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Question
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 P = 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 A 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
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