IB Physics HL Theme C: Wave behaviour questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

Question 1

Paper 2Easy6 marks

A position sensor tracks a component undergoing simple harmonic motion. Its displacement xx from equilibrium is modelled by

x=Acos⁡ϕ,ϕ=ϕ0+2πtT,x=A\cos\phi,\qquad \phi=\phi_0+\frac{2\pi t}{T},

where the phase angle ϕ\phi is in radians. The amplitude is 0.060 m0.060\ \mathrm{m} the period is 0.80 s0.80\ \mathrm{s} and the initial phase angle is π/3\pi/3

(a)

State the phase angle at t=0t=0

[1]
(b)

Calculate the phase angle at t=0.20 st=0.20\ \mathrm{s} in radians.

[2]
(c)

Calculate the displacement of the component at t=0.20 st=0.20\ \mathrm{s}

[2]
(d)

State the increase in phase angle during one complete oscillation, in radians.

[1]

Question 2

Multiple choiceHard1 mark

An oscillator has a period of 2.0 s2.0\ \text{s} Four proposed graphs show its acceleration aa against displacement xx measured from equilibrium. Which trace is consistent with the oscillator undergoing simple harmonic motion?

-0.5-0.4-0.3-0.2-0.100.10.20.30.40.5-6-5-4-3-2-10123456x / ma / m s-2PQRS
Show answer and explanation

Answer: A. For simple harmonic motion, the graph of aa against xx is a straight line through the origin with gradient −ω2-\omega^2 From T=2.0 sT=2.0\ \text{s} ω=2π/T=π rad s−1\omega=2\pi/T=\pi\ \text{rad s}^{-1} so the required gradient is −π2≈−9.87 s−2-\pi^2\approx-9.87\ \text{s}^{-2} Trace P has this negative gradient and passes through the origin.
B: uses a negative gradient through the origin, but its magnitude is about 3.14 s−23.14\ \text{s}^{-2} confusing ω\omega with ω2\omega^2
C: has the required gradient but does not pass through the origin, so acceleration is not proportional to displacement.
D: has a positive gradient, so acceleration is in the direction of displacement rather than towards equilibrium.
Remember: SHM requires acceleration to be proportional to displacement and opposite in direction.

Question 3

Multiple choiceMedium1 mark

A source produces periodic waves in medium X with wavelength λ\lambda period TT and speed vv The source is adjusted to produce waves with period T/2T/2 in medium Y, where their speed is 3v3v What is the wavelength of the waves in medium Y?

Show answer and explanation

Answer: B. Since v=fλv=f\lambda and f=1/Tf=1/T wavelength is λ=vT\lambda=vT The wavelength in medium Y is (3v)(T/2)=3vT/2=3λ/2(3v)(T/2)=3vT/2=3\lambda/2

A: applies the period change but leaves the wave speed unchanged.
C: applies the speed change but leaves the period unchanged.
D: uses 2T2T instead of T/2T/2 then applies the speed factor of 3.

Remember: wavelength is the product of wave speed and period.

Question 4

Multiple choiceMedium1 mark

A travelling wave P has frequency ff and wavelength λ\lambda A second travelling wave Q has the same speed as P and a frequency of 2f2f In Q, particles oscillate parallel to the direction of travel. Which description of Q is correct?

Show answer and explanation

Answer: C. Q is longitudinal because its particles oscillate parallel to the direction of travel. Since the two waves have the same speed, v=fλ=(2f)λQv=f\lambda=(2f)\lambda_Q so λQ=λ2\lambda_Q=\frac{\lambda}{2}

A: reverses the wave classification and treats wavelength as directly proportional to frequency.
B: finds the wavelength correctly but labels parallel oscillations as transverse.
D: identifies the wave correctly but uses a direct rather than inverse frequency–wavelength relationship.

Remember: at fixed wave speed, frequency and wavelength are inversely proportional.

