An observer is moving towards a stationary source of frequency 250 Hz with a velocity of 40 m/s. If the velocity of sound is 330 m/s, the apparent frequency heard by the observer will be:

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  1. 320 Hz
  2. 300 Hz
  3. 280 Hz
  4. None of these

Answer (Detailed Solution Below)

Option 3 : 280 Hz
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Detailed Solution

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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 source(m/s)

CALCULATION:

Given f = 250 Hz, vo = 40 m/s, vs = 0 m/s and v = 330 m/s

  • When the source and the observer moving towards each other the apparent frequency is given as,

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

\(⇒ f'=\frac{(330+40)}{(330- 0)}×250\)

⇒ f' = 280 Hz

  • Hence, option 3 is correct.
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