doppler shifts are involved.)

i) Show that the sound wave arriving at the tree has frequency f ′T = [v/(v − uB)] fB.

ii) Show that the sound wave arriving at the moth has frequency f ′M = [(v−uM)/(v−uB)] fB.

iii) Show that the echo received by the bat from the tree is f ′′T = [(v + uB)/(v − uB)] fB.

iv) Show that the echo received by the bat from the moth is

f ′′M =(

v + uB

v + uM

)(v − uM

v − uB

)fB.

v) Some bats emit a burst of sound waves that changes its frequency from start to finish (ithas a frequency chirp). If uB = 4 m/s, uM = 0.1 m/s and v = 344 m/s what frequenciesemitted by bat would result in echoes from the tree and the moth being received at 10 kHzby the bat.

6. Waves scattered from atoms in a crystal can interfere to make a “diffraction pattern”. This isknown as “Bragg Diffraction”. The details of this diffraction pattern can be used to determinethe crystal’s structure.

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i) Using the figure above, show that the path difference between waves scattered fromatoms in adjacent planes is 2d sin θ. Note that it is customary to measure θ from thelattice plane and not relative to the normal to the surface.

ii) What is the condition for constructive interference of waves at a (distant) point P on adetector?

iii) Lattice planes in crystals are typically spaced by a few Angstroms (1 A = 10−10 m). Whatkind of waves are diffracted by crystals?

The following questions are from a past exam paper (different lecturer). Note: The notation in placesis slightly different from those used in the lecture course. [Numerical Answers: 7i)b)T∗ = 4000 N, c)fbeat = 111 Hz. ii) b)3.44 m, c)0.254 mW/m2]


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Vibrations and Waves (Jan-Feb 2011) Problem Sheet 1 ANSWERS

Vibrations and Waves – Problem Sheet 1 ANSWERS

1. (i) Re[x] = 2 cos(6t)

(ii) Re[x] = −3 sin(5t)

(iii) Re[x] = 2 cos(6t)− 3 sin(6t)

(iv) Re[x] = cos(6t) + 5 sin(6t)

2. (i) Since force is same on each spring F = −k1x1 = −k2x2 ⇒ x1 = k2x2/k1 as required.

(ii) The force applied to the combined system is F and resulting combined extension is x1 + x2.Substituting for x1 and x2 from (i) into the equation given for keff gives F = −keff(−F/k1 −F/k2)⇒ keff = (1/k1 + 1/k2)−1.

(iii) For springs in parallel, the extension is the same for both springs (x1 = x2 = x). The totalforce is the sum of the forces due to each spring, F = F1 + F2 = −k1x + k2x = −(k1 + k2)x.Therefore the effective spring constant is keff = k1 + k2.

3. (i) A = 0.05 m

(ii) ω0 = 7.51/(2π)rad/s (although rad is dimensionless, keeping the rad in the units remindsus that it’s an angular frequency).

(iii) f = ω0/(2π) = 1.20 Hz

(iv) T = 1/ f = 0.84 s

Using ω20 = k/m gives k = mω2

0 = (0.1)(7.51)2N/m = 5.64 N/m.

Initial extension x0 is such that spring restoring force balances gravitational force, so that−kx = −mg. Then x0 = F/k = mg/k = g/ω2

0 = (9.8 m/s2)/(7.51 /s)2 = 0.17 m.

4. v(t) = dx(t)/ dt = −4A sin(4t + φ)

Use A =√

x2 + (x/ω)2 and cos φ = x/A and sin φ = −x/(ωA)

(i) A =√

0.32 + (0)2 = 0.3 m and cos φ = 1, sin φ = 0 ⇒ φ = 0

(ii) A =√

(−0.5)2 + (0)2 = 0.5 m and cos φ = −1, sin φ = 0 ⇒ φ = π

(iii) A =√

(0)2 + (1.2/4)2 = 0.3 m and cos φ = 0, sin φ = (−1.2)/(4×0.3) = −1 ⇒ φ = −π/2

5. (i) When the liquid is displaced by = +x on the RHS (and −x on the LHS) there is a netgravitational restoring force on the liquid equivalent to the weight difference on the two sides.F = −volume× density× g = −2xAρg (minus since it’s a restoring force). Using F = ma, withthe total mass of the liquid given by LAρ, this gives

−2xAρg = LAρd2xdt2

(ii) Rearranging (and cancelling Aρ from both sides) gives

d2xdt2 +

2gL

x = 0

which is the same form as the SHM equation x + ω20x = 0 and therefore has solutions x(t) =

B cos(ω0t + φ) with ω0 =√

2g/L. The amplitude B is determined by initial conditions x(0) =

h, x(0) = 0 which gives B = h, φ = 0 giving x(t) = h cos(ω0t).

