Question

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

• A. $1.2f,\lambda$
• B. $0.8f,0.8\lambda$
• C. $1.2f,1.2\lambda$
• D. $0.8f,1.2\lambda$

Answer 1

Answer: (A)

$1.2f,\lambda$

When an observer moves towards a stationary source of sound then apparent frequency heard by the observer increases. Then the apparent frequency heard in this situation

$f^{\prime }=\left(\frac{v+v_0}{v-v_s}\right)f$

As source is stationary hence $v_s=0$

$f^{\prime }=\left(\frac{v+v_0}{v}\right)f$

given $v_0=\frac{v}{5}$

Substituting in the relation for f' we have

$f^{\prime }=\left(\frac{v+v/5}{v}\right)f=\frac{6}{5}=1.2f$

Motion of observer does not affect the wavelength reaching the observer hence wavelength remains $\lambda .$