Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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A Sources of Sound Operates at 2.0 Khz, 20 W Emitting Sound Uniformly in All Directions. the Speed of Sound in Air is 340 M S−1 and the Density of Air is 1.2 Kg M −3. - Physics

Sum

A sources of sound operates at 2.0 kHz, 20 W emitting sound uniformly in all directions. The speed of sound in air is 340 m s−1 and the density of air is 1.2 kg m −3. (a) What is the intensity at a distance of 6.0 m from the source? (b) What will be the pressure amplitude at this point? (c) What will be the displacement amplitude at this point?

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Solution

Given:
Velocity of sound in air v = 340 ms−1
Power of the source = 20 W
Frequency of the source f = 2,000 Hz
Density of air ρ = 1.2 kgm −3

(a) Distance of the source r = 6.0 m
Intensity is given by:

\[I = \frac{P}{A}\]

where is the area.

\[\Rightarrow I = \frac{20}{4\pi r^2} = \frac{20}{4 \times \pi \times 6^2}      \left( \because r = 6  m \right)\] 

\[ \Rightarrow I = 44  \text { mw/ m }^2\]

(b) As we know,

\[I = \frac{p_0^2}{2\rho v} . \] 

\[ \Rightarrow    P_0  = \sqrt{I \times 2\rho v}\] 

\[ \Rightarrow  P_0  = \sqrt{2 \times 1 . 2 \times 340 \times 44 \times {10}^{- 3}}\] 

\[ \Rightarrow  P_0  = 6 . 0  \text { Pa } \text { or } \text { N/ m }^2\]

(c) As we know, I = 2π2S02v2ρV.
S0 is the displacement amplitude.

\[\Rightarrow    S_0  = \sqrt{\frac{I}{2 \pi^2 v^2 \rho V}}\]

On applying the respective values, we get:
S0 = 1.2 × 10−6 m

Concept: Wave Motion
  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 16 Sound Waves
Q 17 | Page 353
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