(iii) ω0 =√

2g/L.

(iv) v(t) = x(t) = −ω0h sin(ω0t)

(v) a(t) = x(t) = −ω20h cos(ω0t)

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(vi) From the diagram, the change in PE from equilibrium is equivalent to moving volume Axof liquid (from the left to the right side) and raising it a distance x. Therefore PE = (ρAx)gx =

gρAx2 = gρAh2 cos2(ω0t)

(vii) KE = (1/2)mx2 = (1/2)ρLAω20h2 sin2(ω0t). But ω2

0 = 2g/L, and so KE = ρgAh2 sin2(ω0t).

(viii) Total energy TE = PE + KE = ρgAh2.

(ix) KE = TE− PE = ρgAh2 − PE = ρgAh2 − ρgAx2 = ρgA(h2 − x2)

6. Taylor series expansion of U(x) about x0 gives

U(x) = U(x0) + U ′(x0)(x− x0) + (1/2)U ′′(x0)(x− x0)2 + (1/3!)U ′′′(x0)(x− x0)3 + . . .

Because x0 is a stable equilibrium, U ′(x0) = 0 (no net force) and U ′′(x0) > 0 (local minimum ofU.) Therefore the force, F(x) = − dU(x)/ dx is given by

F(x) = −U ′′(x0)(x− x0)− (1/2)U ′′′(x0)(x− x0)2− . . .

For small displacements (x− x0) we can neglect the second and higher terms, so that F(x) =

−U ′′(x0)(x− x0) = −k(x− x0), where k = U ′′(x0) > 0 i.e., a linear restoring force.

Note, small displacements are such that |[(1/2)U ′′′(x0)/U ′′(x0)](x − x0)| << 1 (with similarrelationships for the higher order terms.)

Finally, if you allow the displacements to increase, you should see that there comes a pointwhen the second and higher terms that you have neglected here start to become relevantagain and the system starts to become non-linear. This is an important point: most physicalsystems are not purely linear or purely non-linear, but have ranges in which they can be treatedas linear and ranges in which they are non-linear. Of course the size of the linear range canvary enormously between different systems.



Vibrations and Waves – Problem Sheet 2 ANSWERS

1. Equation of motion is x+γx+ω20x = 0. Substitute in trial solution x = A e iωt gives α2 +γα+ω2

0 =

0. This has oscillatory solutions if γ2 < 4ω20. Critical damping is when γ2 = 4ω2

0. which givesγ = 2ω0.

Use critical damping condition γ = 2ω0, to eliminate γ from the equation of motion and x(t).Then substitute x(t), x(t) and x(t) into the LHS of the equation of motion. All terms cancel eachother so that the LHS is identically zero. Therefore x(t) is a valid solution.

2.x = A e−

γ2 t cos(ωdt + φ)

Differentiate to get

v = x = A e−γ2 t[−γ

2cos(ωdt + φ)− ωd sin(ωdt + φ)

]At t = 0, with the boundary values these give

x(0) = x0 = A cos φ and v(0) = 0 = A[−γ

2cos φ− ωd sin φ

]Equation for v(0) = 0 gives tan φ = −γ/(2ωd). Substituting this result into the equation for x(0)gives

A =x0

cos φ= x0

√γ2

4ω2d

+ 1

3. (i) L→ m, q→ x, R→ b, 1/C → k, V0 → F0

(ii)

q +RL︸︷︷︸γ

q +1

LC︸︷︷︸ω2

0

q =V0

Lcosω ⇒ γ = R/L ω0 =

√1/LC

(iii) Substituting trial solution q = A e iωt into the complex form of the equation of motion, gives

