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  4. A body is walking away from a wall towards an observer at a speed of 1 m/s and blows a whistle whose frequency is 680 Hz. The number of beats heard by the observer per second is (velocity of sound in air = 340 m/s )

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Q. A body is walking away from a wall towards an observer at a speed of $1\, m/s$ and blows a whistle whose frequency is $680 \,Hz$. The number of beats heard by the observer per second is (velocity of sound in air $= 340 \,m/s$ )

Haryana PMTHaryana PMT 2001

Solution:

When source and observer are moving close to each other, then apparent frequency
$n=\left(\frac{v+v_{o}}{v-v_{s}}\right) \times n \ldots(1)$
Here, $v_{s}=1 \,m / s$,
$v_{0}=1 \,m / s$,
$n=680 \,Hz$
So, from equation (1)
$n=\left(\frac{340+1}{340-1}\right) 680=684$
Number of beats per second
$=684-680=4$ beats / sec