Question

A whistling engine is approaching a stationary observer with a velocity of 110m/s. The velocity of sound is 330m/s. The ratio of frequencies as heard by the observer as the engine approaches and recedes is

• A. 4:3
• B. 4:1
• C. 3:6
• D. 2:1

Answer 1

Answer: (D)

2:1

Given : $V_0=0m/s$
$V_s=110m/s$ (train approach )
$V_S=-110m/s$ (train recedes)
$V=330m/s$

By Doppler effect formula:
When train approach
${\delta }_1^1=\delta \left(\frac{300-0}{330-110}\right)$$=\frac{330\delta }{220}$

When train recedes
${\delta }_2^1=\delta \left(\frac{300-0}{330-(-110)}\right)$$=\delta \left(\frac{330}{330+110}\right)$$=\frac{330\delta }{440}$

So, $\frac{{\delta }_1^1}{{\delta }_2^1}=\frac{330\delta /220}{330\delta /440}=\frac{2}{1}$