# IB DP Physics 9.5 – Doppler effect Question Bank SL Paper 3

Question

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 0.5c.

b.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.

## 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.

b.

(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;

Question

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.

b.Determine the speed of the mosquito.

## 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;

b.

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 $$u = 0.45{\text{ m}}\,{{\text{s}}^{ – 1}}$$;

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