An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency?
A
zero
B
`0.5%`
C
`5%`
D
`20%`
Text Solution
Verified by Experts
The correct Answer is:
D
Given `upsilon_(0)=(upsilon)/5rArr upsilon_(0)rArr upsilon_(0)d=320/5 =64 m//s` When observer move towards the stationary source, then `n'=((upsilon+upsilon_(0))/(upsilon))n` `n'=((320+64)/320) n` `n'=(384/320)n` `(n')/n=384/320` Hence, percentage increases, `((n'-n)/n)=((384-320)/320xx100) %` `=(64/320xx100)%` `=20 %`
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