Question

A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?​

Answer 1

Answer:

Ultrasonic beep frequency emitted by the bat, f=40kHz

Velocity of the bat, v

b

=0.03v

where, v= velocity of sound in air

The apparent frequency of the sound striking the wall is given as:

f

′

=(

v−v

b

v

)f

=(

v−0.03v

v

)×40

=

0.97

40

kHz

This frequency is reflected by the stationary wall (f

s

) toward the bat.

The frequency (f

′′

) of the received sound is given by the relation:

f

′′

=(

v

v+v

b

)f

′

=(

v

v+0.03v

)×

0.97

40

=

0.97

1.03×40

=42.47kHz

Answer 2

AnSwer :

Ultrasonic beep frequency emitted by the bat, ν = 40 kHz

Velocity of the bat, v${\sf{_b}}$ = 0.03

Where, v = velocity of sound in air

The apparent frequency of the sound striking the wall is given as:

${ \leadsto{ \sf{v' = \bigg( \dfrac{v}{v - v_b} \bigg)v}}}$

${ \leadsto{ \sf{v' = \bigg( \dfrac{v}{v - 0.03v} \bigg) \times 40}}}$

${ \leadsto{ \sf{v' = \dfrac{40}{0.97}KHz }}}$

The frequency is reflected by the stationary wall (v${\sf{_s}}$ = 0) towards the bat.

The frequency (v${\sf{^n}}$ ) of the received sound is given by the relation :-

${ \leadsto{ \sf{ {v}^{n} = \bigg( \dfrac{v +v_b }{v } \bigg)v'}}}$

${ \leadsto{ \sf{ {v}^{n} = \bigg( \dfrac{v +0.03v }{v } \bigg) \times \dfrac{40}{0.97} }}}$

${ \leadsto{ \sf{ {v}^{n} = \bigg( \dfrac{v (1+0.03)}{v } \bigg) \times \dfrac{40}{0.97} }}}$

${ \leadsto{ \sf{ {v}^{n} = \dfrac{1.03 \times 40}{0.97} }}}$

${ \leadsto{ \sf{ {v}^{n} = 42.47 KHz}}}$