C.2 Wave model – IB Physics exam questions with mark scheme

1Core4 marks

A wave travels along a rope. The graph shows the displacement $y$ of the rope against position $x$ at one instant. The frequency of the wave is $6.0\ \mathrm{Hz}$.

(a) Distinguish between transverse and longitudinal waves. [2]

(b) Use the graph to determine the speed of the wave. [2]

Transverse: oscillations (of particles) are perpendicular to the direction of energy transfer (accept direction of wave travel/propagation)(1)
Longitudinal: oscillations are parallel to the direction of energy transfer (accept direction of wave travel/propagation)(1)
Wavelength $\lambda=0.40$ m read from graph (one full cycle)(1)
$v=f\lambda = 6.0\times0.40 = 2.4\ \mathrm{m\,s^{-1}}$ (2.4)(1)
Answer: (a) transverse: oscillations perpendicular to the direction of energy transfer; longitudinal: oscillations parallel to it (b) 2.4 m s$^{-1}$
Examiner tip: This is a displacement–distance graph, so one full cycle along the $x$-axis is the wavelength, not the period; use $v=f\lambda$.

2Core4 marks

A Wi-Fi router emits electromagnetic waves of frequency $2.4\ \mathrm{GHz}$.

(a) State two properties that all electromagnetic waves have in common. [2]

(b) Calculate the wavelength of the waves emitted by the router. [2]

Any one of: travel at the same speed $c$ / $3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ in a vacuum; can travel through a vacuum / need no medium; are transverse; consist of oscillating electric and magnetic fields (perpendicular to each other)(1)
A second, different property from the list (ignore 'transfer energy' on its own as it applies to all waves; do not accept 'can be reflected' etc. without a property specific to EM waves)(1)
Converts frequency: $f = 2.4\times10^{9}$ Hz and uses $\lambda = c/f$(1)
$\lambda = 3.00\times10^{8}/2.4\times10^{9} = 0.13$ m (accept 0.125 m; award [2] for a bare correct answer; using 2.4 Hz or $2.4\times10^{6}$ Hz scores [0] for this mark)(1)
Answer: (a) e.g. travel at $c$ in a vacuum; transverse (also: need no medium; oscillating E and B fields) (b) 0.13 m (0.125 m)
Examiner tip: Giga means $10^{9}$: a slip in the prefix is the most common way to lose both calculation marks.

3Core4 marks

A buoy floats on the sea as water waves pass it. The graph shows how the vertical displacement of the buoy varies with time.

(a) State the amplitude of the waves. [1]

(b) Determine the frequency of the waves. [2]

Amplitude = 0.60 m (do not accept 1.2 m, the peak-to-trough height)(1)
Period read from graph: $T = 4.0$ s(1)
$f = 1/T = 0.25$ Hz(1)
$\lambda = v/f = 7.5/0.25 = 30$ m (accept $\lambda = vT = 7.5\times4.0$; allow ECF from (b))(1)
Answer: (a) 0.60 m (b) 0.25 Hz (c) 30 m
Examiner tip: This is a displacement–time graph, so one full cycle along the axis gives the period, not the wavelength.

Stuck on C.2 Wave model?

Get it sorted in a free 30-minute session

We work through your weakest questions on this topic on Zoom. I reply the same day.

Free 30-min session on this topic Send me 5 questions a day

4Standard5 marks

A sound wave travels through air from left to right. The graph shows, at one instant, the displacement of air particles against their equilibrium position $x$. A positive displacement is to the right.

(a) State the displacement of the particle whose equilibrium position is $x = 0.20$ m. [1]

(b) Identify, with a reason, one value of $x$ at which there is a centre of a compression. [2]

6 μm to the right (of its equilibrium position) (direction required; accept $+6$ μm with 'right' implied by the sign convention stated)(1)
$x = 0.40$ m or $x = 1.20$ m(1)
Reason: particles to the left of this point are displaced to the right and particles to the right of it are displaced to the left, so particles crowd together there (do not award the reason mark for 'displacement is zero' alone, as this is also true at a rarefaction)(1)
Wavelength read from graph: $\lambda = 0.80$ m(1)
$f = v/\lambda = 340/0.80 = 430$ Hz (accept 425 Hz; allow ECF for wrong $\lambda$)(1)
Answer: (a) 6 μm to the right (b) $x = 0.40$ m (or 1.20 m) (c) 430 Hz (425 Hz)
Examiner tip: Compressions and rarefactions both sit where displacement is zero; you must look at which way the neighbouring particles have moved to tell them apart.

5Standard7 marks

Two students measure the speed of sound using an echo. Student A stands $66.0\ \mathrm{m}$ from a large flat wall, measured with a tape measure, and claps so that each clap coincides with the echo of the previous clap. Student B uses a stopwatch to time 30 intervals between claps and records $12.0\ \mathrm{s}$.

