Topic C. Wave behaviour HL TEST 1

IB Physics: C. Wave Behaviour

Test 1

25 Multiple-Choice Questions and 5 Extended-Response Questions

Suggested time: 90 minutes | Total: 75 marks

Instructions

Answer all questions. Section A carries 25 marks; Section B carries 50 marks.
Choose one option for each MCQ. Award 1 mark for a correct answer; no marks are deducted for an incorrect answer.
Use a calculator and the IB Physics data booklet. Show working, units and clear reasoning in Section B.
Numerical MCQ options are rounded; select the closest option. Give extended-response answers to 2–3 significant figures unless uncertainty determines the precision.
This is an original mixed SL/HL classroom test, not an official IB paper. Additional HL questions and subparts are labelled.
Answers and a suggested markscheme appear only in the box at the bottom.

Use $c=3.00\times10^8\,\mathrm{m\,s^{-1}}$ and
$v_{\mathrm{sound}}=340\,\mathrm{m\,s^{-1}}$ unless stated otherwise.
For SHM, amplitude is $x_0$ and displacement is $x$ .
For double slits, fringe spacing is $s$ , slit separation $d$ and screen distance $D$ ; individual slit width is $b$ .

Section A — Multiple Choice [25 marks]

1. C.1

Which statement gives the defining condition for simple harmonic motion?

A. Acceleration is constant and opposite to velocity.
B. Acceleration is proportional to displacement and in the same direction.
C. Acceleration is proportional to displacement and directed towards equilibrium.
D. Acceleration is proportional to velocity and opposite to velocity.

2. C.1

For an oscillator, $a=-64x$ , where $a$ is in $\mathrm{m\,s^{-2}}$ and $x$ is in metres.
What is its period?

A. $0.125\,\mathrm{s}$
B. $0.393\,\mathrm{s}$
C. $0.785\,\mathrm{s}$
D. $1.57\,\mathrm{s}$

3. C.1

The mass attached to an ideal spring is increased by a factor of four.
The spring constant is unchanged. What happens to the oscillation period?

A. It doubles.
B. It halves.
C. It is unchanged.
D. It increases by a factor of four.

4. HL C.1

An oscillator has displacement $x=x_0\sin(2\pi t/T)$ .
At what time after $t=0$ does its acceleration first reach its most negative value?

A. $T/8$
B. $T/4$
C. $T/2$
D. $3T/4$

5. HL C.1

An ideal oscillator has total mechanical energy $E_T$ .
What is its kinetic energy at displacement $x=x_0/2$ ?

A. $E_T/4$
B. $E_T/2$
C. $3E_T/4$
D. $E_T$

6. C.2

A sound wave travels through still air in the positive horizontal direction.
How do the air particles oscillate?

A. Back and forth parallel to the propagation direction.
B. Up and down perpendicular to the propagation direction.
C. Continuously forward at the sound speed.
D. In stationary circular orbits perpendicular to propagation.

7. C.2

A wave travels at $12\,\mathrm{m\,s^{-1}}$ and has period $0.040\,\mathrm{s}$ . What is its wavelength?

A. $0.0033\,\mathrm{m}$
B. $0.030\,\mathrm{m}$
C. $0.30\,\mathrm{m}$
D. $0.48\,\mathrm{m}$

8. C.2

A displacement–distance snapshot has adjacent crests at positions $0.12\,\mathrm{m}$ and $0.52\,\mathrm{m}$ .
Crest displacement is $+3.0\,\mathrm{mm}$ and trough displacement
is $-3.0\,\mathrm{mm}$ . Which pair gives amplitude and wavelength?

A. $x_0=6.0\,\mathrm{mm},\quad\lambda=0.40\,\mathrm{m}$
B. $x_0=3.0\,\mathrm{mm},\quad\lambda=0.20\,\mathrm{m}$
C. $x_0=3.0\,\mathrm{mm},\quad\lambda=0.40\,\mathrm{m}$
D. $x_0=6.0\,\mathrm{mm},\quad\lambda=0.20\,\mathrm{m}$

9. C.2

A wave crosses a stationary boundary into a medium where its speed is twice as large. What happens to its frequency and wavelength?

A. Frequency doubles; wavelength is unchanged.
B. Frequency is unchanged; wavelength doubles.
C. Frequency halves; wavelength doubles.
D. Frequency doubles; wavelength halves.

