Consider both sources emitting sound. When is at R and approaches the detector with a speed 4 , the beat frequency measured by the detector is __________ Hz. [2023]

Question

$S_{1}$ and $S_{2}$ are two identical sound sources of frequency $656 Hz$ . The source $S_{1}$ is located at O and $S_{2}$ moves anticlockwise with a uniform speed $4 \sqrt{2} {ms}^{- 1}$ on a circular path around O, as shown in the figure. There are three points P, Q and R on this path such that P and R are diametrically opposite while Q is equidistant from them. A sound detector is placed at point P. The source $S_{1}$ can move direction OP.

[Given: The speed of sound in air is $324 {ms}^{- 1}$ ]

Q. Consider both sources emitting sound. When $S_{2}$ is at R and $S_{1}$ approaches the detector with a speed 4 ${ms}^{- 1}$ , the beat frequency measured by the detector is __________ Hz. [2023]

[Given: The speed of sound in air is $324 {ms}^{- 1}$ ]

Q. Consider both sources emitting sound. When $S_{2}$ is at R and $S_{1}$ approaches the detector with a speed 4 ${ms}^{- 1}$ , the beat frequency measured by the detector is __________ Hz. [2023]

Smart Achievers · Accepted Answer

(8.2)

$S_{2}$

$v = 324 m/s$

$Frequency heard due to movement of (S_{1})$

$f_{1} = (\frac{v}{v - u_{s}}) f_{0}$

$f_{1} = \frac{324}{320} \times 656$

$And frequency heard due to movement of (S_{2})$

$f_{2} = 656 Hz$

$∴$ $Beat frequency Δ f = f_{1} - f_{2}$

$= 656 (\frac{324}{320} - 1)$

$Δ f = 8.2$

Solution

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Solution

Ans. (8.2) f0=656 Hz v=324 m/s Frequency heard due to movement of (S1) f1=(vv-us)f0 f1=324320×656 And frequency heard due to movement of (S2) f2=656 Hz ∴ Beat frequency Δf=f1-f2 =656(324320-1) Δf=8.2

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