Question

A bat emits a sound at a frequency of 30.0 kHz as it approaches a wall. The bat detects beats such that the frequency of the echo is 900 Hz higher than the frequency the bat is emitting. The speed of sound in air is 340 m/s at emits a sound at a frequency of 30.0 kHz

(a) What is the speed of the bat?
(b) What is the wavelength of the sound that the bat hears15?

Answer 1

a) the speed of the bat is 5.02 m/s

b) the wavelength of the sound that the bat hears is 0.011 m

Step-by-step explanation:

Given the data in the question;

Frequency of sound emitted by a bat f = 30.0 kHz = 30000 Hz

detected frequency by the bat δf = 900 Hz

speed of sound in air c = 340 m/s

Let speed of sound and speed of bat be c and
`v_s` respectively;

Now, frequency of the sound that is coming from the bat towards the wall due to DROPPLER EFFET will be;

f₁ = ( c / ( c -
`v_s` ) )f ----- let this be equ 1

Also, frequency does not change after deflection. The bat becomes an observer as the dropper is shifted because the reflected sound wave is coming towards it;

Hence, Doppler shifted frequency will be;

f₂ = ( (c +
`v_s` ) / c )f₁

from equ 1, f₁ = ( c / ( c -
`v_s` ) )f, so we substitute

f₂ = ( (c +
`v_s` ) / c ) × ( c / ( c -
`v_s` ) )f

f₂ = ( (c +
`v_s` ) / ( c -
`v_s` ) )f

∴ beat frequency will be;

δf = f₂ - f = ( (c +
`v_s` ) / ( c -
`v_s` ) )f - f

δf = ( 2
`v_s` / c -
`v_s` )f

δf = ( 2
`v_s` / c -
`v_s` )f

2f/δf = c -
`v_s` /
`v_s`

2f/δf = c/
`v_s` -
`v_s` /
`v_s`

2f/δf = c/
`v_s` - 1

c/
`v_s` = 2f/δf + 1

`v_s` = c / (2f/δf + 1)

now, we substitute in our values;

`v_s` = 340 / ((2×30000 / 900 ) + 1)

`v_s` = 340 / (66.6666 + 1)

`v_s` = 340 / 67.6666

`v_s` = 5.02 m/s

Therefore, the speed of the bat is 5.02 m/s

b) the wavelength of the sound that the bat hears

frequency of reflected wave is;

f₂ = f + δf = 30000 + 900 = 30900 Hz

λ₂ = c / f₂

we substitute

λ₂ = 340 / 30900

λ₂ = 0.011 m

Therefore, the wavelength of the sound that the bat hears is 0.011 m