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.

Type

Multiple choice Written

Difficulty

Easy Medium Hard

Question 1

Paper 2Easy6 marks

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

x=A\cos\phi,\qquad \phi=\phi_0+\frac{2\pi t}{T},

where the phase angle $\phi$ is in radians. The amplitude is $0.060\ \mathrm{m}$ the period is $0.80\ \mathrm{s}$ and the initial phase angle is $\pi/3$

(a)

State the phase angle at $t=0$

[1]

(b)

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

[2]

(c)

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

[2]

(d)

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

[1]

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

Show answer and explanation

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

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

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

Question 3

Multiple choiceMedium1 mark

A travelling wave P has frequency $f$ and wavelength $\lambda$ A second travelling wave Q has the same speed as P and a frequency of $2f$ 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, $v=f\lambda=(2f)\lambda_Q$ so $\lambda_Q=\frac{\lambda}{2}$

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 4

Multiple choiceMedium1 mark

A monochromatic beam illuminates two identical slits, each of width $a$ whose centres are separated by $d$ The ratio $d/a$ 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, $a\sin\theta=\lambda$ A missing interference maximum occurs at the same angle, where $d\sin\theta=n\lambda$ Equating the angles gives $n=d/a$

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 5

Paper 2Medium9 marks

Two loudspeakers, $S_1$ and $S_2$ 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 $0.01\ \text{m}$

(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 $n$ of constructive interference at R. Show your reasoning.

[3]

Question 6

Paper 2Easy9 marks

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

(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 $0.600\ \mathrm{m}$ and the wave speed on it is $120\ \mathrm{m\,s^{-1}}$ 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 $340\ \mathrm{m\,s^{-1}}$ in a pipe of length $0.600\ \mathrm{m}$ 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 7

Multiple choiceMedium1 mark

Two sinusoidal waves, each of amplitude $a$ and wavelength $\lambda$ travel in opposite directions along a string and form a standing wave. There is a displacement node at $x=0$ Points $P$ and $Q$ are at $x=\lambda/12$ and $x=\lambda/3$ respectively. What is the ratio of the oscillation amplitude at $P$ to that at $Q$ 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 $x=0$ the amplitude at position $x$ is $2a|\sin(2\pi x/\lambda)|$ Thus $A_P=a$ and $A_Q=\sqrt{3}a$ giving $A_P/A_Q=1/\sqrt{3}$ Both points lie between the same pair of adjacent nodes, so their oscillations are in phase and their phase difference is $0$

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 $x_P/x_Q=1/4$

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

Question 8

Multiple choiceEasy1 mark

A source emits light with wavelength 600 nm and moves directly away from an observer at $1.5 \times 10^6\ \text{m s}^{-1}$ Take the speed of light as $3.0 \times 10^8\ \text{m s}^{-1}$ 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 $v/c$ Thus $v/c=(1.5 \times 10^6)/(3.0 \times 10^8)=5.0 \times 10^{-3}=0.50\%$

A: converts the fractional change to a percentage using a factor of 10 rather than 100.
B: introduces an erroneous factor of $1/2$
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.

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