A train approaches stationary observer, the velocity of train being `1/20` of the velocity of sound. A sharp blast is blown with the whistle of the engine at equal intervals of 1s. The interval between the successive blasts as heard by the observer

Options

A. `1/20` s
B. `1/2` min
C. `19/20` s
D. `19/20` min

Answer

C

Explanation

From Doppler's effect, in the sound the apparent frequency of the source due to a relative motion between source and observer is
`n'=n((v-v_(0))/(v-v_(s)))`
where, v is velocity of sound, `v_(0)` the velocity of observer and `v_(s)` the velocity of source.
Given, `v_(0)=0` (observer stationary)
`v_(s)=v/20`
`n=1` Hz (as blast is blown at an interval of 1s)
`:. n'=v/(v-v/20)xx1=20/19 Hz`
Observed time interval between two successive blasts
`=19/20 s`