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 ?
Given, the frequency of sound emitting from a bat is 40 kHz . The bat is moving at 0.03 times the speed of sound in air.
The apparent frequency of sound is given by the equation,
ν ′ = ( v v − v s ) ν
Substituting the values in the above equation, we get:
ν ′ = ( v v − 0.03 v ) 40 kHz = 41 .24 kHz
This frequency strikes the wall and returns after being reflected by the wall.
The frequency of the sound wave which is received is given by the equation,
ν ″ = ( ( v + v l ) v ) ν ′
Substituting the values in the above equation, we get:
ν ″ = ( ( v + 0.03 v ) v ) 41 .24 kHz = 42 .47 kHz
Thus, the frequency that the bat hears is 42.47 kHz .
Given, the frequency of sound emitting from a bat is
The apparent frequency of sound is given by the equation,
Substituting the values in the above equation, we get:
This frequency strikes the wall and returns after being reflected by the wall.
The frequency of the sound wave which is received is given by the equation,
Substituting the values in the above equation, we get:
Thus, the frequency that the bat hears is
