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) Figure-6.89 shows the corresponding situation. Source gets closest to observer when it is at position B but at this instant observer receives those sound waves which source had emitted from an earlier position A shown in figure-6.89, so that when source reaches A to B, in the same time sound reaches A to O. Thus the apparent frequency heard by observer is
`n_(ap) = n_(0) [(v)/(v - v_(s)costheta)]`
`costheta = (AB)/(AO)`
If sound takes `t_(1)` time from A to O, we have
`AB = v_(s)t_(1) = 0.8t_(1)`
`AO = vt_(1)`
Now from equation -(6.274)
`cos theta = 0.8`
Now from equation (6.273)
`n_(ap) = n_(0) 0[(v)/(v - (0.8)^(2)v]] = n_(0) [(1)/(0.36)]` = `(1.8 xx 10^(3))/(0.36) = 5kHz`
(b) Observer receives a frequency equal to original frequency of source corresponding to the waves emitted by source from position B, as in this case velocity component of source is there in the direction ofobserver.
In this case the time taken by sound to reach observer from B is
`t_(0) = (l)/(v)`
In this direction distance travelled by source is
`S = v_(s)t = 0.8vt = 0.8l`
At the instant when observer receives this sound the separation between source and observer can be given as
`x = sqrt(l^(2) + (0.8l)^(2)) = sqrt(1.64) xx l`
= `sqrt(1.16) xx 250`
= 320 m