Question 5

Paper 2Hard10 marks

A cord and a spring each carry a sinusoidal travelling disturbance along the equilibrium xx-axis. Figure 1 shows the displacement of particles from equilibrium at t=0t=0 and t=0.20 st=0.20\ \mathrm{s} plotted against each particle’s equilibrium coordinate xx In the upper panel, yy is the cord-particle displacement perpendicular to xx with positive yy upwards. In the lower panel, uu is the spring-particle displacement along xx with positive uu in the +x+x direction. Each disturbance retains its profile as it travels.

y / mmu / mmx = 0.20 mx = 0.20 m+50+40+30+20+100-10-20-30-40-50+25+20+15+10+50-5-10-15-20-250.000.200.400.600.801.001.201.401.60x / m0.000.200.400.600.801.001.201.401.60x / mt = 0t = 0.20 s
(a)

Determine the direction of propagation and the speed of the disturbance on the cord.

[2]
(b)

State the wavelength of the disturbance on the cord.

[1]
(c)

At x=0.20 mx=0.20\ \mathrm{m} determine the instantaneous velocity of the cord particle at t=0.20 st=0.20\ \mathrm{s} Give your answer with its direction.

[4]
(d)

The spring disturbance is now launched in the opposite direction, with its speed and wavelength unchanged. Describe how its profile at t=0.20 st=0.20\ \mathrm{s} is displaced relative to its profile at t=0t=0 State whether the disturbance is still longitudinal, and explain your answer.

[3]

Question 6

Multiple choiceMedium1 mark

A monochromatic beam illuminates two identical slits, each of width aa whose centres are separated by dd The ratio d/ad/a is an integer greater than 1. Which order of the two-slit interference pattern is absent because it coincides with the first minimum of the single-slit diffraction envelope?

Show answer and explanation

Answer: B. At the first single-slit minimum, asin⁡θ=λa\sin\theta=\lambda A missing interference maximum occurs at the same angle, where dsin⁡θ=nλd\sin\theta=n\lambda Equating the angles gives n=d/an=d/a

A: reverses the ratio of slit width to centre separation.
C: uses the first diffraction-minimum index as the interference order without matching the angles.
D: uses twice the first-minimum condition, introducing an extra factor of 2.

Remember: an interference order is missing when its maximum coincides with a diffraction minimum.

Question 7

Paper 2Medium9 marks

Two loudspeakers, S1S_1 and S2S_2 are connected to the same signal generator and emit sound in phase. Figure 1 shows measured distances from each loudspeaker to three positions on a detector’s path. The path lengths are measured to the nearest 0.01 m0.01\ \text{m}

S2S1RQPS2R = 1.54 mS2Q = 1.70 mS2P = 2.18 mS1R = 0.34 mS1Q = 0.80 mS1P = 1.58 m
(a)

Explain why the sources must be coherent for the maxima and minima of the interference pattern to remain in fixed positions.

[2]
(b)

P and Q are consecutive positions of maximum intensity.

(i)

Using Figure 1, calculate the path difference at P and at Q.

[2]
(ii)

Determine the wavelength of the sound.

[2]
(c)

Determine the order nn of constructive interference at R. Show your reasoning.

[3]

Question 8

Paper 2Hard10 marks

In a laboratory, a laser illuminates both slits of a double-slit plate at normal incidence. The separation of adjacent bright fringes is measured on a screen parallel to the plate. A reference laser has wavelength 600 nm600\ \mathrm{nm} With the original plate and a screen distance of 1.20 m1.20\ \mathrm{m} the measured fringe separation is 3.00 mm3.00\ \mathrm{mm}

(a)

Show that the slit separation of the original plate is approximately 0.240 mm0.240\ \mathrm{mm}

[2]
(b)

The screen is moved to a distance of 1.80 m1.80\ \mathrm{m} with the original plate and reference laser unchanged. Determine the fringe separation.

[1]
(c)

A replacement plate is used with the reference laser and the screen at 1.80 m1.80\ \mathrm{m} The measured fringe separation is 3.00 mm3.00\ \mathrm{mm} Determine the slit separation of the replacement plate.

[1]
(d)

The replacement plate is used with an unknown laser wavelength. The screen is at 1.20 m1.20\ \mathrm{m} and the fringe separation is 2.40 mm2.40\ \mathrm{mm} Determine the wavelength of this laser.

[2]
(e)

In a further run, an unknown laser gives a fringe separation of 3.00 mm3.00\ \mathrm{mm} with the screen at 1.20 m1.20\ \mathrm{m} but the identity of the slit plate used was not recorded. A technician claims that the laser wavelength is the same as that of the reference laser. Determine whether this claim is justified, supporting your answer quantitatively.

[4]

Question 9

Paper 2Easy9 marks

A string on a musical instrument is fixed at both ends and vibrates in the mode shown in Figure 1.

00.10.20.30.40.50.6-2.5-2-1.5-1-0.500.511.522.5position, x / mdisplacement, y / mmAB
(a)

State the number of displacement nodes and the number of antinodes shown. Count the fixed ends as nodes.

[2]
(b)

The string has length 0.600 m0.600\ \mathrm{m} and the wave speed on it is 120 m s−1120\ \mathrm{m\,s^{-1}} For the mode shown, calculate the wavelength and frequency. Give both answers to 3 significant figures.

[3]
(c)

Compare the oscillation amplitudes at A and B, and state their phase difference.

[2]
(d)

Sound travels at 340 m s−1340\ \mathrm{m\,s^{-1}} in a pipe of length 0.600 m0.600\ \mathrm{m} The pipe is closed at one end and open at the other. Calculate the wavelength and frequency of the third harmonic. Ignore end correction.

[2]

Question 10

Multiple choiceMedium1 mark

Two sinusoidal waves, each of amplitude aa and wavelength λ\lambda travel in opposite directions along a string and form a standing wave. There is a displacement node at x=0x=0 Points PP and QQ are at x=λ/12x=\lambda/12 and x=λ/3x=\lambda/3 respectively. What is the ratio of the oscillation amplitude at PP to that at QQ and what is the phase difference between their oscillations?

Show answer and explanation

Answer: C. For two equal waves forming a standing wave with a node at x=0x=0 the amplitude at position xx is 2a∣sin⁡(2πx/λ)∣2a|\sin(2\pi x/\lambda)| Thus AP=aA_P=a and AQ=3aA_Q=\sqrt{3}a giving AP/AQ=1/3A_P/A_Q=1/\sqrt{3} Both points lie between the same pair of adjacent nodes, so their oscillations are in phase and their phase difference is 00

A: reverses the requested amplitude ratio.
B: finds the ratio correctly but treats points in the same segment between nodes as oscillating in antiphase.
D: assumes amplitude is proportional to distance from the node, giving xP/xQ=1/4x_P/x_Q=1/4

Remember: points in the same segment between adjacent nodes oscillate in phase.

Question 11

Multiple choiceEasy1 mark

A source emits light with wavelength 600 nm and moves directly away from an observer at 1.5×106 m s−11.5 \times 10^6\ \text{m s}^{-1} Take the speed of light as 3.0×108 m s−13.0 \times 10^8\ \text{m s}^{-1} What is the magnitude of the relative change in wavelength, expressed as a percentage?

Show answer and explanation

Answer: C. For light, the magnitude of the relative wavelength change is approximately v/cv/c Thus v/c=(1.5×106)/(3.0×108)=5.0×10−3=0.50%v/c=(1.5 \times 10^6)/(3.0 \times 10^8)=5.0 \times 10^{-3}=0.50\%

A: converts the fractional change to a percentage using a factor of 10 rather than 100.
B: introduces an erroneous factor of 1/21/2
D: shifts the decimal point one place too far when converting the fraction to a percentage.

Remember: multiply a fractional change by 100 to express it as a percentage.

Question 12

Paper 2Hard10 marks

A sound emitter moves uniformly through still air. At one instant, the distance between adjacent wavefronts is 0.646 m ahead of the emitter and 0.714 m behind it. The speed of sound in the air is 340 m s⁻¹. In a second arrangement, the emitter is stationary and a detector moves towards it at the speed found in (a). The emitter has the same frequency in both arrangements.

(a)

Determine the emitted frequency and the speed of the emitter. Show your reasoning.

[4]
(b)

Sketch wavefront diagrams for (i) the sound emitter moving towards a stationary detector and (ii) a stationary emitter with the detector moving towards it. Indicate the directions of motion and the relative spacing of the wavefronts.

[2]
(c)

Calculate the frequency detected in each sound arrangement. Give your answers to three significant figures.

[2]
(d)

A light beacon moves towards an observer. State how the detected frequency and wavelength compare with those emitted, and state one relevant difference between light waves and sound waves.

[2]

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