To permit measurement of her speed a sky diver carries a buzzer emitting a steady tone at 1800 Hz. A friend on the ground at the landing site directly below listens to the amplified sound he receives. Assume the air is calm and the sound speed is 343`(m)/(s)` independent of altitude. While the sky diver is falling at terminal speed, his friend on the ground receives waves of frequency 2150 Hz. (a) What is the sky diver's speed of descent?
(b) Suppose the sky diver can hear the sound of the buzzer reflected from the ground. What frequency does she receive?

Explanation

Sky divers typically reach a terminal speed of about 150mi/h, so this sky diver should also fall near this rate. Since her friend receives a higher frequency as a result of the Doppler shift, the sky diver should detect a frequency with twice the Doppler shift, at approximately
`f'=1800Hz+2(2 150Hz-1800Hz)=2500Hz`
We can use the equation for the Doppler effect to answer both (a) and (b) Let `f_e=1800Hz` represent the emitted frequency, `v_e` be the speed of the sky diver and `f_g=2150Hz` be the frequency of the wave crests reaching the groung.
(a). The sky diver source is moving towards the stationary ground, so we rearrange the equation as follows:
`f_g=f_e((v)/(v-v_0))`
`impliesv_e=v(1-(f_e)/(f_g))=(343(m)/(s))(1-(1800Hz)/(2150Hz))=55.8(m)/(s)`
(b). The ground now becomes a stationary source, reflecting crests with the 2150 Hz frequency at which they reach the ground, and sending them to a moving observer, who receives them at the rate
`f_(e2)=f_g((v+v_e)/(v))=(2150Hz)((343(m)/(s)+55.8(m)/(s))/(343(m)/(s)))=2500Hz`
The answers appear to be consistent with our predictions, although the sky diver is falling somewhat slower than expected The doppler effect can be used to find the speed of many different types of moving objects, like raindrops (with Doppler radar) and cars (with police radar).