A point source of power 50π watts is producing sound waves of frequency 1875 Hz. The velocity of sound is 330 m/s, atmospheric pressure is 1.0 × 105 Pa, density of air is 40099πkgm-3. Then the displacement amplitude at r = 330 m from the point source is

I=P4πr2=2πρvf2A2 ⇒ A=I2πfrP2πρvHence A=12π×1875×330×(50π)2π×40099π×330 m = 1.13×10-6 m = 1.13 μm.

A point source of power 50π watts is producing sound waves of frequency 1875 Hz. The velocity of sound is 330 m/s, atmospheric pressure is 1.0 × 105 Pa, density of air is 40099πkgm-3. Then the displacement amplitude at r = 330 m from the point source is

I=P4πr2=2πρvf2A2 ⇒ A=I2πfrP2πρvHence A=12π×1875×330×(50π)2π×40099π×330 m = 1.13×10-6 m = 1.13 μm.

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A point source of power 50 $π$ watts is producing sound waves of frequency 1875 Hz. The velocity of sound is 330 m/s, atmospheric pressure is $1.0 \times 10^{5} Pa$ , density of air is $\frac{400}{99 π} {kgm}^{- 3}$ . Then the displacement amplitude at $r = \sqrt{330} m$ from the point source is

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$0.5 μm$

$0.2 μm$

1 $μm$

2 $μm$

answer is C.

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Detailed Solution

$I = \frac{P}{4 π r^{2}} = 2 π ρ v f^{2} A^{2} \Rightarrow A = \frac{I}{2 π f r} \sqrt{\frac{P}{2 π ρ v}}$

Hence $A = \frac{1}{2 π \times 1875 \times \sqrt{330}} \times \sqrt{\frac{(50 π)}{2 π \times (\frac{400}{99 π}) \times 330}} m = 1.13 \times 10^{- 6} m = 1.13 μ m$ .

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