(−ω2 + iωγ + ω20)A��e iωt =

V0

L��e iωt ⇒ A =

(V0/L)ω2

0 − ω2 + iωγ

A = |A| = (V0/L)√(ω2

0 − ω2)2 + ω2γ2tan φ =

−ωγω2

0 − ω2

(iv) Pav = 12 RωA2

(v) Substituting for A from (iii) into Pav gives

Pav =RV2

02L2

ω2

(ω20 − ω2)2 + ω2γ2 =

RV20

2L2γ2

1(ω2

0 − ω2)2

ω2γ2 + 1

The factor in the square bracket has maximum value 1 when (ω2

0 − ω2) = 0 which givesω = ω0. In this case,

Pav =RV2

02L2γ2 =

SRV20

2��L2(RA2/��L2)=

V20

2R

At resonance, as R decreases, the amplitude and hence the current go up in inverseproportion. Power dissipated in the resistor is proportional to current-squared. The sameargument applies for the mechanical system but with R → b and I → v. (However, youmight like to think about what happens when R or b disappear completely!)

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(vi) From the expression for Pav in (v), the power is at half max for (ω20 − ω2)2/ω2γ2 = 1,

which gives(ω2

0 − ω2)2 = (ωγ)2 ⇒ ω2 ± γω− ω20 = 0

Solving gives

ω =γ ±

√γ2 + 4ω2

0

2and ω =

−γ ±√γ2 + 4ω2

0

2Taking the two positive values of ω gives

ω1 =

√ω2

0 +γ2

4− γ

2ω2 =

√ω2

0 +γ2

4+γ

2

∆ω = ω2 − ω1 = γ and hence ω0/∆ω = ω0/γ = Q. [Alternatively, we do not need todiscard the negative ω, but recognise that they are equivalent to the positive solutionssince cosωt = cos(−ωt).]

(vii) The instantaneous power supplied by driving voltage is

P = VI = Vq = V0 cosωt × [−ωA sin(ωt + φ)]

Using sin(ωt + φ) = sinωt cos φ + cosωt sin φ, this gives

P = −V0ωA(cosωt sinωt cos φ + cos2 ωt sin φ)

On taking the time average, cosωt sinωt averages to zero and cos2 ωt averages to 1/2over a complete cycle. This gives the time average power as

Pav =−V0ωA sin φ

2

Using the results from (iii),

A sin φ = Im[A e iφ] =−(V0/L)ωγ

(ω20 − ω2)2 + ω2γ2

which gives for the power,

Pav =−V0ω

2−(V0/L)ωγ

(ω20 − ω2)2 + ω2γ2 =

V20γ

2Lω2

(ω20 − ω2)2 + ω2γ2

In the lectures we worked out the power required by a driven mechanical system atsteady state by considering the power dissipated by the damping force. This came out tobe

Pav =12

F20γ

mω2

(ω20 − ω2)2 + ω2γ2

Making the usual identities F0 → V0, b → R, m → L, this can be seen to be identical tothe expression we have just calculated by considering the driving force directly.

4. (i) from the figure FWHM is 0.5Hz and the peak is at 3Hz. This gives Q = ω0/∆ω ≈(2π× 3.0)/(2π× 0.5) ≈ 6.

(ii) ω20 = k/m which gives k = mω2 = 1.0× (2π× 3.0)2 = 355 N/m.

γ = b/m which gives b = mγ = mω0/Q = 3.14 Ns/m.

5. (i)LAρx = −2Aρgx− 8πηLx

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(ii) Rearranging the equation of motion into standard form lets us identify the standard quan-tities γ and ω0, thus:

x +8πηAρ︸︷︷︸γ

x +2gL︸︷︷︸ω2

0

x = 0

Then, substituting into the formulae gives

ωd =

√ω2

0 −γ2

4=

√2gL− 16π2η2

A2ρ2

ωres =

√ω2

0 −γ2

2=

√2gL− 32π2η2

A2ρ2

Q =ω0

γ=

√2gL

Aρ8πη

[Note: in Classwork 3 you will be asked to show that ωres =

√ω2

0 −γ2

2 ]

(iii) For light damping Q > 0.5. Substituting in values for water gives Q = 3.94 which implieslight damping. Substituting values into the above formula gives ωd = 1.964 s−1.

(iv) Similarly, substituting values for oil gives Q = 0.0177 which implies heavy damping.

(v) Using the above formula for ωres with the material values for water gives ωres = 1.948.The steady state amplitude is

A =F0/m√

(ω20 − ω2)2 + ω2γ2

and the time average power dissipation is

Pav =12

bω2A2 =12

mγω2A2 =γF2

02m

ω2

(ω20 − ω2)2 + ω2γ2

Use m = ρAL = 0.25 kg and F0 = AP0 = 0.25 N and substitute values into the aboveexpressions to obtain values for ω0 = 1.980 s−1 and γ = 0.503 s−1. Then substituting intothe above expressions gives

A(ωres) = 1.013 m Pav(ωres) = 0.245 W

A(ω0) = 1.005 m Pav(ω0) = 0.249 W

Power is not just a function of the amplitude, but it also depends on the angular fre-quency ω (through the velocity). So, a maximum in amplitude will not be a maximum inamplitude×ω and one would expect the peak power to be at a slightly higher frequencythan the peak amplitude, which it is.



Vibrations and WavesProblem Sheet 3 ANSWERS

1. String has length L, mass density µ, tension T . v =√

T/µ, but from question, T =

153Lµg ⇒ T/µ = 153Lg. Substituting into the equation for v gives v =√

153LgAlso, v = L/t, where t is the time for a wave to travel length L. Equating these gives

L/t =√

153Lg⇒ L = 153gt2= 15.0 m

2. (i) f = 1.5 Hz

(ii) Writing ψ = cos(ωt − kz + φ0). From ψ1, and ψ2, k(z2 − z1) = π/8 + 2nπ ⇒ k =

(π/16 + nπ)/m where n is an integer. So the wavelength is any of λ = 2π/k =

2π/(π/16 + nπ) = 32/(1 + 16n)m (λ > 0).

(iii) v = ω/k = 3π/(π/16 + nπ). So velocity is any of 48/(1 + 16n)zm/s .

3. If two functions ψ1 and ψ2 are solutions of the wave equation, then we need to showthat any superposition aψ1 + bψ2 is also a solution. Writing the wave equation

∂2ψ

∂z2 −1v2

∂2ψ

∂t2 = 0

and substituting the superposition into the LHS gives

∂2(aψ1 + bψ2)∂z2 − 1

v2

∂2(aψ1 + bψ2)∂t2 =

∂2(aψ1)∂z2 +

∂2(bψ2)∂z2 − 1

v2

∂2(aψ1)∂t2 − 1

v2

∂2(bψ2)∂t2

= a(∂2ψ1

∂z2 −1v2

∂2ψ1

∂t2

)+ b(∂2ψ2

∂z2 −1v2

∂2ψ2

∂t2

)But since ψ1 and ψ2 satisfy the wave equation, both brackets in the last expressionare zero, and so the superposition satisfies the wave equation.

4. The wave equation is∂2ψ

∂z2 −1v2

∂2ψ

∂t2 = 0

Writing f ′(y) =d f (y)

dyand f ′′(y) =

d2 f (y)dy2 ,

then writing y = z− vt,∂ f (y)∂z

= f ′(y)dydz

= f ′(y) and∂2 f (y)∂z2 =

d f ′(y)dy

dydz

= f ′′(y).

Similarly,∂ f∂t

= f ′(y)dydt

= −v f ′(y) and∂2 f (y)∂z2 =

d[−v f ′(y)]dy

dydz

= v2 f ′′(y).

Substituting into the LHS of the wave equation gives

∂2 f (z− vt)∂z2 − 1

v2

∂2 f (z− vt)∂t2 = f ′′(z− vt)− 1

v2 [v2 f ′′(z− vt)] = 0

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5. (i)

1.0 0.5 0.0 0.5 1.0z/m

0.0

0.2

0.4

0.6

0.8

1.0

1.2

f/m

(ii)

0 1 2 3 4 5 6z/m

0.0

0.2

0.4

0.6

0.8

1.0

1.2

f/m

t=1s t=5s

Function has moved +4 m in 4 s giving a velocity of +1 m/s

(iii)

6 5 4 3 2 1 0z/m

0.0

0.2

0.4

0.6

0.8

1.0

1.2

f/m

t=1st=5s

Function has moved −4 m in 4 s giving a velocity of −1 m/s

6. (i) f (z− vt) = A exp[−(z− vt)2/w2] where v = 5 m/s.

