A source of sound having frequency 300 Hz and a receiver are located along the same line normal to the wall as shown in the figure. Both the source and the receiver are stationary and the wall recedes from the source with a velocity of 20 m s- 1 . If the beat frequency registered by the receiver is (240/x) Hz , then what is the value of x ? [velocity of sound = 330 m s- 1 ]

Question

A source of sound having frequency 300 Hz and a receiver are located along the same line normal to the wall as shown in the figure. Both the source and the receiver are stationary and the wall recedes from the source with a velocity of 20 m s- 1 . If the beat frequency registered by the receiver is (240/x) Hz , then what is the value of x ? [velocity of sound = 330 m s- 1 ] <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-fvv2z8wkzhv3.png" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

As ; f1=f (for direct sound) Now for reflected sound f2=((V - 20/V + 20))f If b is the beat frequency, <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-fvv2z8wk8zch.png" /> ∴ b=f1-f2 ∴ f-((V - 20/V + 20))f=(f × 40/V + 20) =(300 × 40/350)=(240/7)Hz

As ; f1​=f (for direct sound) Now for reflected sound f2​=(V+20V−20​)f If b is the beat frequency, ∴b=f1​−f2​ ∴f−(V+20V−20​)f=V+20f×40​ =350300×40​=7240​Hz

As ; $f_{1} = f$ (for direct sound)
Now for reflected sound $f_{2} = (\frac{V - 20}{V + 20}) f$
If $b$ is the beat frequency,

$∴ b = f_{1} - f_{2}$
$∴ f - (\frac{V - 20}{V + 20}) f = \frac{f \times 40}{V + 20}$
$= \frac{300 \times 40}{350} = \frac{240}{7} Hz$