Question

Two hawks fly toward one another. The first flies at 15 m/s and the other flies at 20 m/s. They screech at each other; the first emits a frequency of 3200 Hz and the other emits a frequency of 3800 Hz. What frequencies do they each receive if the speed of sound is 330 m/s that day

Answer 1

Answer:When two objects are moving towards each other, the apparent frequency of sound waves they emit towards each other increases, while the wavelength decreases. This is due to the Doppler effect.

The formula for the Doppler effect is:

f' = f(v +/- vr)/(v +/- vs)

where:

f' = the observed frequency

f = the emitted frequency

v = the speed of sound

vr = the relative velocity between the two objects

vs = the velocity of the source (i.e., the hawk emitting the sound)

In this case, the relative velocity between the two hawks is:

vr = (15 m/s + 20 m/s) = 35 m/s

For the first hawk emitting a frequency of 3200 Hz, the observed frequency received by the other hawk is:

f' = 3200 Hz * (330 m/s + 35 m/s)/(330 m/s - 20 m/s) = 4073 Hz

For the second hawk emitting a frequency of 3800 Hz, the observed frequency received by the first hawk is:

f' = 3800 Hz * (330 m/s + 35 m/s)/(330 m/s + 15 m/s) = 4139 Hz

Therefore, the first hawk receives a frequency of 4073 Hz, while the second hawk receives a frequency of 4139 Hz.

Step-by-step explanation: