# IBDP Physics C.2 Wave model IB Style Question Bank : HL Paper 2

### Question

(a) A transverse water wave travels to the right. The diagram shows the shape of the surface of the water at time $$t=0 . \mathrm{P}$$ and $$\mathrm{Q}$$ show two corks floating on the surface.

(i) State what is meant by a transverse wave.[1]

(ii) The frequency of the wave is $$0.50 \mathrm{~Hz}$$. Calculate the speed of the wave.[1]

(iii) Plot on the diagram the position of $$\mathrm{P}$$ at time $$t=0.50 \mathrm{~s}$$.[1]

(iv) Show that the phase difference between the oscillations of the two corks is $$\pi$$ radians.[1]

(b) Monochromatic light is incident on two very narrow slits. The light that passes through the slits is observed on a screen. $$M$$ is directly opposite the midpoint of the slits. $$x$$ represents the displacement from $$\mathrm{M}$$ in the direction shown.

A student argues that what will be observed on the screen will be a total of two bright spots opposite the slits. Explain why the student’s argument is incorrect. [2]

(c) The graph shows the actual variation with displacement $$x$$ from $$M$$ of the intensity of the light on the screen. $$I_0$$ is the intensity of light at the screen from one slit only.

(i) Explain why the intensity of light at $$x=0$$ is $$4 I_0$$.[2]

(ii) The slits are separated by a distance of $$0.18 \mathrm{~mm}$$ and the distance to the screen is $$2.2 \mathrm{~m}$$. Determine, in $$\mathrm{m}$$, the wavelength of light.[2]

(iii) The two slits are replaced by many slits of the same separation. State one feature of the intensity pattern that will remain the same and one that will change.[2]
Stays the same:…………………………………………………………………………………………………….
Changes:……………………………………………………………………………………………………………….

(d) (i) Two sources are viewed though a single slit. The graph shows the diffraction pattern of one source.

Sketch, on the axes, the diffraction pattern of the second source when the images of the two sources are just resolved according to the Rayleigh criterion.[1]
(ii) Centaurus $$\mathrm{A}$$ is a galaxy a distance of $$1.1 \times 10^{23} \mathrm{~m}$$ away. A radio telescope of diameter $$300 \mathrm{~m}$$ operating at a wavelength of $$3.2 \mathrm{~cm}$$ is used to observe the galaxy. Determine the minimum size of the radio emitting region of the galaxy that can be resolved by this telescope.[2]

Ans:

a i «A wave where the displacement of particles/oscillations of particles/movement of particles/vibrations of particles is perpendicular/normal to the direction of energy transfer/wave travel/wave velocity/wave movement/wave propagation

a ii $$V=\alpha 0.50 \times 16=» 8.0 \ll \mathrm{ms}^{-1}$$

iv ALTERNATIVE 1
Phase difference is $$\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}$$ $$\alpha=\pi_b$$
ALTERNATIVE 2
One wavelength/period represents «phase difference» of $$2 \pi$$ and «corks» are $$1 / 2$$ wavelength/period apart so phase difference is $$\pi / O W T T E \checkmark$$

b.light acts as a wave «and not a particle in this situation» light at slits will diffract / create a diffraction pattern light passing through slits will interfere / create an interference pattern \&creating bright and dark spots»

c i The amplitude «at $$x=0$$ will be doubled $$\checkmark$$ intensity is proportional to amplitude squared / I $$\propto A^2 \checkmark$$

c ii Use of $$s=\frac{\lambda D}{d} \Rightarrow \lambda=\frac{s d}{D}$$ OR $$s=\frac{n \lambda D}{d} \Rightarrow \lambda=\frac{s d}{n D}$$
$\lambda=\ll \frac{0.567 \times 10^{-2} \times 0.18 \times 10^{-3}}{2.2}=» 4.6 \times 10^{-7} \ll \mathrm{mm}$

iii Stays the same: Position/location of maxima/distance/separation between maxima «will be the same» / OWTTE $$\checkmark$$

Changes: Intensity/brightness/width/sharpness «of maxima will changew/ OWTTE $$\checkmark$$

d i . Maximum coinciding with first minimum $$A N D$$ minimum coinciding with maximum $$\checkmark$$

d ii ALTERNATIVE 1
\begin{aligned} & \frac{d}{D}=1.22 \times \frac{\lambda}{b} \text { therefore } d=\frac{1.22 \times \lambda \times D}{b} \\ & \varangle d \approx 1.22 \times \frac{3.2 \times 10^{-2} \times 1.1 \times 10^{23}}{300},=1.4 \times 10^{+9} \alpha \mathrm{m} * \end{aligned}
ALTERNATIVE 2
\begin{aligned} & \theta=\kappa 1.22 \frac{\lambda}{b}=1.22 \times \frac{3.2 \times 10^{-2}}{300}=» 1.3 \times 10^{-4} \text { «radians } » \\ & \mathrm{~d}=\alpha\left(1.1 \times 10^{23}\right)\left(1.3 \times 10^{-4}\right)=» 1.4 \times 10^{19} \ll \mathrm{m} » \end{aligned}

### Question

The diagram shows the direction of a sound wave travelling in a metal sheet.

1. Particle P in the metal sheet performs simple harmonic oscillations. When the displacement of P is 3.2 μm the magnitude of its acceleration is 7.9 m s2. Calculate the magnitude of the acceleration of P when its displacement is 2.3 μm. [2]

2. The wave is incident at point Q on the metal–air boundary. The wave makes an angle of 54° with the normal at Q. The speed of sound in the metal is 6010 m s–1 and the speed of sound in air is 340 m s–1. Calculate the angle between the normal at Q and the direction of the wave in air. [2]

3. The frequency of the sound wave in the metal is 250 Hz. Determine the wavelength of the wave in air. [1]

4. The sound wave in air in (c) enters a pipe that is open at both ends. The diagram shows the displacement, at a particular time T, of the standing wave that is set up in the pipe.

A  particular air molecule has its equilibrium position at the point labelled M. On the diagram, at time T,

1. draw an arrow to indicate the acceleration of this molecule. [1]

2. label with the letter C a point in the pipe that is at the centre of a compression. [1]

e.       Sound of frequency f = 2500 Hz is emitted from an aircraft that moves with speed v = 280 ms–1 away from a stationary observer.

The speed of sound in still air is c = 340 ms–1.

1. frequency heard by the observer. [2]

2. wavelength measured by the observer. [1]

Ans:

a

Expression or statement showing acceleration is proportional to displacement

so « 7.9 × $$\frac{2.3}{3.2}$$»= 5.7 «ms-2»

b

sinθ = $$\frac{340}{6010}$$ × sin54°

θ = 2.6°

c

λ = « $$\frac{340}{250}$$ =» 1.36 ≈ 1.4 «m»

d i horizontal arrow «at M» pointing left

d ii

any point labelled C on the vertical line shown below

e i

f’ = 2500 × $$\frac{340}{340+280}$$

f ′ = 1371 ≈1400  «Hz »

e ii λ = $$\frac{340}{1371}$$ ≈ 0.24 / 0.25 «m »

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