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:

• A. 1.2f, 1.2λ
• B. 1.2f, λ
• C. f, 1.2λ
• D. 0.8f, 0.8λ

Answer 1

Answer: (B)

1.2f, λ

Answer 2

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