# IBDP Physics Unit 4. Waves- 4.5 Standing waves-IB Style Question Bank SL Paper 2

Question

A loudspeaker emits sound towards the open end of a pipe. The other end is closed. A standing wave is formed in the pipe. The diagram represents the displacement of molecules of air in the pipe at an instant of time. X and Y represent the equilibrium positions of two air molecules in the pipe. The arrow represents the velocity of the molecule at Y.

The loudspeaker in (a) now emits sound towards an air–water boundary. A, B and C are parallel wavefronts emitted by the loudspeaker. The parts of wavefronts A and B in water are not shown. Wavefront C has not yet entered the water. a.i.

Outline how the standing wave is formed.

a.ii.

Draw an arrow on the diagram to represent the direction of motion of the molecule at X.

a.iii.

Label a position N that is a node of the standing wave.

a.iv.

The speed of sound is 340 m s–1 and the length of the pipe is 0.30 m. Calculate, in Hz, the frequency of the sound.

b.i.

The speed of sound in air is 340 m s–1 and in water it is 1500 m s–1.

The wavefronts make an angle θ with the surface of the water. Determine the maximum angle, θmax, at which the sound can enter water. Give your answer to the correct number of significant figures.

b.ii.

Draw lines on the diagram to complete wavefronts A and B in water for θ < θmax.

## Markscheme

a.i.

the incident wave «from the speaker» and the reflected wave «from the closed end»

superpose/combine/interfere

Do not allow meet/interact[1 mark]

a.ii.

Horizontal arrow from X to the right

MP2 is dependent on MP1

Ignore length of arrow[1 mark]

a.iii.

P at a node [1 mark]

a.iv.

wavelength is λ = «$$\frac{{4 \times 0.30}}{3}$$ =» 0.40 «m»

f = «$$\frac{{340}}{{0.40}}$$» 850 «Hz»

Award  for a bald correct answer

Allow ECF from MP1[2 marks]

b.i.

$$\frac{{\sin {\theta _c}}}{{340}} = \frac{1}{{1500}}$$

θc = 13«°»

Award  for a bald correct answer

Award  for a bald answer of 13.1

Answer must be to 2/3 significant figures to award MP2

b.ii.

correct orientation

greater separation

Do not penalize the lengths of A and B in the water

Do not penalize a wavefront for C if it is consistent with A and B

MP1 must be awarded for MP2 to be awarded [2 marks

Question

A large cube is formed from ice. A light ray is incident from a vacuum at an angle of 46˚ to the normal on one surface of the cube. The light ray is parallel to the plane of one of the sides of the cube. The angle of refraction inside the cube is 33˚. Each side of the ice cube is 0.75 m in length. The initial temperature of the ice cube is –20 °C.

a.i.

Calculate the speed of light inside the ice cube.

a.ii.

Show that no light emerges from side AB.

a.iiii.

Sketch, on the diagram, the subsequent path of the light ray.

b.i.

Determine the energy required to melt all of the ice from –20 °C to water at a temperature of 0 °C.

Specific latent heat of fusion of ice = 330 kJ kg–1
Specific heat capacity of ice = 2.1 kJ kg–1 k–1
Density of ice = 920 kg m–3

b.ii.

Outline the difference between the molecular structure of a solid and a liquid.

## Markscheme

a.i.

«v = c$$\frac{{{\text{sin }}i}}{{{\text{sin }}r}}$$ =» $$\frac{{3 \times {{10}^8} \times {\text{sin}}\left( {33} \right)}}{{{\text{sin}}\left( {46} \right)}}$$

2.3 x 108 «m s–1»

a.ii.

light strikes AB at an angle of 57°

critical angle is «sin–1$$\left( {\frac{{2.3}}{3}} \right)$$ =» 50.1°

49.2° from unrounded value

angle of incidence is greater than critical angle so total internal reflection

OR

light strikes AB at an angle of 57°

calculation showing sin of “refracted angle” = 1.1

statement that since 1.1>1 the angle does not exist and the light does not emerge[Max 3 marks]

a.iiii.

total internal reflection shown

ray emerges at opposite face to incidence

Judge angle of incidence=angle of reflection by eye or accept correctly labelled angles

With sensible refraction in correct direction

b.i.

mass = «volume x density» (0.75)3 x 920 «= 388 kg»

energy required to raise temperature = 388 x 2100 x 20 «= 1.63 x 107

energy required to melt = 388 x 330 x 103 «= 1.28 x 108

1.4 x 108 «J» OR 1.4 x 105 «kJ»

Accept any consistent units

Award [3 max] for answer which uses density as 1000 kg–3 (1.5× 108 «J»)

b.ii.

in solid state, nearest neighbour molecules cannot exchange places/have fixed positions/are closer to each other/have regular pattern/have stronger forces of attraction

in liquid, bonds between molecules can be broken and re-form

OWTTE

Accept converse argument for liquids[Max 1 Mark]

Question

A student investigates how light can be used to measure the speed of a toy train. Light from a laser is incident on a double slit. The light from the slits is detected by a light sensor attached to the train.

The graph shows the variation with time of the output voltage from the light sensor as the train moves parallel to the slits. The output voltage is proportional to the intensity of light incident on the sensor. a.

Explain, with reference to the light passing through the slits, why a series of voltage peaks occurs.

b.i.

The slits are separated by 1.5 mm and the laser light has a wavelength of 6.3 x 10–7 m. The slits are 5.0 m from the train track. Calculate the separation between two adjacent positions of the train when the output voltage is at a maximum.

b.ii.

Estimate the speed of the train.

c.

In another experiment the student replaces the light sensor with a sound sensor. The train travels away from a loudspeaker that is emitting sound waves of constant amplitude and frequency towards a reflecting barrier. The sound sensor gives a graph of the variation of output voltage with time along the track that is similar in shape to the graph shown in the resource. Explain how this effect arises.

## Markscheme

a.

«light» superposes/interferes

pattern consists of «intensity» maxima and minima
OR
consisting of constructive and destructive «interference»

voltage peaks correspond to interference maxima

b.i.

«$$s = \frac{{\lambda D}}{d} = \frac{{6.3 \times {{10}^{ – 7}} \times 5.0}}{{1.5 \times {{10}^{ – 3}}}} =$$» 2.1 x 10–3 «m»

If no unit assume m.

b.ii.

correct read-off from graph of 25 m s

v = «$$\frac{x}{t} = \frac{{2.1 \times {{10}^{ – 3}}}}{{25 \times {{10}^{ – 3}}}} =$$» 8.4 x 10–2 «m s–1»

Allow ECF from (b)(i)

c.

ALTERNATIVE 1

«reflection at barrier» leads to two waves travelling in opposite directions

mention of formation of standing wave

maximum corresponds to antinode/maximum displacement «of air molecules»
OR
complete cancellation at node position