Question

A source of sonic oscillations with frequency `v_(0)=1700Hz` and a receiver are located at the same point. At the moment `t=0` the source starts receding from the receiver with constant acceleration `w=10.0m//s^(2)`. Assuing the velocity of sound to be equal to `v=340 m//s`, find the oscillation frequency registered by the stationary receiver `t=10.0s `after the start of motion.

Answer 1

If should be noted that the frequency emitted by the source at time `t` could not be received at the same moment by the receiver, because till that time source will cover the distance `(1)/(2) w^(2)` and sound wave will take the further time `(1)/(2) wt^(2) //v` to reach the receiver. Therefore the frequency noted by the receiver at time `t` should be emitted by the source at the time `t_(1) lt t`. Therefore
`t_(1)+((1)/(2) w t_(1)^(2)//v)=t ....(1)`
and the frequency noted by the receiver
`v=v_(0)(v)/( v+wt_(1)) ...(2)`
Solving Eqns `(1)` and `(2)`, we get
`v=(v_(0))/( sqrt(1+(2wt)/( v)))1.35kHz`.