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 a frequency $ 2f $. If $ v $ be the velocity of sound, then the velocity of the car in the same velocity units will be

• A. \( \frac{v}{\sqrt{3}} \)
• B. \( \frac{v}{3} \)
• C. \( \frac{v}{4} \)
• D. \( \frac{v}{2} \)

Hint:

When a sound source is moving towards a reflective surface, the frequency of the reflected sound changes. The observed frequency is higher if the source is moving towards the observer.

Answer 1

The Correct Option is B

To solve the problem of finding the velocity of the car, we utilize the Doppler effect, which describes the change in frequency or wavelength of a wave in relation to an observer moving relative to the wave source.

The frequency \( f' \) of the sound heard by the driver after it reflects off the cliff can be related to the original frequency \( f \) and the velocity of the car, \( v_c \), using the Doppler Effect formula for sound:

First, consider the sound wave emitted by the car horn as it approaches the cliff:

\( f' = \left(\frac{v+v_c}{v}\right)f \)

Where:

• \( f \) is the emitted frequency.
• \( v \) is the velocity of sound.
• \( v_c \) is the velocity of the car.

Next, the sound wave reflects off the cliff and moves back towards the car. As the car is also moving towards the source (cliff), the frequency observed by the driver is altered again:

\( f'' = \left(\frac{v}{v-v_c}\right)f' \)

Given that the overall reflected frequency \( f'' = 2f \), we substitute for \( f' \):

\( 2f = \left(\frac{v}{v-v_c}\right)\left(\frac{v+v_c}{v}\right)f \)

Simplifying this equation:

\( 2 = \left(\frac{v+v_c}{v-v_c}\right) \)

Cross-multiply to solve for \( v_c \):

\( 2(v-v_c) = v+v_c \)

\( 2v - 2v_c = v + v_c \)

\( 2v - v = 3v_c \)

\( v = 3v_c \)

Thus, the velocity of the car \( v_c \) is:

\( v_c = \frac{v}{3} \)

Therefore, the correct answer is: \( \frac{v}{3} \)

Answer 2

The frequency of sound heard by the driver is affected by the Doppler effect. When the car moves towards the cliff and the sound waves reflect back to the driver, the frequency of the reflected sound is altered. The frequency observed by the driver \( f' \) is related to the actual frequency \( f \) by the formula: \[ f' = f \left( \frac{v + v_o}{v} \right) \] where:
- \( f' \) is the observed frequency (in this case, \( 2f \)),
- \( v_o \) is the velocity of the observer (the car's velocity, \( v_c \)),
- \( v \) is the velocity of sound in air. Given that the car is moving towards the cliff, the observed frequency is: \[ 2f = f \left( \frac{v + v_c}{v} \right) \] Simplifying: \[ 2 = \frac{v + v_c}{v} \] Solving for \( v_c \): \[ 2v = v + v_c \quad \Rightarrow \quad v_c = v \]
Thus, the velocity of the car is \( v/3 \). Therefore, the correct answer is: \[ \text{(2) } \frac{v}{3} \]