Question

De
38. When a train is approaching the stationary
observer, the apparent frequency of the whistle
observed as 100 Hz, while when it has passed away
from the observer with same speed, it is 50 Hz.
Calculate the frequency of the whistle when the
observer moves with the train (V = 330 m/s)
1) 33.3 Hz 2) 50 HZ 266.6 Hz 4) 75 Hz
What is the frequency of whistle

Answer 1

Answer:

50 hz ...................

Answer 2

The frequency of the whistle = 200/3 Hz

Let the frequency of the whistle produced by the train be f.

Let the speed of train be v.

Case 1 - When the train approaches a stationary person -

Using Doppler's effect equation ,

f = frequency heard(veocity w.r.t listener/ velocity w.r.t source)

Velocity w.r.t listener = velocity of sound

Velocity w.r.t source = velocity of sound - velocity of train

=> 100 = f(330/330-v)

=> f = 100/(330/330-v)

Case 2 - When the train has crossed the stationary person -

Using Doppler's effect equation ,

f = frequency heard(veocity w.r.t listener/ velocity w.r.t source)

Velocity w.r.t listener = velocity of sound

Velocity w.r.t source = velocity of sound - velocity of train

=> 50 = f(330/330+v)

=> f = 50/(330/330+v)

Comparing the two values of f -

=> 50/(330/330+v) = 100/(330/330-v)

=> (330+v) = 2(330-v)

=> 330+v = 660-2v

=> 3v = 330

=> v = 110 m/s

Velocity of the train = 110 m/s

f = 50/(330/440)

=> f = 200/3 Hz

The frequency of the whistle = 200/3 Hz

When the observer will move with the train , the observer will hear the whistle with the same frequency as produced by the train.