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NEET PHYSICSMedium

A train moving at a speed of 220 ms−1220 \text{ ms}^{-1} towards a stationary object, emits a sound of frequency 1000 Hz1000 \text{ Hz}. Some of the sound reaching the object gets reflected back to the train as an echo. The frequency of the echo as detected by the driver of the train is (speed of sound in air is 330 ms−1330 \text{ ms}^{-1})

A

3500 Hz3500 \text{ Hz}

B

4000 Hz4000 \text{ Hz}

C

5000 Hz5000 \text{ Hz}

D

3000 Hz3000 \text{ Hz}

Step-by-Step Solution

  1. Frequency received by the stationary object: The train acts as a source moving towards a stationary observer (the object). The apparent frequency f′f' received by the object is given by the Doppler effect formula: f′=f(vv−vs)f' = f \left( \frac{v}{v - v_s} \right) . Substitute the given values (f=1000 Hzf = 1000 \text{ Hz}, v=330 m/sv = 330 \text{ m/s}, vs=220 m/sv_s = 220 \text{ m/s}): f′=1000(330330−220)=1000(330110)=1000×3=3000 Hzf' = 1000 \left( \frac{330}{330 - 220} \right) = 1000 \left( \frac{330}{110} \right) = 1000 \times 3 = 3000 \text{ Hz}
  2. Frequency of the echo heard by the driver: The object now acts as a stationary source reflecting the sound of frequency f′=3000 Hzf' = 3000 \text{ Hz}. The driver acts as an observer moving towards this stationary source with a velocity vo=220 m/sv_o = 220 \text{ m/s}. The frequency f′′f'' detected by the driver is: f′′=f′(v+vov)f'' = f' \left( \frac{v + v_o}{v} \right) . Substitute the values into the formula: f′′=3000(330+220330)=3000(550330)=3000×53=5000 Hzf'' = 3000 \left( \frac{330 + 220}{330} \right) = 3000 \left( \frac{550}{330} \right) = 3000 \times \frac{5}{3} = 5000 \text{ Hz} Therefore, the frequency of the echo detected by the driver is 5000 Hz5000 \text{ Hz}.
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