A source of sound with natural frequency `f_0=1800Hz` moves uniformly along a straight line separated from a stationary observer by a distance `l=250m`. The velocity of the source is equal to `eta=0.80` fraction of the velocity of the sound.
Q. Find the frequency of osund received by the observer at the moment when the source gets closest to him.

Explanation

`(a)` When the observer receives the sound, the source is closest to him. It means, that frequency is emitted by the source sometimes before `( ` figure `)`. Figure shows that the source approaches the stationary observer with velocity `v_(s) cos theta`.
Hence the frequency noted by the observer
`v=v_(0)((v)/( v-v_(s) cos theta))`
`=v_(0)((v)/( v-etav cos theta))=(v_(0))/(1- eta cos theta ) (1)`
But `(x)/(v_(s))=(sqrt(l^(2)+x^(2)))/( v), `So,` (x)/(sqrt(l^(2)+x^(2)))=(v_(s))/(v)=eta`
or `cos theta = eta ....(2)`
Hence from Eqns. `(1)` and `(2)` the sought frequency
`v=(v_(0))/(1-eta^(2))=5 kHz`
`(b)` When the source is right in from of `O`, the sound emitted byy it will not be Doppler shifted because `theta=90^(@)`. This sound will be received at `O` at time `t=(l)/( v)` after the source has passed it. The source will by then have moved ahead by a distance `v_(s) t= l eta`. The distance between the source and the observer at this time will be `lsqrt(1+ eta^(2))=0.32km.`