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  4. An object, moving in a straight line with velocity 100 ms-1, goes past a stationary observer. If the object emits note of 400 Hz while moving, the change in the frequency note by the observer as the object goes past him is (speed of sound in air = 300 ms-1)

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Q. An object, moving in a straight line with velocity $100 \,ms^{-1}$, goes past a stationary observer. If the object emits note of $400 \,Hz$ while moving, the change in the frequency note by the observer as the object goes past him is
(speed of sound in air $= 300 \,ms^{-1})$

KEAMKEAM 2020

Solution:

Given $f=400 \,H_{2} ; c=300\, m / s, v=100 \,m / c$
Let $f_{1}$ and $f_{2}$ are the frequencies observed before and after train passes the observer.
Then,
$f_{1}=f \frac{C}{C-v}=400 \times \frac{300}{200}=600\, H _{2}$
$f_{2}=f \frac{C}{c+v}=400 \times \frac{300}{400}=300 \,H _{2} $
$\therefore \Delta f=f_{1}-f_{2}=300 \,Hz$