Two engines cross each other in opposite direction. One engine produce 540 Hz sound. People sitting in the second engine will hear sound of which frequency before and after passing from one another? The velocity of both the engines is 40 m/s. Velocity of sound is 340 m/s.

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Solution

Given, $v_{s} = v_{0} = 40 m s^{- 1}; v = 340 m s^{- 1}$

$n = 540 H z; n "= ?$

Before they cross each other

Ref. image 1
Apparent frequency

$n "= \frac{v + v_{0}}{v - v_{s}} = \frac{(340 + 40)}{340 - 40)} \times 540$

$= \frac{380}{300} \times 540 = 38 \times 18$

Or $n " 684 H z$

After crossing,
Apparent frequency,

Ref. image 2

$n "= \frac{v - v_{0}}{v + v_{s}} . n$

$= (\frac{340 - 40}{340 + 40}) \times 540$

$= \frac{300}{380} \times 540$

or $n "= 426.3 H z$

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Two engines cross each other in opposite direction. One engine produce 540 Hz sound. People sitting in the second engine will hear sound of which frequency before and after passing from one another? The velocity of both the engines is 40 m/s. Velocity of sound is 340 m/s.

Given, vs=v0=40ms−1; v=340ms−1n=540Hz ; n"=?Before they cross each otherRef. image 1Apparent frequencyn"=v+v0v−vs=(340+40)340−40)×540=380300×540=38×18Or n"684HzAfter crossing,Apparent frequency,Ref. image 2n"=v−v0v+vs.n=(340−40340+40)×540=300380×540or n"=426.3Hz