A bat watcher detects the sound of a hunting greater horseshoe bat at 76 kHz . She knows
that this species produces echolocation calls at 80 kHz . What is the speed of this bat with respect to the batwatcher? Is the bat receding or approaching?
Fr = ((Vs-Vr)/(Vs+Vb))*Fb
Fr = ((343-0)/(343+Vb))*80 = 76 kHz
(343/(343+Vb))*80 = 76
27,440/(343+Vb) = 76
26,068+76Vb = 27,440
76Vb = 27440 - 26,068 = 1372
Vb = 18 m/s. = Velocity of the bat.
The bat is receding, because the freq.
heard is lower than the freq. emitted by the bat.
To solve this question, we need to apply the Doppler effect formula. The Doppler effect is a phenomenon that explains the change in frequency of a wave when there is relative motion between the source of the wave and the observer.
The formula for the Doppler effect is as follows:
f' = f × (v + vo) / (v + vs)
Where:
f' is the perceived frequency
f is the actual frequency emitted by the source
v is the speed of sound in air
vo is the velocity of the observer (bat-watcher) relative to the medium (air)
vs is the velocity of the source (bat) relative to the medium (air)
In this case, we know that the actual frequency emitted by the bat is 80 kHz (f = 80 kHz) and the perceived frequency detected by the bat-watcher is 76 kHz (f' = 76 kHz).
We need to find the speed of the bat (vs) with respect to the bat-watcher (observer). Since the observer is stationary, the velocity of the observer (vo) is zero.
Using the formula, we can rewrite it as:
f' = f × (v + 0) / (v + vs)
76 kHz = 80 kHz × v / (v + vs)
Now we need to solve for vs, the velocity of the source (bat) relative to the medium (air).
Rearranging the equation, we have:
76 kHz (v + vs) = 80 kHz × v
76 kHz v + 76 kHz vs = 80 kHz × v
76 kHz vs = 80 kHz × v - 76 kHz v
76 kHz vs = 4 kHz × v
vs = 4 kHz × v / 76 kHz
Now, we know the actual frequency of the bat (80 kHz) and the perceived frequency (76 kHz). We need to substitute the speed of sound in air (v), which is approximately 343 m/s. Let's convert the frequencies to Hz:
f' = 76 kHz = 76,000 Hz
f = 80 kHz = 80,000 Hz
Now we can calculate the speed of the bat (vs):
vs = 4 kHz × 343 m/s / 76 kHz
vs = 0.0181 m/s (rounded to 4 decimal places)
Therefore, the speed of the bat with respect to the bat-watcher is approximately 0.0181 m/s.
To determine if the bat is receding or approaching, we need to know the sign of the velocity. In this case, since the detected frequency (76 kHz) is lower than the emitted frequency (80 kHz), the bat is receding from the bat-watcher.