A stationary source sends forth monochromatic sound. A wall approaches it with velocity 33 cm/s. The propagation velocity of sound in the medium is `c=330(m)/(s)`. How much, in per cent , does the wavelength of sound change on reflection from the wall?

Explanation

Frequency of sound when it reached the wall
`v^(')=v(v+u)/(v)`
wall will reflect the sound with same frequency `v^(')`. Thus frequency noticed by a stationary observer after reflection from wall.
`v^('') =v^(') (v)/( v-u)` , since wall behaves as a source of frequency `v^(')`
Thus, `v^('') =v( v+u)/( v). (v)/(v-u)=v(v+u)/(v-u)`
or,` lambda^('') =lambda(v-u)/( v+u)` or `(lambda^(''))/( lambda)=(v-u)/( v+u)`
So, `1-(lambda^(''))/( lambda)=1-(v-u)/( v+u) = ( 2u )/( v+u)`
Hence the sought percentage change in wavelength
`=(lambda-lambda^(''))/( lambda)xx100=(2u)/(v+u) xx100%=0.2 % decrease. `