Question

A stationary source is emitting sound at a fixed frequency ${{f}_{0}}$, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of ${{f}_{0}}$. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is 330 m/s.

Answer 1

Hint: When a source of wave is moving with respect to an observer, there is an apparent shift in the frequency of the wave as observed by the observer. This effect of apparent change in frequency of the wave is known as Doppler’s effect. The apparent shift in frequency can be calculated by using Doppler shift formula.
Formula used:
Doppler shift formula, $f'=\left( \dfrac{v\pm {{v}_{o}}}{v\pm {{v}_{s}}} \right){{f}_{0}}$

Complete answer:
Doppler effect, also known as Doppler shift, is the phenomenon that is observed when the source of a wave is moving with respect to an observer.
If ${{f}_{0}}$ and v are the frequency and velocity of the wave produced by the source, then the apparent frequency is given by
$f'=\left( \dfrac{v\pm {{v}_{o}}}{v\pm {{v}_{s}}} \right){{f}_{0}}$
${{v}_{o}}$ is the velocity of observer and is taken positive if the observer moves towards the source and negative if it moves away from the source
${{v}_{s}}$ is the velocity of source and it is positive if the source moves away from the observer and negative if it moves in the opposite direction.
Let us assume cars are moving with speeds ${{v}_{1}}$ and ${{v}_{2}}$. The cars will be treated as an observer which is approaching the source. Then, these will be treated as a source, which is moving in the direction of sound.
Then frequency received by car 1 from the source is
${{f}_{1}}=\dfrac{v+{{v}_{1}}}{v-{{v}_{1}}}{{f}_{0}}$
The frequency received by car 2 from the source is
${{f}_{2}}=\dfrac{v+{{v}_{2}}}{v-{{v}_{2}}}{{f}_{0}}$
The difference in frequencies reflected from car 1 and car 2 is
${{f}_{1}}-{{f}_{2}}=\left( \dfrac{v+{{v}_{1}}}{v-{{v}_{1}}}-\dfrac{v+{{v}_{2}}}{v-{{v}_{2}}} \right){{f}_{0}}=\dfrac{2v({{v}_{1}}-{{v}_{2}})}{(v-{{v}_{1}})(v-{{v}_{2}})}{{f}_{0}}$
Since, ${{v}_{1}}$ and ${{v}_{2}}$ are much smaller than the speed of sound $v$ which is 330 m/s, we can take $v-{{v}_{1}}=v-{{v}_{2}}\approx v$.
Therefore,
$\dfrac{1.2}{100}{{f}_{0}}=\dfrac{2({{v}_{1}}-{{v}_{2}})}{v}{{f}_{0}}$
$\Rightarrow {{v}_{1}}-{{v}_{2}}=0.006v=0.006\times 330m/s=1.98m/s$
$1.98m{{s}^{-}}1=7.128km\,h{{r}^{-1}}\approx 7\,km/hr$
The difference between speed of the cars is 7 km/hr.

Note:
If the motion of the source of the wave relative to the observer is towards it then the Doppler shift is positive i.e. the apparent frequency is higher than the actual frequency. If the source moves in the opposite direction with respect to the observer, there is a downward shift in observed frequency.
It must be noted here that Doppler shift is not due to real shift in frequency of the source.