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

Question

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 (A)  (1/20)s  (B)  (1/20) min  (C)  (19/20)s  (D)  (19/20) min

Tardigrade Admin · Accepted Answer

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^{'} = n (\frac{ν - v _{o}}{v - v _{s}})

Given,

v_{o} = 0, v_{s} = \frac{v}{20}

n = 1 Hz

(as blast is blown at an interval of

1 s

)

∴ n^{'} = \frac{v}{ν - \frac{v}{20}} \times 1 = \frac{20}{19} Hz

Observed time interval between two successive blasts

= \frac{19}{20} s

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

$\frac{1}{20} s$

$\frac{1}{20} min$

$\frac{19}{20} s$

$\frac{19}{20} min$

Q. A train approaches a stationary observer, the velocity of train being 201​ 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

201​s

201​min

2019​s

2019​min

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

$\frac{1}{20} s$

$\frac{1}{20} min$

$\frac{19}{20} s$

$\frac{19}{20} min$