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

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

Velocity of the bat, v_b = 0.03 v

Where, v = velocity of sound in air

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

`v' = (v/(v - v_b))v`

`= (v/(v - 0.03v)) xx 40`

`= 40/0.97` kHz`

This frequency is reflected by the stationary wall (`v_s = 0`) toward the bat.

The frequency (v") of the received sound is given by the relation:

`v" = ((v + v_b)/v) v'`

`= ((v+0.03v)/v) xx 40/0.97`

`= (1.03 xx 40)/ 0.97 = 42.47 kHz`

Answer 2

Here, the frequency of sound emitted by the bat, υ = 40 kHz. Velocity of bat, υ_s = 0.03 υ, where υ is velocity of sound. Apparent frequency of sound striking the wall

v' = v/(v - v_s) xx v = v/(v - 0.03v) xx 40 kHz

= `40/0.97 kHz`

This frequency is reflected by the wall and is received by the bat moving towards the wall

So `v_s = 0`

`v_L = 0.03 v`

`v" = (v+v_L)/v xx v'= (v + 0.03v)/v (40/0.97)`

`= 1.03/0.97 xx 40 kHz = 42.47 kHz`

 

`:. y = (1,1) = 7.5 sin (732.81^@)` 

`= 7.5 sin (90 xx 8 + 12.81^@) = 7.5 sin 12.81^@`

= 7.5 xx 0.2217

`= 1.6629 ~~ 1.663 cm`

The velocity of the oscillation at a given point and time is given as:

`v =d/(dt) y (x,t) = d/dt [7.5 sin(0.0050x + 12t + pi/4)]`

`= 7.5 xx 12 cos (0.0050x + 12t + pi/4)`

At x = 1 cm and t = 1 s.

`v = y(1,1)= 90 cos(0.0050x + 12t + pi/4)`

At x = 1 cm and t = 1 s

`v = y(1,1) = 90 cos (12.005 + pi/4)`

= 90 cos(732.81^@) = 90 cos(90xx8+ 12.81)

`= 90 cos(12.81^@)`

`= 90 xx 0.975 = 87.75 "cm/s"`

Now the equation of a propagating wave is given by

`y(x,t) = a sin(kx + wt + phi)`

Where

k = (2pi)/lambda

`:. lambda = (2pi)/k`

And `omega = 2piv`

`:. v = omega/(2pi)`

`Speed, v = vlambda = omega/k`