(a) Calculate the speed of sound obtained by the students. [2]

(b) Explain why the students timed 30 intervals rather than one. [2]

(c) The uncertainty in the time is $\pm0.2\ \mathrm{s}$ and the uncertainty in the distance is $\pm0.5\ \mathrm{m}$. Determine the absolute uncertainty in the speed of sound. [2]

(d) Suggest why this method would not work well if the wall were only 10 m away. [1]

Time for one interval $= 12.0/30 = 0.400$ s and distance travelled by sound in this time $= 2\times66.0 = 132$ m(1)
$v = 132/0.400 = 330\ \mathrm{m\,s^{-1}}$ (a value of $165\ \mathrm{m\,s^{-1}}$ from omitting the factor of 2 scores [1] max)(1)
Human reaction time / starting and stopping the stopwatch gives an uncertainty of about 0.1–0.2 s that is the same however many intervals are timed(1)
So timing many intervals makes the (percentage/fractional) uncertainty in the time for one interval much smaller (accept: the uncertainty is spread over 30 intervals)(1)
Fractional uncertainties added: $0.2/12.0 + 0.5/66.0 = 0.0167 + 0.0076 = 0.024$ (accept 2.4%)(1)
$\Delta v = 0.024\times330 = \pm8\ \mathrm{m\,s^{-1}}$ (accept $\pm7.9$ or $\pm8.0$; allow ECF from (a))(1)
The interval between claps would be only about 0.06 s, too short to clap at that rate / to distinguish the echo from the clap (accept: echo returns before the clap has finished)(1)
Answer: (a) 330 m s$^{-1}$ (b) reaction-time uncertainty is fixed, so its percentage effect on one interval is reduced (c) $\pm8$ m s$^{-1}$ (d) interval about 0.06 s: too short to clap at / echo merges with clap
Examiner tip: Sound goes to the wall and back, so the distance is $2\times66.0$ m; forgetting this halves the answer.

6Standard6 marks

A transverse wave of frequency $5.0\ \mathrm{Hz}$ travels to the right along a stretched string. The graph shows the displacement $y$ of the string against distance $x$ at one instant. Point P is at a crest and point Q is at the equilibrium position.

(a) State the direction in which Q is moving at this instant. [1]

(b) Calculate the speed of the wave. [1]

(d) Calculate the total distance moved by P in 1.0 s. [2]

Upwards (accept towards positive $y$ / perpendicular to the string, up the page; do not accept 'to the right')(1)
$v = f\lambda = 5.0\times0.80 = 4.0\ \mathrm{m\,s^{-1}}$(1)
$T = 1/f = 0.20$ s and Q needs a quarter of a cycle (accept: crest at P must travel 0.20 m to reach Q)(1)
$t = T/4 = 0.050$ s (accept $0.20/4.0 = 0.050$ s; allow ECF from (b))(1)
Number of oscillations in 1.0 s = 5.0 and each oscillation moves P a distance of $4A = 6.0$ cm(1)
Distance $= 5\times6.0 = 30$ cm $= 0.30$ m (answers of 0 m, from displacement, or 4.0 m, from wave speed, score [0])(1)
Answer: (a) upwards (b) 4.0 m s$^{-1}$ (c) 0.050 s (d) 30 cm (0.30 m)
Examiner tip: Particles on a transverse wave move up and down only: P does not travel along the string, it covers $4A$ in every cycle.

7Stretch6 marks

A transverse wave travels along a long rope that lies along the $x$-axis. Graph 1 shows the displacement $y$ of the rope against position $x$ at time $t = 0$. Graph 2 shows how the displacement of the particle at $x = 0$ varies with time.

(a) Determine the speed of the wave. [2]

(b) Deduce, using both graphs, the direction in which the wave is travelling. [2]

$\lambda = 1.2$ m from Graph 1 and $T = 0.30$ s from Graph 2(1)
$v = \lambda/T = 1.2/0.30 = 4.0\ \mathrm{m\,s^{-1}}$(1)
Graph 2 shows the particle at $x = 0$ is moving upwards (displacement increasing from zero) at $t = 0$(1)
In Graph 1 the rope just to the left of $x = 0$ would be displaced upwards / the crest behind it must move towards it, so the wave travels in the positive $x$ direction (to the right) (accept equivalent argument: shifting Graph 1 slightly to the right makes $y$ at $x=0$ positive)(1)
In 0.075 s the wave moves $4.0\times0.075 = 0.30$ m (accept: 0.075 s is $T/4$), so the particle at 0.60 m now has the displacement that the particle at 0.30 m had at $t = 0$(1)
$y = -4.0$ cm (4.0 cm below equilibrium / at a trough) (allow ECF for direction in (b))(1)
Answer: (a) 4.0 m s$^{-1}$ (b) positive $x$ direction (to the right) (c) $-4.0$ cm
Examiner tip: Do not confuse the two graphs: read the wavelength only from the $y$–$x$ graph and the period only from the $y$–$t$ graph.

8Standard5 marks

Yellow light from a sodium lamp has a wavelength of $589\ \mathrm{nm}$ in a vacuum.

