Question

When an engine passes near to a stationary observer then its apparent frequencies occurs in the ratio 53. The velocity of engine is (velocity of sound is 340 m/s)?

Answer 1

n1=(v/v-a)n

n2=(v/v+a)n

n1/n2=(v+a)/(v-a)

5/3=(340+a)/(340-a)

5(340-a)=3(340+a)

(5-3)340=3a+5a

2×340=8a

a= (2×340)/8

a=85m/s

Answer 2

Dear Student,

◆ Answer -

v = 85 m/s

● Explanation -

Let v be velocity of the engine. Given that ratio of apparent frequencies is 5/3.

f1/f2 = (340+v)/(340-v)

5/3 = (340+v)/(340-v)

5 (340-v) = 3 (340+v)

1700 - 5v = 1020 + 3v

5v + 3v = 1700 - 1020

8v = 680

v = 85 m/s

Hence, velocity of engine is 85 m/s.

Thanks dear. Hope this helps you...