Question

At what speed should a source of sound move away from a stationary observer so that observer finds the Apparent frequency equal to the half of the orginal frequency​

Answer 1

Answer:

v' = \frac{c}{c+v_{s}}v_{o} \Rightarrow \frac{v_{o}}{2} = \frac{c}{c+v_{s}}v_{o} \Rightarrow v_{s} = v

Frequency when observer is stationary and source is moving away from observer -

\nu {}'= \nu _{0}.\frac{C}{C+V_{s}}

- wherein

C= speed of sound

V_{s}= speed of source

\nu _{0}= original frequency

\nu {}'= apparent frequency

SO THAT UR ANSWER IS =V

Explanation:

HOPE IT IS WELL CLEARED TO UH