Question

A bat is flying towards a stationary wall at a constant speed. The bat emits a sound of frequency 125 kHz. The speed of the sound in the air is 343 m/s. If the frequency of the sound that is reflected back by the wall is 135 kHz, then the speed of the bat is:
1. 11.2 m/s
2. 9.2 m/s
3. 10 m/s
4. 13.2 m/s

Answer 1

Correct Answer - Option 4 : 13.2 m/s

CONCEPT:

Doppler effect:

• 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 v = 343 m/s, f = 125 kHz, and f' = 135 kHz

• The apparent frequency is given as,

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

• Here the source (bat) and the observer (bat), both are the same,

\(⇒ f'=\frac{(v+ v_{b})}{(v- v_{b})}× f\)

\(⇒ 135\times10^3=\frac{(343+ v_{b})}{(343- v_{b})}× 125\times10^3\)

⇒ 260v_b = 3430

⇒ vb = 13.2 m/s

• Hence, option 4 is correct.