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?

#### 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* = 2π^{2}*S*_{0}^{2}*v*^{2}*ρV**.**S*_{0} is the displacement amplitude.

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

On applying the respective values, we get:*S*_{0} = 1.2 × 10^{−6} m