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?

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

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 respectively;

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

f₁ = ( c / ( c - ) )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 + ) / c )f₁

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

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

f₂ = ( (c + ) / ( c - ) )f

∴ beat frequency will be;

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

δf = ( 2 / c - )f

δf = ( 2 / c - )f

2f/δf = c - /

2f/δf = c/ - /

2f/δf = c/ - 1

c/ = 2f/δf + 1

=  c / (2f/δf + 1)

now, we substitute in our values;

=  340 / ((2×30000 / 900 ) + 1)

=  340 / (66.6666 + 1)

=  340 / 67.6666

=  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

answer; how far out of place which the object finished relative to the starting location;