Question

An engine blowing whistle, is approaching a stationary observer with a velocity of $110m{s}^{-1}$. The ratio of frequencies heard by the observer while engine approaching and receding away from him will be ($ifv=330m{s}^{-1}$)

• A. 1 : 4
• B. 4 : 1
• C. 2 : 1
• D. 1 : 2

Answer 1

Answer: (C)

2 : 1

when source is approaching to the observer,

$n_1=\frac{330}{330-110}n=\left(3/2\right)n$

when source is receding from the observer,

$n_2=\frac{330}{330+110}n=\left(3/4\right)n$

$\frac{n_1}{n_2}=\frac{\left(3/2\right)}{\left(3/4\right)}=\frac{2}{1}$.

option "C" is correct.