A bat moving at 5.6 m/s is chasing a flying
insect. The bat emits a 60 kHz chirp and
receives back an echo at 60.63 kHz.
At what speed is the bat gaining on its
prey? Take the speed of sound in air to be
338 m/s.
Answer in units of m/s
To find the speed at which the bat is gaining on its prey, we need to calculate the difference between the emitted frequency and the received frequency and then convert it to velocity.
Step 1: Calculate the frequency difference:
The difference in frequency (Δf) can be found using the equation:
Δf = f1 - f2
Where f1 is the emitted frequency (60 kHz) and f2 is the received frequency (60.63 kHz).
Δf = 60 kHz - 60.63 kHz
Δf = -0.63 kHz
Step 2: Convert the frequency difference to velocity:
The Doppler Effect equation relates the frequency difference to the velocity:
Δf/f0 = v/c
Where Δf is the frequency difference, f0 is the emitted frequency (60 kHz), v is the velocity of the bat relative to the prey, and c is the speed of sound in air (338 m/s).
Solving for v:
v = Δf * c / f0
Substituting the known values:
v = (-0.63 kHz) * (338 m/s) / (60 kHz)
Step 3: Convert the result to m/s:
The result obtained from the above calculation is in kHz * m/s units. To convert it to m/s, we need to multiply it by 1000.
v_in_mps = v * 1000
Now, let's calculate the value:
v = (-0.63 kHz) * (338 m/s) / (60 kHz) = -3.54 m/s
v_in_mps = -3.54 m/s * 1000 = -3540 m/s
The negative sign indicates that the bat is moving away from its prey at a relative speed of 3540 m/s.