Question

A car is moving towards a high cliff. The car driver sounds a horn of frequency \[f\]. The reflected sound heard by the driver has the frequency \[2f\]. If v be the velocity of sound, then the velocity of the car, in the same velocity units will be:
\[A)\dfrac{v}{\sqrt{2}}\]
\[B)\dfrac{v}{2}\]
\[C)\dfrac{v}{3}\]
\[D)\dfrac{v}{4}\]

Answer 1

Hint: Here, the car driver sounds the horn and a stationary cliff reflects the sound. And the frequencies of sound produced and reflected sounds are given in the question. Here, we can use the equation for frequency heard by an observer at rest. And substitute that frequency in the reflected frequency equation. Since the car and driver has the same velocities, we can modify the equation accordingly and find the velocity of the car.
Formula used:
\[{{f}^{'}}=\left( \dfrac{v}{v-{{v}_{s}}} \right)f\]
\[{{f}^{''}}=\left( \dfrac{v+{{v}_{0}}}{v} \right){{f}^{'}}\]

Complete answer:
Initially, when the car driver sounds the horn, the car acts as the source. And the source is moving towards an observer. Then, for an observer at rest, the frequency is given by,
\[{{f}^{'}}=\left( \dfrac{v}{v-{{v}_{s}}} \right)f\] ------------ 1
Where,
\[{{v}_{s}}\] is the velocity of the source/car
\[v\] is the velocity of the sound
\[f\] is the real frequency
\[{{f}^{'}}\] is the apparent frequency
Here, the frequency of sound produced by a car is \[f\].
Then, the sound is reflected by the cliff. Hence, the cliff acts as a source and the car driver is the observer. Here, the car driver is approaching the stationary cliff. Hence, the frequency heard by the car driver is given by,
\[{{f}^{''}}=\left( \dfrac{v+{{v}_{0}}}{v} \right){{f}^{'}}\] ------------ 2
Where,
\[{{v}_{0}}\] is the velocity of observer/driver
Given that, reflected sound frequency heard by the driver, \[{{f}^{''}}=2f\]
Substitute it in equation 2. We get,
\[2f=\left( \dfrac{v+{{v}_{0}}}{v} \right){{f}^{'}}\] ----------- 3
Substitute equation 1 in 3. We get,
\[2f=\left( \dfrac{v+{{v}_{0}}}{v} \right)\left( \dfrac{v}{v-{{v}_{s}}} \right)f\]
\[2f=\left( \dfrac{v+{{v}_{0}}}{v-{{v}_{s}}} \right)f\]
\[2v-2{{v}_{s}}=v+{{v}_{0}}\]
Here, since the car driver is the observer, velocity of driver is same as the velocity of car \[{{v}_{0}}={{v}_{s}}\]
Then,
\[2v-2{{v}_{s}}=v+{{v}_{s}}\]
Velocity of car, \[{{v}_{s}}=\dfrac{v}{3}\]

So, the correct answer is “Option C”.

Note:
The Doppler effect for sound is the decrease or increase in sound wave frequency as the sound of an object moves towards or away an observer. It is the effect produced by a moving source of sound waves in which there is an apparent downward shift in frequency for the observer from whom the source is receding and an apparent upward shift in frequency for an observer to whom the source is approaching. This concept can be applied to even light.