(a) Calculate the frequency of the light. [1]

(b) The light passes from air into a glass block, where its speed is $2.00\times10^{8}\ \mathrm{m\,s^{-1}}$. State which of the frequency or the wavelength of the light stays the same, and explain why. [2]

(d) An X-ray tube produces radiation of frequency $6.0\times10^{18}\ \mathrm{Hz}$. Calculate its wavelength in a vacuum. [1]

$f = c/\lambda = 3.00\times10^{8}/589\times10^{-9} = 5.09\times10^{14}$ Hz (accept $5.1\times10^{14}$ Hz)(1)
The frequency stays the same(1)
Frequency is determined by the source / the number of waves arriving at the boundary each second must equal the number leaving it (do not accept 'frequency never changes' without a reason)(1)
$\lambda = v/f = 2.00\times10^{8}/5.09\times10^{14} = 3.93\times10^{-7}$ m (accept 393 nm or 390 nm; allow ECF from (a))(1)
$\lambda = 3.00\times10^{8}/6.0\times10^{18} = 5.0\times10^{-11}$ m(1)
Answer: (a) $5.09\times10^{14}$ Hz (b) frequency; it is set by the source (waves in per second = waves out per second) (c) 393 nm (d) $5.0\times10^{-11}$ m
Examiner tip: When light changes medium the speed and wavelength change together; the frequency is fixed by the source.

9Hard6 marks

Sound waves and light waves both transfer energy from one place to another.

Compare and contrast the nature of sound waves in air with the nature of light waves. Your answer should refer to what oscillates, the direction of the oscillations, the need for a medium and the speed of the waves. [6]

Similarity: both transfer energy (and information) without a net transfer of matter(1)
Similarity: both obey $v = f\lambda$; both can be reflected, refracted, diffracted and show interference(1)
Sound: air molecules / particles oscillate about their equilibrium positions(1)
Light: electric and magnetic fields oscillate, perpendicular to each other(1)
Sound is longitudinal: oscillations parallel to the direction of energy transfer, producing compressions and rarefactions (pressure variations)(1)
Light is transverse: oscillations perpendicular to the direction of energy transfer(1)
Sound is a mechanical wave and needs a medium, so cannot travel through a vacuum; light is electromagnetic and can travel through a vacuum(1)
Speed of sound in air is about $340\ \mathrm{m\,s^{-1}}$; light travels at $3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ in a vacuum (and slightly less in air), roughly a million times faster(1)
Speed of sound depends on the medium (faster in liquids and solids); light travels more slowly in denser optical media such as glass(1)

Level	Marks	What it looks like
Level 3	5–6	Clear comparison covering what oscillates, direction of oscillation, medium and speed for both waves, with at least one valid similarity; correct physics terminology (longitudinal, transverse, compression/rarefaction, electromagnetic) and approximate speeds given.
Level 2	3–4	Correct points about both waves on at least three of the four aspects, but the comparison is partly one-sided or a similarity is missing; minor inaccuracies in terminology.
Level 1	1–2	Isolated correct statements, e.g. sound is longitudinal or light can travel through a vacuum, with little or no comparison.
No marks	0	No relevant content.

Answer: Both transfer energy and obey $v=f\lambda$. Sound: air particles oscillate parallel to energy transfer (longitudinal), needs a medium, about 340 m s$^{-1}$. Light: perpendicular E and B fields oscillate perpendicular to energy transfer (transverse), travels through a vacuum at $3.00\times10^{8}$ m s$^{-1}$.
Examiner tip: 'Compare and contrast' needs similarities as well as differences, and each difference should be stated for both waves side by side.

10Hard6 marks

An earthquake produces two kinds of seismic wave that travel through the Earth's crust: P-waves, which are longitudinal, and S-waves, which are transverse. In the crust, P-waves travel at $8.0\ \mathrm{km\,s^{-1}}$ and S-waves at $4.5\ \mathrm{km\,s^{-1}}$.

(a) S-waves cannot travel through the Earth's liquid outer core but P-waves can. Explain this observation. [2]

(b) A seismometer records the first S-wave arriving $25$ s after the first P-wave. Determine the distance from the seismometer to the earthquake. [3]

In a transverse wave the particles oscillate perpendicular to the direction of travel, which needs a shear / sideways restoring force between neighbouring layers(1)
A liquid cannot sustain a shear force / has no rigidity, but it can be compressed, so longitudinal (compression) waves can pass through it(1)
Sets up $\dfrac{d}{4.5} - \dfrac{d}{8.0} = 25$ (with consistent units)(1)
$d\left(0.222 - 0.125\right) = 25$, so $d = 25/0.0972$(1)
$d = 260$ km (accept 257 km; award [3] for a bare correct answer; using a single speed scores [0])(1)
$\lambda = v/f = 8.0\times10^{3}/2.0 = 4.0\times10^{3}$ m (4.0 km)(1)
Answer: (a) S-waves are transverse and need a medium that resists shear; liquids do not, but can transmit compressions (b) 260 km (257 km) (c) 4.0 km
Examiner tip: Both waves cover the same distance; write a time for each and subtract, rather than multiplying 25 s by one of the speeds.

Stuck on this topic?

Book a free 30-minute session on WhatsApp and we'll find exactly which step is costing you the marks. Or build a timed paper on it.

Free 30-minute session