10. C.2

An electromagnetic wave in vacuum has frequency $6.0\times10^{14}\,\mathrm{Hz}$ .
In which region of the spectrum does it lie?

A. Microwave
B. Visible
C. Infrared
D. X-ray

11. C.3

Light of frequency $f_0$ and wavelength $\lambda_0$ in vacuum enters a material with refractive index $1.25$ through a
stationary boundary at normal incidence. Which set gives its frequency, speed and wavelength in the material?

A. $0.80f_0,\quad0.80c,\quad\lambda_0$
B. $f_0,\quad1.25c,\quad1.25\lambda_0$
C. $1.25f_0,\quad0.80c,\quad0.64\lambda_0$
D. $f_0,\quad0.80c,\quad0.80\lambda_0$

12. C.3

Light travels from a transparent material of refractive index $\sqrt2$ towards air. Its incidence angle is $45^\circ$ .
What happens?

A. The refracted ray travels along the boundary.
B. Total internal reflection occurs.
C. The refracted ray bends towards the normal.
D. The refraction angle is 30°.

13. C.3

Two coherent sources emit in phase with wavelength $0.20\,\mathrm{m}$ . Their distances to P are $3.10\,\mathrm{m}$ and $3.40\,\mathrm{m}$ .
Each wave has amplitude $x_0$ at P.
What is the resultant amplitude there?

A. $2x_0$
B. $x_0$
C. $0$
D. $x_0/2$

14. C.3

In a double-slit experiment, wavelength is doubled, screen distance is tripled and slit separation is doubled. The original fringe spacing was $s$ .
What is the new spacing?

A. $s/3$
B. $2s/3$
C. $2s$
D. $3s$

15. HL C.3

A diffraction grating has 500 lines per millimetre. Light of wavelength $650\,\mathrm{nm}$ is incident normally.
What is the angle of the second-order maximum?

A. $19.0^\circ$
B. $40.5^\circ$
C. $49.5^\circ$
D. $65.0^\circ$

16. C.4

A string of length $0.90\,\mathrm{m}$ is fixed at both ends.
A standing-wave mode has three antinodes. What is the wavelength?

A. $0.30\,\mathrm{m}$
B. $0.45\,\mathrm{m}$
C. $0.90\,\mathrm{m}$
D. $0.60\,\mathrm{m}$

17. C.4

Two non-node points lie in adjacent loops of a standing wave.
What is their phase difference?

A. $\pi\,\mathrm{rad}$
B. $0\,\mathrm{rad}$
C. $\pi/2\,\mathrm{rad}$
D. It depends on their separation within those loops.

18. C.4

An air column is closed at one end and open at the other.
Its first-harmonic frequency is $170\,\mathrm{Hz}$ .
Neglecting end correction, what is its next allowed resonance frequency?

A. $255\,\mathrm{Hz}$
B. $340\,\mathrm{Hz}$
C. $510\,\mathrm{Hz}$
D. $680\,\mathrm{Hz}$

19. C.4

A lightly damped oscillator is driven by a force of fixed amplitude. Damping is increased while it remains lightly damped.
How does its amplitude–frequency resonance peak change?

A. It becomes higher and narrower.
B. It becomes lower and broader.
C. It becomes higher and broader.
D. It becomes lower and narrower.

20. C.4

Which damping condition gives the fastest return to equilibrium without oscillation after a system is displaced and released?

A. No damping
B. Light damping
C. Heavy damping
D. Critical damping

21. C.5

An observer moves towards a stationary loudspeaker through still air.
What happens to the frequency heard and the wavelength in the air?

A. Heard frequency increases; wavelength in the air is unchanged.
B. Heard frequency increases; wavelength in the air decreases.
C. Heard frequency decreases; wavelength in the air is unchanged.
D. Heard frequency increases; wavelength in the air increases.

22. C.5

A stellar spectral line has rest wavelength $400.0\,\mathrm{nm}$ and is observed at $400.4\,\mathrm{nm}$ .
Assuming a low-speed Doppler shift, what is the star’s radial motion?

