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Question
CBSEENPH11026266

An engine moving towards a wall with a velocity 50 m/s  emits a note of 1.2 kHz. Speed of sound in air is 350 m/s. The frequency of the note after reflection from the wall as heard by the driver of the engine is

  • 2.4 kHz

  • 0.24 kHz

  • 1.6 KHz

  • 1.2 kHz

Solution

C.

1.6 KHz

The reflected sound appears to propagate in a direction opposite to that of moving engine. Thus, the source and the observer can be presumed to approach each other with same velocity. 

The Doppler effect is observed whenever the source of waves is moving with respect to an observer. 

So according to Doppler effect

      v' = v ( v + vo)v - vs

                = v v + vsv - vs                        ( vo = vs)

⇒         v' = 1.2 350 + 50350 - 50

                = 1.2 × 400300

           v' = 1.6 kHz