Question

A bat flies at a steady speed of 4 ms-1 emitting a sound of f = 90 ×103 hz. It is flying horizontally towards a vertical wall. The frequency of the reflected sound as detected by the bat will be _____. (take velocity of sound in air as 330 ms-1).

Answer 1

your answer will be 9.27khz
f = 90 ×103 = 9,270=9,270/1000=9.27khz
Solution:
Doppler Effect equation:
n
$n'=( \frac{v + v_{0} }{v - v _{0}} )n = 9.27khz.( \frac{330 \frac{m}{s} + \frac{4m}{s} }{330 \frac{m}{s} - \frac{4m}{s} } = 9.27khz$

Answer 2

Given :

Speed of bat , v = 4 m/s .

Frequency emitted by bat , f = 90000 Hz = 90 KHz .

To Find :

The frequency of the reflected sound as detected by the bat .

Solution :

We know , frequency of reflected sound is given by :

$f'=(\dfrac{v_c+v_o}{v_c-v_o}) f$

Here , $v_c$ is velocity of sound in air and $v_c=330\ m/s$ .

Putting all given values we get :

$f'=(\dfrac{330+4}{330-4})\times 90\ KHz\\\\f'=92.2\ KHz$

Learn More :

Doppler's Effect

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