This question is about the Doppler effect.

The diagram shows wavefronts in air produced by a stationary source S of sound. The distance between successive wavefronts is equal to the wavelength of the sound. The speed of sound is *c*.

a.

On the diagram, sketch three successive wavefronts produced when S is moving to the left at a speed of 0.5*c*.

A source of X-rays rotates on a turntable. Radiation of wavelength 7.5 nm is emitted by the source and undergoes a maximum shift of 0.50 fm. The distance between the source and the detector is large in comparison to the diameter of the turntable.

(i) Determine the speed of a point on the edge of the turntable.

(ii) State the assumption you made in your answer to (b)(i).

**Answer/Explanation**

## Markscheme

a.

3 circular wavefronts;

2 centres/sources of wavefronts move left (by one box);

*Drawn circular wavefronts may be larger as in diagram here, or could be equal sized. Both are acceptable. *

(i) \(v = \frac{{5 \times {{10}^{ – 16}} \times 3 \times {{10}^8}}}{{7.5 \times {{10}^{ – 9}}}}\);

20(ms–1);*Use of sound equation not acceptable. *

(ii) assume speed of X-rays \( = \) *c */ assume speed of turntable << *c*;

This question is about the Doppler effect.

Georgia carries out an experiment to measure the speed of mosquitoes. She sets up a microphone to record the sounds of passing mosquitoes.

One mosquito is moving in a straight line with constant speed and passes very close to the microphone as seen in the diagram. The mosquito produces a sound of constant frequency.

a.

The speed of sound in air is \({\text{340 m}}\,{{\text{s}}^{ – {\text{1}}}}\).

The maximum frequency recorded is 751 Hz and the minimum frequency recorded is 749 Hz. Explain this observation.

Determine the speed of the mosquito.

**Answer/Explanation**

## Markscheme

a.

as the mosquito approaches the wavelength perceived by Georgia is shorter and therefore the perceived frequency is higher;

as the mosquito is moving away, the wavelength perceived is longer than the emitted and therefore the perceived frequency is lower;

due to the Doppler effect;

approaching \(751 = f \times \frac{{340}}{{340 – u}}\);

moving away \(749 = f \times \frac{{340}}{{340 + u}}\);

to produce \(u = 0.45{\text{ m}}\,{{\text{s}}^{ – 1}}\);

*or*

emitted frequency is \(\frac{{751 + 749}}{2} = 750{\text{ Hz}}\);

applying the Doppler effect for approach (or recession), \(751 = 750\frac{{340}}{{340 – u}}\)\(\,\,\,\)** or**\(\,\,\,\)\(749 = 750\frac{{340}}{{340 + u}}\);

to produce \(u = 0.45{\text{ m}}\,{{\text{s}}^{ – 1}}\);

## Examiners report

a.

In (a), many candidates drew the classic diagram with wavefronts closer in front of the source and further apart behind. However, very few related this to the situation and the values given.

(b) was a more difficult question with few working through to the solution.

This question is about the Doppler effect.

A child on a carousel (merry-go-round) moves with a speed of \({\text{6.5 m}}\,{{\text{s}}^{ – 1}}\) along a horizontal circular path ABCDA. A stationary observer is at a large distance from the carousel.

The child blows a whistle while moving from position B to position D. The whistle emits sound of frequency 850 Hz. The speed of sound in air is \({\text{330 m}}\,{{\text{s}}^{ – 1}}\).

a.

Describe what is meant by the Doppler effect.

(i) Determine the minimum frequency of the sound heard by the observer.

(ii) Describe the variation of the frequency of the sound heard by the observer.

**Answer/Explanation**

## Markscheme

a.

change in the observed/perceived frequency;

when there is relative motion between the source and the observer / when either the source or observer is moving;

(i) \(f = 850\left[ {\frac{{330}}{{330 + 6.5}}} \right]\);

\(f = 830{\text{ (Hz)}}\);

*Award **[1 max] **for correct answer if moving observer equation is used with 6.5 in the numerator or if the approximate formula is used.*

(ii) frequency increases;

(from 830 (Hz)) to 870 (Hz);

## Examiners report

a.

(a) was usually answered correctly, but it is worth emphasising that it is the perceived frequency that changes.

In (i) there were many correct answers, but sometimes the wrong Doppler equation was used or an incorrect sign convention chosen. Having found the minimum frequency (at position B) the vast majority then stated, in (ii), that frequency decreased from B to C. This is a question that needs careful discussion in class with future candidates.

This question is about radio telescopes.

A distant galaxy emits radio waves of frequency 6.0×10^{9} Hz and is moving with speed 6.0×10^{6} ms^{–1} directly away from an observer on Earth.

a.

Determine the wavelength of the radio wave as measured by the observer on Earth.

The radio signals from two stars on opposite sides of the galaxy are detected on Earth using a radio telescope. The telescope has a circular receiving dish.

(i) State the Rayleigh criterion for the images of two point sources to be just resolved.

(ii) The galaxy is 2.0×10^{21}m from Earth and the stars are separated by 5.0×10^{19}m. Determine the minimum size of the telescope dish required to resolve the images of the two stars at a wavelength of 5.1×10^{–2}m.

**Answer/Explanation**

## Markscheme

a.

