Answer:
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 [tex]v_s[/tex] respectively;
Now, frequency of the sound that is coming from the bat towards the wall due to DROPPLER EFFET will be;
f₁ = ( c / ( c - [tex]v_s[/tex] ) )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 + [tex]v_s[/tex] ) / c )f₁
from equ 1, f₁ = ( c / ( c - [tex]v_s[/tex] ) )f, so we substitute
f₂ = ( (c + [tex]v_s[/tex] ) / c ) × ( c / ( c - [tex]v_s[/tex] ) )f
f₂ = ( (c + [tex]v_s[/tex] ) / ( c - [tex]v_s[/tex] ) )f
∴ beat frequency will be;
δf = f₂ - f = ( (c + [tex]v_s[/tex] ) / ( c - [tex]v_s[/tex] ) )f - f
δf = ( 2[tex]v_s[/tex] / c - [tex]v_s[/tex] )f
δf = ( 2[tex]v_s[/tex] / c - [tex]v_s[/tex] )f
2f/δf = c - [tex]v_s[/tex] / [tex]v_s[/tex]
2f/δf = c/[tex]v_s[/tex] - [tex]v_s[/tex] / [tex]v_s[/tex]
2f/δf = c/[tex]v_s[/tex] - 1
c/[tex]v_s[/tex] = 2f/δf + 1
[tex]v_s[/tex] = c / (2f/δf + 1)
now, we substitute in our values;
[tex]v_s[/tex] = 340 / ((2×30000 / 900 ) + 1)
[tex]v_s[/tex] = 340 / (66.6666 + 1)
[tex]v_s[/tex] = 340 / 67.6666
[tex]v_s[/tex] = 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
