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?

Answer 1

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.
$$\begin{gathered} \Rightarrow I=\frac{20}{4\mathrm{\pi }{r}^2}=\frac{20}{4\times \mathrm{\pi }\times 6^2}\left(\because r=6\mathrm{m}\right) \\ \Rightarrow I=44\mathrm{mw}/{\mathrm{m}}^2 \end{gathered}$$

(b) As we know,
$$\begin{gathered} I=\frac{p_0^2}{2\rho v}. \\ \Rightarrow P_0=\sqrt{I\times 2\rho v} \\ \Rightarrow P_0=\sqrt{2\times 1.2\times 340\times 44\times {10}^{-3}} \\ \Rightarrow P_0=6.0\mathrm{Pa}\mathrm{or}\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$
(c) As we know, I = 2π²S₀²v²ρV.
S₀ is the displacement amplitude.
$\Rightarrow S_0=\sqrt{\frac{I}{2{\mathrm{\pi }}^2{v}^2\rho V}}$
On applying the respective values, we get:
S₀ = 1.2 × 10⁻⁶ m