In: Physics
A bird flying directly toward a stationary bird-watcher emits a cry at a frequency of 1,030Hz. The bird-watcher hears a frequency of 1120 Hz. The speed of sound on this day is 341m/s. a,What is the speed of the bird? b, Now there are two birds emitting a cry at 1,030Hz. Both are flying east, one on each side of the birdwatcher. One is flying at 23m/s while the other is flying at 33m/s. Is there anything the birdwatcher can do to hear the same frequency from the two birds at once (answer “yes” or “no” in the space below)? If so, describe what he can do; if not, explain why this is not possible. Include formulas if it is appropriate to do so.
The bird is flyting directly towards the stationary watcher and emitting 1030 Hz. The higher observed frequency 1120 Hz can be explained by Doppler effect.
vsource = 27.4 m/s
(b)
In another situation one bird in moving awar from watcher (Lets call it bird A) and another bird is coming towards him (Lets call it bird B). Both emit sound of same frequency. Due to Doppler effect, the observed frequency by watcher from A will be lower than 1030 Hz, and that from B will be higher than 1030 Hz.
If the watcher moves towards A, the observed frequncy from A starts to increase while that from B starts to decrease (as he moves away from it) from earlier observed frequencies. There will be a velocity of the watcher for which the observed frequencies from A and b will be same.
Although it is not clear in part (b) which bird is ahead. Lets assume A (speed = 23m/s) is ahead and B (speed = 33m/s) is behind. And watcher is between them. Watcher moves towards A with speed vo.
Formula of Doppler effect for both source and observer moving
Observed frequency from A
Observed frequency from B
(+ and - signs have to be put according to the dirction of motion of source and observer.
As per the question
fAO = fBO
fA = fB = 1030 Hz
vo = 28.41 m/s