Question:

An observer moves towards a stationary source of sound with a speed \(\frac {1}{5}^{th}\) of the speed of sound. The wavelength and frequency of the source emitted are λ and f respectively. The apparent frequency and wavelength recorded by the observer are respectively:

Updated On: Oct 21, 2023
  • 1.2f, 1.2λ
  • 1.2f, λ
  • f, 1.2λ
  • 0.8f, 0.8λ
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The Correct Option is B

Solution and Explanation

Apparent frequency:

\(f ^′ = \frac {v+v_s}{v }f\)

\(f ^′ = \frac {v+(\frac 15)v}{v }f\)

\(f ^′ = (1+\frac 15)f\)

\(f ^′ = \frac 65f\)

\(f ^′ = 1.2f\)

Since, Source is stationary 
⇒ λ = constant

Motion of observer does not affect the wavelength reaching the observer hence wavelength remains unchanged means wavelength will be λ.

So, the correct option is (B): 1.2f, λ

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Concepts Used:

Doppler Effect

The Doppler effect is a phenomenon caused by a moving wave source that causes an apparent upward shift in frequency for observers who are approaching the source and a visible downward change in frequency for observers who are retreating from the source. It's crucial to note that the impact isn't caused by a change in the source's frequency.

 

 

 

 

 

 

 

 

 

 

The Doppler effect may be seen in any wave type, including water waves, sound waves, and light waves. We are most familiar with the Doppler effect because of our encounters with sound waves