\(\left( {{\rm{since}}\frac{v}{c} \ll 1} \right)\)\(\Delta f\left( { = \frac{v}{c}f} \right) = \frac{{6.0 \times {{10}^6}}}{{3.0 \times {{10}^8}}}6.0 \times {10^9}\left( { = 0.12 \times {{10}^9}{\rm{Hz}}} \right)\);

\(f’ = f – \Delta f\left( { = 5.9 \times {{10}^9}{\rm{Hz}}} \right)\);

\(\lambda ‘\left( { = \frac{c}{{f’}} = \frac{{3.00 \times {{10}^8}}}{{5.9 \times {{10}^9}}}} \right) = 5.1 \times {10^{ – 2}}\left( {\rm{m}} \right)\);

*Award [2 max] if ƒ’=ƒ+Δƒ used to give *λ

*‘*=4.9×10

^{-2}(m).

(i) the two (point-like) sources generate diffraction patterns with central maxima;

the central maximum of one pattern overlaps with the first minimum of the second diffraction pattern;

(ii) \(\theta \approx \frac{d}{D} = \frac{{5.0 \times {{10}^{19}}}}{{2.0 \times {{10}^{21}}}} = 0.025({\rm{rad}})\)

\(\left( {b > 1.22\frac{{5.1 \times {{10}^{ – 2}}}}{{0.025}} = } \right)2.5\left( {\rm{m}} \right)\);

*Allow [1 max] for solution that omits 1.22.*

## Examiners report

a.

was well done by some, although many used the equation appropriate for sound not the approximation for light where c >> v.

In (b) (i) candidates often got the first mark by implication when gaining second mark. There was a lot of confused algebra in (b)(ii).

This question is about the Doppler effect.

a.

Describe the Doppler effect.

A spectral line from a source on Earth has a frequency of 4.672×10^{14} Hz. When this same line is observed from a distant galaxy it is found to have shifted to 4.669×10^{14} Hz.

(i) State the direction of the motion of the galaxy relative to Earth.

(ii) Deduce the speed of the galaxy relative to Earth.

**Answer/Explanation**

## Markscheme

a.

(apparent) shift in frequency of wave;

when relative motion between source of waves and observer;

approach gives increase in frequency / recession decrease in frequency;*Allow answer in terms of red or blue shifts.*

(i) galaxy moving away from Earth;

(ii) rearranges data book equation thus \(v \approx \frac{{\Delta f}}{f}c\);

\(v = \frac{{3.0 \times {{10}^{11}} \times 3 \times {{10}^8}}}{{4.672 \times {{10}^{14}}}}\);

\(v = 1.93 \times {10^5}{\rm{m}}{{\rm{s}}^{ – 1}}\);

This question is about the Doppler effect in sound.

a.

A fire engine is travelling at a constant velocity towards a stationary observer. Its siren emits a note of constant frequency. As the engine passes close to the observer, the frequency of the note perceived by the observer decreases. Explain this decrease in terms of the wavefronts of the note emitted by the siren.

The frequency of the note emitted by the siren is 400 Hz. After the fire engine has passed, the frequency of the note detected by the observer is 360 Hz. Calculate the speed of the fire engine. (Take the speed of sound in air to be 340 ms^{–1}.)

**Answer/Explanation**

## Markscheme

a.

diagram showing (non concentric) wavefronts closer together in front/further apart behind source;

frequency is higher as source approaches (because more wavefronts are received per unit of time);

frequency is lower as source recedes (because fewer waverfronts are received per unit of time);

\(360 = 400\left( {\frac{{340}}{{340 + {u_{\rm{s}}}}}} \right)\);

u_{s}=38ms^{–1};*Award [2] for a bald correct answer.*

This question is about the Doppler effect.

a.

Describe what is meant by the Doppler effect as it relates to sound.

An ambulance is travelling at a speed of 28.0 ms^{–1} along a straight road. Its siren emits a continuous sound of frequency 520 Hz. The ambulance is approaching a stationary observer. The observer measures the frequency of the note to be 566 Hz. Determine the speed of sound.

**Answer/Explanation**

## Markscheme

a.

observed/perceived change in pitch/frequency;

when there is relative motion between source and observer;

recognize that \(f’ = \left[ {\frac{v}{{v – {u_{\rm{s}}}}}} \right]f\);

\(566 = \left[ {\frac{v}{{v – 28}}} \right]520\);

to give *v*=345(ms^{-1}) ;*Award [0] for use of the moving observer Doppler equation.*

*Award*

**[2 max]**for the use of +28 to give -345(ms^{-1}).*Otherwise award only the first marking point for substitution of the incorrect*

*values in the correct equation.*

This question is about the Doppler effect.

The sound emitted by a car’s horn has frequency ƒ , as measured by the driver. An observer moves towards the stationary car at constant speed and measures the frequency of the sound to be ƒ ‘.

a.

Explain, using a diagram, any difference between ƒ and ƒ*‘*.

The frequency ƒ is 3.00×10^{2}Hz. An observer moves towards the stationary car at a constant speed of 15.0ms^{−1}. Calculate the observed frequency ƒ’ of the sound. The speed of sound in air is 3.30×10^{2}ms^{−1}.

**Answer/Explanation**

## Markscheme

a.

circular wavefronts around source, equally spaced;

moving observer intercepts more wavefronts per unit time / the time between intercepting successive wavefronts is less;

hence observes a higher frequency / ƒ′ > ƒ;**or**

circular wavefronts around source, equally spaced;

the velocity of the sound waves with respect to the observer is greater;

since \(f’ = \frac{{v’}}{\lambda }\), observed frequency is also greater;

\(f’ = f\left( {\frac{{v + {u_o}}}{v}} \right) = 300\left( {\frac{{330 + 15}}{{330}}} \right)\);

=314 Hz; *Award [0] for use of moving source formula.*

*Award*

**[1]**for use of v-u_{o}to give 286 Hz.