Question

A siren emitting a sound of frequency $800\, Hz$ moves away from an observer towards a cliff at a speed of $15\,m\,s^{-1}$. Then, the frequency of sound that the observer hears in the echo reflected from the cliff is : (Take velocity of sound in air = $330 \, ms^{-1}$)

• A. 800 Hz
• B. 838 Hz
• C. 885 Hz
• D. 765 Hz

Answer 1

The Correct Option is B

 To find the frequency of the sound that the observer hears in the echo reflected from the cliff, we use the Doppler effect. The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source.

1. First, we calculate the frequency of the sound when it is directly observed coming from the siren moving away from the observer:
   • Let \(f_s = 800 \, \text{Hz}\) be the frequency of the siren.
   • The speed of the siren \(v_s = 15 \, \text{m/s}\).
   • Velocity of sound in air \(v = 330 \, \text{m/s}\).
   • When the source is moving away, the observed frequency \(f'= \left( \frac{v}{v + v_s} \right) f_s\).
   • Substitute the values: \(f' = \left( \frac{330}{330 + 15} \right) \times 800 = \left( \frac{330}{345} \right) \times 800 \approx 765.22 \, \text{Hz}\).
2. The siren's sound hits the cliff and reflects back. Here, the cliff acts as a new source of sound with frequency \(f'\).
3. The echo is then heard by the observer. Now, the roles are reversed: the observer acts as the source moving towards the new source (the cliff):
   • The frequency received by the source moving towards the observer \(f'' = \left( \frac{v + v_s}{v} \right) f'\).
   • Substitute the value of \(f'\): \(f'' = \left( \frac{330 + 15}{330} \right) \times 765.22 \approx \left( \frac{345}{330} \right) \times 765.22 \approx 837.65 \, \text{Hz}\).
4. Rounding to the nearest integer, the frequency is \(838 \, \text{Hz}\).

Therefore, the frequency of sound that the observer hears in the echo reflected from the cliff is 838 Hz, which is the correct answer.