A. Approaching at $300\,\mathrm{km\,s^{-1}}$ .
B. Receding at $30\,\mathrm{km\,s^{-1}}$ .
C. Receding at $300\,\mathrm{km\,s^{-1}}$ .
D. Receding at $3000\,\mathrm{km\,s^{-1}}$ .

23. HL C.5

A source emits $640\,\mathrm{Hz}$ while approaching a stationary observer at $20\,\mathrm{m\,s^{-1}}$ .
Sound speed is $340\,\mathrm{m\,s^{-1}}$ .
What frequency is observed?

A. $604\,\mathrm{Hz}$
B. $680\,\mathrm{Hz}$
C. $640\,\mathrm{Hz}$
D. $1280\,\mathrm{Hz}$

24. HL C.5

A stationary source emits $800\,\mathrm{Hz}$ .
An observer moves away at $17\,\mathrm{m\,s^{-1}}$ through still air.
Sound speed is $340\,\mathrm{m\,s^{-1}}$ .
What frequency is observed?

A. $840\,\mathrm{Hz}$
B. $842\,\mathrm{Hz}$
C. $800\,\mathrm{Hz}$
D. $760\,\mathrm{Hz}$

25. C.5

Two stars emit the same rest spectral line. Star Y recedes with twice the radial speed of star X. Both speeds are much smaller than $c$ .
Compared with X, what is the approximate magnitude of Y’s fractional wavelength shift?

A. Twice as large
B. Half as large
C. The same
D. Four times as large

Section B — Extended Responses [50 marks]

26. C.1 — A mass–spring oscillator [10 marks]

A mass of $0.200\,\mathrm{kg}$ is attached to an ideal horizontal spring of spring constant $80.0\,\mathrm{N\,m^{-1}}$ .
It oscillates without friction with amplitude $x_0=0.0500\,\mathrm{m}$ .
At $t=0$ , it is released from rest at $x=+x_0$ .

(a) State the defining condition for simple harmonic motion, including the direction of acceleration. [2]

(b) Calculate the oscillation period. [2]

(c) HL
Calculate the velocity when $x=+0.0300\,\mathrm{m}$ and the mass is
moving towards equilibrium. Include its sign. [2]

(d) HL
Calculate the total mechanical energy and the kinetic energy at $x=+0.0300\,\mathrm{m}$ . [2]

(e) HL
Determine a phase constant $\phi$ and write a displacement equation in the form $x=x_0\sin(\omega t+\phi)$ for these initial conditions.
[2]

27. C.2 — Lab-style wave-speed investigation [10 marks]

A student uses a vibration generator to produce small-amplitude travelling waves on a long string at constant tension. A photograph is taken at each driving
frequency. The measured distance L is from the first to the fifth of five consecutive crests.

$f\,/\,\mathrm{Hz}$	$L\,/\,\mathrm{m}$
40.0	3.20
50.0	2.56
60.0	2.13
70.0	1.83
80.0	1.60

Each measurement of $L$ has absolute uncertainty $\pm0.02\,\mathrm{m}$ . Frequency uncertainty is negligible.
Assume wave speed is constant over this frequency range.

(a) Express the wavelength in terms of $L$ . [1]

(b) Using the $60.0\,\mathrm{Hz}$ row, calculate the wavelength and wave speed. [2]

(c) State the horizontal-axis quantity that should be plotted against $L$ on the vertical axis to obtain a straight line through
the origin. Explain how the gradient relates to wave speed. [2]

(d) Use the first and last rows to calculate this gradient and hence determine the wave speed. [2]

(e) Use the $60.0\,\mathrm{Hz}$ row to estimate the percentage uncertainty in wave speed and express the speed with its absolute
uncertainty. [2]

(f) Suggest one experimental precaution that reduces interference from reflections at the far end of the string. [1]

28. C.3 — Double-slit interference [10 marks]

Monochromatic light of wavelength $520\,\mathrm{nm}$ illuminates two narrow slits from the same source. The slit separation is
$d=0.260\,\mathrm{mm}$ and the screen distance is $D=2.00\,\mathrm{m}$ . Use the small-angle approximation.

