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  4. A train approaches a 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 a second. The interval between the successive blasts as heard by the observer is

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Q. A train approaches a stationary observer, the velocity of train being $ \frac{1}{20} $ of the velocity of sound. A sharp blast is blown with the whistle of the engine at equal intervals of a second. The interval between the successive blasts as heard by the observer is

Rajasthan PMTRajasthan PMT 2007Electromagnetic Waves

Solution:

From Doppler's effect in sound the apparent change in frequency of the source due to a relative motion between source and observer is
$n^{\prime}=n\left(\frac{\nu-v_{o}}{v-v_{s}}\right)$
Given, $v_{o}=0, v_{s}=\frac{v}{20}$
$n=1 Hz$ (as blast is blown at an interval of $1 s$ )
$\therefore n^{\prime}=\frac{v}{\nu-\frac{v}{20}} \times 1=\frac{20}{19} Hz$
Observed time interval between two successive blasts $=\frac{19}{20} s$