the frequency changes by 10 % as the source approaches a stationary observer with constant speed `v_(s)`. What would be the percentage change in frequency as the sources reaccedes the observer with the same speed ? Given , that `v_(s)lt lt v `(v= speed pf sound in air )

Options

A. `14.3%`
B. `20%`
C. `16.7%`
D. `10%`

Answer

D

Explanation

When the source approaches the observer
`f_1=f((v)/(v-v_s))=f(1-(v_s)/(v))^(-)=f(1+(v_s)/(v))`
or `((f_1-f)/(f))xx100=(v_s)/(v)xx100=10`
In the second case, when the source recedes from the observer
`f_2=f((v)/(v+v_s))=f(1+(v_s)/(v))^(-)=f(1-(v_s)/(v))`
`((f_2-f)/(f))xx100=-(v_s)/(v)xx100=-10`
In the first case, observed frequency increases by `10%` while in the second case, observed frequency decreases by `10%`.