(a) State what coherence means and explain why illuminating both slits from the same source helps produce stable fringes. [2]

(b) Calculate the fringe spacing. [2]

(c) Determine the path difference at the third-order bright fringe and state the interference condition there. [2]

(d) In a separate arrangement, the screen distance is $1.50\,\mathrm{m}$ and the slit separation is $0.520\,\mathrm{mm}$ . Wavelength is unchanged.
Calculate the new fringe spacing. [2]

(e) HL
Starting again from the original arrangement, the width of each slit is halved while slit separation is unchanged. State what happens to the interference fringe spacing and to the width of the central diffraction envelope. [2]

29. C.4 — Standing waves and resonance [10 marks]

A string of length $0.800\,\mathrm{m}$ is fixed at both ends. Wave speed on the string is $120\,\mathrm{m\,s^{-1}}$ .
The string vibrates in its second harmonic.

(a) Explain how a standing wave forms from travelling waves. [2]

(b) Sketch the second-harmonic pattern and label all nodes and antinodes. [2]

(c) Calculate the wavelength and frequency of this mode. [2]

(d) An air column of length $0.500\,\mathrm{m}$ is closed at one end and open at the other. Determine its first-harmonic frequency
and its next allowed resonance frequency.
Use $340\,\mathrm{m\,s^{-1}}$ and neglect end correction.
[2]

(e) A lightly damped oscillator is driven by a force of fixed amplitude. State two changes in its amplitude–frequency resonance peak when
damping is increased but remains light. [2]

30. C.5 — Sound and light Doppler shifts [10 marks]

A moving horn emits $800\,\mathrm{Hz}$ .
A stationary observer hears $850\,\mathrm{Hz}$ as the horn approaches along the source–observer line through still air.
Take the sound speed as $340\,\mathrm{m\,s^{-1}}$ .

(a) Explain the increased observed frequency in terms of wavefront spacing and wave speed. [2]

(b) HL
Calculate the speed of the source. [2]

(c) HL
Determine the frequency heard when the source recedes from the stationary observer at the same speed. [2]

(d) HL
In an independent experiment, the source is stationary and emits $800\,\mathrm{Hz}$ . An observer moves away from it at
$10.0\,\mathrm{m\,s^{-1}}$ through still air.
Calculate the observed frequency. [2]

(e) A stellar spectral line has rest wavelength $500.0\,\mathrm{nm}$ and observed wavelength $500.5\,\mathrm{nm}$ . Assuming a low-speed Doppler interpretation, determine the direction and magnitude of the star’s radial velocity. [2]

Answers and Suggested Markscheme

Total: 75 marks. Accept equivalent correct reasoning and
appropriate rounding. In extended responses, allow follow-through from an earlier
numerical error when the later method is correct.
The allocations below are a suggested classroom markscheme.

Section A — MCQ answers [25 marks]

Question	Answer	Reason
1	C	The defining relation is $a=-\omega^2x$ .
2	C	$\omega=8.0\,\mathrm{rad\,s^{-1}}$ , so $T=2\pi/8=0.785\,\mathrm{s}$ .
3	A	$T\propto\sqrt m$ , so the period increases by $\sqrt4=2$ .
4	B	The most negative acceleration occurs at $x=+x_0$ , first reached at $T/4$ .
5	C	$E_p/E_T=x^2/x_0^2=1/4$ , so $E_k=3E_T/4$ .
6	A	Sound in air is longitudinal; particles oscillate parallel to the direction of energy transfer.
7	D	$\lambda=vT=12(0.040)=0.48\,\mathrm{m}$ .
8	C	Amplitude is measured from equilibrium; wavelength is the distance between adjacent crests.
9	B	Frequency is unchanged at a stationary boundary; $\lambda=v/f$ therefore doubles.
10	B	$\lambda=c/f=5.0\times10^{-7}\,\mathrm{m}=500\,\mathrm{nm}$ , which is visible light.
11	D	Frequency is unchanged; $v=c/1.25=0.80c$ and $\lambda=\lambda_0/1.25$ .
12	A	$\theta_c=\sin^{-1}(1/\sqrt2)=45^\circ$ ; at the critical angle, refraction is at 90°.
13	C	$\Delta=0.30\,\mathrm{m}=1.5\lambda$ ; equal-amplitude waves arrive in antiphase and cancel.
14	D	$s=\lambda D/d$ , so the factor is $2\times3/2=3$ .
15	B	$d=2.00\times10^{-6}\,\mathrm{m}$ and $\sin\theta=2\lambda/d=0.650$ .
16	D	Three loops give $L=3\lambda/2$ , so $\lambda=2L/3=0.60\,\mathrm{m}$ .
17	A	Adjacent loops oscillate in antiphase; points within one loop oscillate in phase.
18	C	Only odd harmonics occur for this boundary condition; the next allowed mode is the third harmonic.
19	B	Greater damping dissipates more energy and produces a lower, broader resonance peak.
20	D	Critical damping is the boundary between oscillatory and non-oscillatory return and gives the fastest non-oscillatory return.
21	A	Observer motion changes the rate of encountering wavefronts but does not change their spacing in the air.
22	C	Longer wavelength indicates recession; $v\approx c(0.4/400.0)=3.0\times10^5\,\mathrm{m\,s^{-1}}$ .
23	B	$f\prime=640(340/(340-20))=680\,\mathrm{Hz}$ .
24	D	$f\prime=800((340-17)/340)=760\,\mathrm{Hz}$ .
25	A	The low-speed fractional shift is proportional to radial speed: $\|\Delta\lambda\|/\lambda\approx v/c$ .

