Question

A whistling engine is approaching a stationary observer with a velocity of 110 m/s. The velocity of sound is 330 m/s. The ratio of frequencies as heard by an observer at the time of approaching and passing of engine is:
1. 3 : 6
2. 4 : 1
3. 4 : 3
4. 2 : 1

Answer 1

Correct Answer - Option 4 : 2 : 1

CONCEPT:

• Doppler effect in physics is defined as the increase (or decrease) in the frequency of sound, light, or other waves as the source and the observer move towards (or away from) each other.
  • Waves emitted by a source traveling towards an observer gets compressed. In contrast, waves emitted by a source traveling away from an observer get stretched out.
  • Doppler Effect (Doppler Shift) was first proposed by Christian Johann Doppler in 1842.
• Doppler effect formula: When the source and the observer moving towards each other

​\(⇒ f'=\frac{(v+ v_{o})}{(v- v_{s})}× f\)

Where f' = apparent frequency(Hz), f = actual frequency(Hz), v = velocity of the sound wave(m/s), vo = velocity of the observer(m/s), and vs = velocity of the sound(m/s)

CALCULATION:

Given vo = 0 m/s, v_s = 110 m/s and v = 330 m/s

The apparent frequency when the observer is stationary is given as,

\(⇒ f'=\frac{v}{(v- v_{s})}× f\)     -----(1)

• When an engine is approaching the observer,(v_s will be taken positive)

\(⇒ f'_{1}=\frac{v}{(v- v_{s})}× f\)

\(⇒ f'_{1}=\frac{330}{(330-110)}× f\)

\(⇒ f'_{1}=\frac{3}{2}f\)     -----(2)

• When the engine is passing to the observer,(vs will be taken negatively)

\(⇒ f'_{2}=\frac{v}{v- (-v_{s})}× f\)

\(⇒ f'_{2}=\frac{330}{(330+110)}× f\)

\(⇒ f'_{2}=\frac{3}{4}f\)     -----(3)

By equation 2 and equation 3,

\(\Rightarrow \frac{f_{1}^{'}}{f_{2}^{'}}=\frac{2}{1}\)

Hence, option 4 is correct.