A sound source emitting sound of frequency 450 Hz is approaching a stationary observer with velocity 30 m s- 1 and another identical source is going away from the observer with the same velocity. If the velocity of sound is 330 m s- 1 , then the difference of frequencies heard by observer is

Question

A sound source emitting sound of frequency 450 Hz is approaching a stationary observer with velocity 30 m s- 1 and another identical source is going away from the observer with the same velocity. If the velocity of sound is 330 m s- 1 , then the difference of frequencies heard by observer is (A) 103.5Hz (B) 82.5 Hz (C) 33.5 Hz (D) 92.5 Hz

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-npiqirjow2rk.png" /> f1=((v/v - vs))f0=(330/330 - 30)× 450 f1=(330/300)× 450=11× 45 f1=495 Hz f2=(330/360)× 450 f2=412.5 Hz Δ f=82.5 Hz

$103.5 Hz$

$82.5 Hz$

$33.5 Hz$

$92.5 Hz$

$f_{1} = (\frac{v}{v - v _{s}}) f_{0} = \frac{330}{330 - 30} \times 450$
$f_{1} = \frac{330}{300} \times 450 = 11 \times 45$
$f_{1} = 495 Hz$
$f_{2} = \frac{330}{360} \times 450$
$f_{2} = 412.5 Hz$
$Δ f = 82.5 Hz$

Q. A sound source emitting sound of frequency 450Hz is approaching a stationary observer with velocity 30ms−1 and another identical source is going away from the observer with the same velocity. If the velocity of sound is 330ms−1 , then the difference of frequencies heard by observer is

103.5Hz

82.5Hz

33.5Hz

92.5Hz

f1​=(v−vs​v​)f0​=330−30330​×450 f1​=300330​×450=11×45 f1​=495Hz f2​=360330​×450 f2​=412.5Hz Δf=82.5Hz

$103.5 Hz$

$82.5 Hz$

$33.5 Hz$

$92.5 Hz$

$f_{1} = (\frac{v}{v - v _{s}}) f_{0} = \frac{330}{330 - 30} \times 450$
$f_{1} = \frac{330}{300} \times 450 = 11 \times 45$
$f_{1} = 495 Hz$
$f_{2} = \frac{330}{360} \times 450$
$f_{2} = 412.5 Hz$
$Δ f = 82.5 Hz$