1.
To measure her speed, a skydiver carries a buzzer emitting a steady tone at 1 800 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 speed of sound is independent of altitude. While the skydiver is falling at
terminal speed, her friend on the ground receives waves of frequency 2 238 Hz. (Use 343 m/s as
the speed of sound.)
(a) What is the skydiver's speed of descent?
m/s
(b) Suppose the skydiver can hear the sound of the buzzer reflected from the ground. What
frequency does she receive?
Hz
Solution
a) From Doppler eqn we have
f' = f*(v)/(v - vS)
Here vS is the descent speed
So vS = v - f*v/f' = 343 – (1800*343)/2238= 67.13 m/s
b) Now the source frequency f is 2238Hz,the source speed vS is 0 and the listener speed vL is
67.13m/s
So f' (the frequency heard by the skydiver) is f' = f*(v + vL)/v = 2238*(343 + 67.13)/343 =
2676 Hz