(ii) f (z + vt) = [1 + (z + vt)2/w2]−1 where v = 2 m/s.

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7.

Using the lengths as noted on the diagram, from symmetry the modes can be iden-tified immediately. In the lowest mode (mode 1) both pendulums have the sameamplitude, are in phase and oscillate as if the rod were not present. The angular fre-quency is ω1 =

√g/(`1 + `2). In the higher mode (mode 2) the pendulums have equal

and opposite motion, the rod is stationary and the pendulums have angular frequencydetermined by the length of the string from the rod to the mass, ω2 =

√g/`2. Beat

frequency will be ∆ f = f2 − f1 = (ω1 − ω2)/(2π). The higher the rod is placed, thelonger the beat period.

8. (i) In deriving the waves on a string we applied F = ma to a small segment of thestring. In the lecture we considered only force due to the tension in the string.

To that force we need to add the damping force which is -β∂ψ

∂z∆z. The resulting

equation of motion for a segment ∆z is

T∂2ψ

∂z2 ��∆z− β∂ψ

∂z��∆z = (µ��∆z)

∂2ψ

∂t2

which gives∂2ψ

∂t2 = v2∂2ψ

∂z2 − Γ∂ψ

∂zv =

√Tµ

Γ =β

µ

(ii) Substituting ψ(z, t) into the equation gives ω2 = v2k2− iωΓ so ψ is a solution withthis dispersion relation.

(iii) From the previous part, k2 = ω2/v2 + iωΓ/v2, which is complex, and so thereforeso is k.

(iv) Writing k = K − iκ in the travelling wave solution ψ gives

ψ(z, t) = A exp[ iωt − iKz]︸︷︷︸travelling wave

exp[−κz]︸︷︷︸decay

The damping force is attenuating the oscillations by dissipating energy as thewave progresses through the medium in exactly the same was as the dampingterm in a single damped oscillator attenuates the oscillation with time throughdissipation.



Vibrations and WavesProblem Sheet 4 ANSWERS

1. i) ψ = A[cos(10t − 5z) + cos(9t − 4z)]

ii) Beat frequency, ∆ f = |ω1 − ω2|/(2π) = 1/(2π)Hz

iii) Group velocity vg = ∆ω/∆k = (10− 9)/(5− 4)m/s = 1 m/s

2. i)dω2

d(k2)=

dω2

dωdωdk

dkd(k2)

= 2ωvg12k

= vgvp.

ii) vp =ω

k=

cn

=c√

1− ω2p/ω

2⇒

(1−

ω2p

ω2

)=

c2k2

ω2 ⇒ ω2 = c2k2 + ω2p

iii) From (ii),dω2

d(k2)= c2, but from (i)

dω2

d(k2)= vgvp, ⇒ vg =

c2

vp= c√

1− ω2p/ω

2.

iv) vp can exceed c, vg cannot. Phase velocity decribes the translation of a pureharmonic (cosinusoidal) wave. However, a pure harmonic wave cannot transporta signal, so there are no causality implications of this velocity exceeding c.

3. i) (Using SI units) [S ] = [v2pρλ] = m2s−2kgm−3m = kgs−2 = N/m

ii) vp = ω/k ⇒ ω = kvp = k3/2√S/ρ.

0.0 0.2 0.4 0.6 0.8 1.0k

0.0

0.2

0.4

0.6

0.8

1.0

ω/√ S

/ρ

This is anomalous dispersion (vg > vp)

iii) vg =dωdk

= 32

√kS/ρ = 1.5vp

iv) Use the group velocity. Choose the centre wavelength to calculate when themiddle of wavepacket will reach other side, λ = 0.55 cm. Then vg = 0.43 m/s.Time taken is 2 m/(0.43 m/s) = 4.6 s. Other choices of λ would be valid to cal-culate the time for the start or the end of wavepacket. [Aside: that the differentparts of the wavepacket have different group velocities causing the wavepacketto spread out is an example of group velocity dispersion]

4. i) Applying Snell’s law from medium n0 to medium n1 gives n0 sin θ0 = n1 sin θ1.Then applying it from n1 to n2 gives n1 sin θ1 = n2 sin θ2. Combining these two

equations gives, n2 sin θ2 = n0 sin θ0 ⇒ θ2 = sin−1(

n0

n2sin θ0

).

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ii) From (i), the angle in the bottom layer of the three media is independent of theintervening layer. This must extend to multiple layers.

iii) Using the previous result, the refraction at sea level depends only on the re-fractive index at sea level and not the intervening layers of the atmosphere

n sin θ = sin θ0 ⇒ θ = sin−1(

sin θ0

n

)(where we have used n0 = 1 for

vacuum)

iv)

0 10 20 30 40 50 60 70 80 90θ0 / degrees

1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0.0

0.2θ−θ 0/

degre

es

v) Using part (iii), θ − θ0 = 1.4◦ for θ0 = 90◦, which would mean the Sun wouldappear at 1.4◦ above the horizon. This will be an overestimate of the deviationbecause the atmosphere is not flat, but curved (due to the curvature of the earth)so at very low angles near the horizon, the incident angle on the upper layers ofthe atmosphere is less than the 90◦ assumed here.

5. Use fR = (v− vR)/(v− vS) fS

vS = ub, vR = 0, fS = fB ⇒ f ′T = v/(v− uB) fBi)ii) as (i) except vR = um ⇒ f ′M = (v− uM)/(v− uB) fB

iii) vR = −uB, vS = 0, fR = f ′′T , fS = f ′T,⇒ f ′′T = (v + uB)/v f ′T = [(v + uB/v][v/(v− uB)] fB = (v + uB)/(v− uB) fB

iv) vS = −uM, vR = −uB, fS = f ′M⇒ f ′′M = (v + uB)/(v + uM) f ′M = [(v + uB)/(v + uM)][(v− uM)/(v− uB)] fB

v) Rearrange (iii) to get fB = (v− uB)/(v + uB) f ′′T = 9770 Hz,Rearrange (iv) to get fB = [(v + uM)/(v + uB)][(v− uB)/(v− uM)] = 9776 Hz

6. i)

From the geometry of the diagram the extra path (marked in bold) is 2d sin θ.

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ii) 2d sin θ = nλ (i.e., a whole number of wavelengths in the path differences forscattering from adjacent crystal planes.)

iii) X-ray, electrons, neutrons, etc (waves with λ smaller than 2d)

7. (i) (a) Superposing gives

y = y+ + y− = 2A sin(ωt + φ) sin kx

Applying boundary conditions: y = 0 at x = 0 is already satisfied by sin kx;y = 0 at x = L−md requires kx = 2π(L−md)/λ = nπ ⇒ λn = 2(L−md)/n

(b) v = λ f ⇒ T ∗ = σ(λ f )2. λ = 2L = 2 m. Therefore, T ∗ = 10−3(2 × 1000)2 =

4000 N.(c) Frequencies are ν0 and ν0L/(L−2d) giving a beat frequency νbeat = ν0(L/(L−

2d)− 1) = ν02d/(L− 2d) = 111 Hz.

(ii) (a) Constructive interference occurs when the path difference x2−x1 ≈ a sin θ =

pλ for integer p, which gives sin θmax = pλ/a = pv/(aν)(b) ∆(sin θmax) = v/(aν) ⇒ ∆y ≈ s∆(sin θmax) = sv/aν = 3.44 m(c) For waves, energy ∝ amplitude. The sound intensity (which is sound power

per unit area) for one guitar is I0 = cA20, where A0 is amplitude and c is a

constant. For constructive interference, the amplitude is A = 2A0, thereforethe intensity is I = c(2A0)2 = 4I0. But I0 = (power)/(area) = 1.0 W/(2πs2).Therefore the maximum sound intensity heard is I = (4 × 1.0 W)/[2π ×(50 m)2] = 0.254 mW/m2.


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