26. Mass–spring oscillator [10 marks]

(a) [2]
Acceleration is proportional to displacement from equilibrium [1] and directed towards equilibrium, opposite to displacement [1]:
$a=-\omega^2x$ .

(b) [2]
Correct period relation/substitution [1]; correct numerical period [1].

$\displaystyle T=2\pi\sqrt{\frac{m}{k}} =2\pi\sqrt{\frac{0.200}{80.0}} =0.314\,\mathrm{s}$

$\displaystyle \omega=\sqrt{\frac{k}{m}}=20.0\,\mathrm{rad\,s^{-1}}$

(c) [2]
Correct speed magnitude [1]; negative sign for motion towards equilibrium from positive displacement [1].

$\displaystyle v=-\omega\sqrt{x_0^2-x^2} =-20.0\sqrt{0.0500^2-0.0300^2} =-0.800\,\mathrm{m\,s^{-1}}$

(d) [2]
Total energy [1]; kinetic energy at the stated displacement [1].

$\displaystyle E_T=\frac12m\omega^2x_0^2=0.100\,\mathrm{J}$

$\displaystyle E_p=\frac12m\omega^2x^2=0.0360\,\mathrm{J}$

$\displaystyle E_k=E_T-E_p=0.0640\,\mathrm{J}$

(e) [2]
$\phi=\pi/2$ radians, or an equivalent value modulo
$2\pi$ [1]; displacement equation [1].

$\displaystyle x=0.0500\sin\left(20.0t+\frac{\pi}{2}\right)\,\mathrm{m}$

Here t is in seconds. The equivalent cosine equation describes the same motion, but the requested sine form is shown.

27. Lab-style wave-speed investigation [10 marks]

(a) [1]
Five consecutive crests span four wavelengths:
$\lambda=L/4$ .

(b) [2]
Wavelength [1]; wave speed [1].

$\displaystyle \lambda=\frac{2.13}{4} =0.5325\,\mathrm{m} \approx0.533\,\mathrm{m}$

$\displaystyle v=f\lambda=60.0(0.5325) =31.95\,\mathrm{m\,s^{-1}} \approx32.0\,\mathrm{m\,s^{-1}}$

(c) [2]
Plot $1/f$ on the horizontal axis [1].
Since $v=fL/4$ , $L=4v(1/f)$ ;
the gradient is $4v$ [1].

(d) [2]
Correct gradient with units [1]; divide by four to obtain speed [1].

$\displaystyle G=\frac{3.20-1.60}{1/40.0-1/80.0} =128\,\mathrm{m\,s^{-1}}$

$\displaystyle v=\frac{G}{4}=32.0\,\mathrm{m\,s^{-1}}$

(e) [2]
Percentage uncertainty [1]; correctly expressed speed with absolute
uncertainty [1].

$\displaystyle \frac{\delta v}{v}\times100\% =\frac{0.02}{2.13}\times100\% =0.94\%$

$\displaystyle \delta v=31.95\left(\frac{0.02}{2.13}\right) =0.30\,\mathrm{m\,s^{-1}}$

$\displaystyle v=(32.0\pm0.3)\,\mathrm{m\,s^{-1}}$

(f) [1]
Use an absorbing or damped termination at the far end, so reflected waves are reduced. Accept taking the photograph before a reflected wave returns to the measured region.

