| Date | May 2011 | Marks available | 2 | Reference code | 11M.3.SL.TZ1.2 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 1 |
| Command term | Calculate | Question number | 2 | Adapted from | N/A |
Question
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 '.
Explain, using a diagram, any difference between ƒ and ƒ'.
The frequency ƒ is 3.00×102Hz. 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 3.30×102ms−1.
Markscheme
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-uo to give 286 Hz.
