A source of sound of frequency 1000 H_(Z) moves unifornly along a straight line with velocity 0.8 times velocity of sound . An observer is located at a distance l= 250 m from this line. Find (a) the frequency of the sound at instant when the source is closest to the observer. (b) the distance of the source when he observer no change in the frequency.
Answers
The frequency of the sound is 2777.7 HZ and the distance of source is 320 m.
Explanation:
(a) suppose the pulse which is emitted when the source is at S reaches the observer O in the same time in which the source reaches from S to S' , then
cosθ = SS' / SO = υst / υt = υs / υ = 0.8
Now, f = (υ / υ − υscosθ) f
= {υ / υ − (0.8υ) (0.8)} (1000)
= (1 / 1 − 0.64) (1000)
= 2777.7 HZ
(b) The observer will observer no change in the frequency when the source is at S as shown in figure. In the time when the wave the pulse reaches from S to O , the source will reach from S to s' .
t = SO / υ = SS' / υs
SS' = (υs / υ) SO
(0.8)(250) = 200 m
Therefore, distance of observer from source at this instant is
S'O = √ (SO)^2+(SS')^2
S'O = √ (250)^2 + (200)^2
S'O ≈320 m
Thus the frequency of the sound is 2777.7 HZ and the distance of source is 320 m.