28. Double-slit interference [10 marks]

(a) [2]
Coherent waves have the same frequency and a constant phase difference [1].
Light from a common source gives the two slits a stable relative phase, enabling stable fringes [1].

(b) [2]
Correct relation/substitution [1]; fringe spacing [1].

$\displaystyle s=\frac{\lambda D}{d} =\frac{520\times10^{-9}(2.00)}{0.260\times10^{-3}} =4.00\times10^{-3}\,\mathrm{m} =4.00\,\mathrm{mm}$

(c) [2]
Correct path difference [1]; constructive interference/in-phase arrival [1].

$\displaystyle \Delta=3\lambda =3(520\times10^{-9}) =1.56\times10^{-6}\,\mathrm{m}$

(d) [2]
Correct relation/substitution [1]; new fringe spacing [1].

$\displaystyle s\prime=\frac{520\times10^{-9}(1.50)}{0.520\times10^{-3}} =1.50\times10^{-3}\,\mathrm{m} =1.50\,\mathrm{mm}$

(e) [2]
Fringe spacing remains unchanged at $4.00\,\mathrm{mm}$ [1].
Central diffraction-envelope width doubles because
$W\propto1/b$ [1].

29. Standing waves and resonance [10 marks]

(a) [2]
Two waves of the same frequency, wavelength and amplitude travel in opposite directions [1]. Their superposition produces fixed nodes and antinodes [1].

(b) [2]
A sketch with two loops/antinodes [1] and three labelled nodes, including both fixed ends [1]. Node positions are
$0,\;0.400,\;0.800\,\mathrm{m}$ ;
antinodes are at $0.200$ and $0.600\,\mathrm{m}$ .
Adjacent loops have opposite displacement at a general instant.

(c) [2]
Wavelength [1]; frequency [1].

$\displaystyle L=2\frac{\lambda}{2} \quad\Rightarrow\quad \lambda=L=0.800\,\mathrm{m}$

$\displaystyle f=\frac{v}{\lambda} =\frac{120}{0.800} =150\,\mathrm{Hz}$

(d) [2]
First-harmonic frequency [1]; next allowed frequency [1].

$\displaystyle f_1=\frac{v}{4L} =\frac{340}{4(0.500)} =170\,\mathrm{Hz}$

$\displaystyle f_3=3f_1=510\,\mathrm{Hz}$

(e) [2]
The maximum amplitude decreases [1] and the peak broadens [1].
A shift of the peak towards a lower driving frequency is also an acceptable qualitative change; award a maximum of two marks for two correct changes.

30. Sound and light Doppler shifts [10 marks]

(a) [2]
Successive wavefronts are emitted from source positions closer to the observer, so wavefront spacing ahead of the source is smaller [1].
Sound speed through unchanged air stays constant, so wavefronts reach the stationary observer more frequently [1].

(b) [2]
Correct approaching-source relation/rearrangement [1]; source speed [1].

$\displaystyle f\prime=f\frac{v}{v-u_s} \quad\Rightarrow\quad u_s=v\left(1-\frac{f}{f\prime}\right)$

$\displaystyle u_s=340\left(1-\frac{800}{850}\right) =20.0\,\mathrm{m\,s^{-1}}$

(c) [2]
Correct receding-source sign/relation [1]; observed frequency [1].

$\displaystyle f\prime=800\left(\frac{340}{340+20.0}\right) =755.56\,\mathrm{Hz} \approx756\,\mathrm{Hz}$

(d) [2]
Correct receding-observer sign/relation [1]; observed frequency [1].

$\displaystyle f\prime=800\left(\frac{340-10.0}{340}\right) =776.47\,\mathrm{Hz} \approx776\,\mathrm{Hz}$

(e) [2]
Longer wavelength gives a redshift and radial recession [1];
correct speed [1].

$\displaystyle v_{\mathrm{radial}} \approx c\frac{|\Delta\lambda|}{\lambda_0} =3.00\times10^8\left(\frac{0.5}{500.0}\right) =3.0\times10^5\,\mathrm{m\,s^{-1}}$

Answer:
Receding at approximately $300\,\mathrm{km\,s^{-1}}$ .

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

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