Question

A sound wave traveling in \(20\;{\rm{^\circ C}}\) air has pressure amplitude of\(0.5\;{\rm{Pa}}\). What is the intensity of the wave?

Short answer

The intensity is \(2.83 \times {10^{ - 4}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\).

Step-by-step solution

Given Data

The pressure amplitude is \(P = 0.50\;{\rm{Pa}}\).

The temperature is \(T = 20 + 273\;{\rm{K}} = 293\;{\rm{K}}\).

Idea of Sound intensity

The sound intensity depends on the pressure amplitude, the density of the medium, and the sound wave’s speed. The greater the pressure, the intensity also becomes more remarkable.

Calculation of the intensity

The speed of sound is given by,

\(v = 331\sqrt {\frac{T}{{273}}} \)

Substituting the values we get,

\(\begin{align}v &= 331\sqrt {\frac{{293}}{{273}}} \\ &= 343\;{\rm{m/s}}\end{align}\)

The intensity is given by,

\(I = \frac{{{P^2}}}{{2\rho v}}\)

Here, \(\rho = 1.29\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\) is the density of air.

So, the intensity is,

\(\begin{align}I &= \frac{{{{\left( {0.5} \right)}^2}}}{{2 \times 1.29 \times 343}}\\ &= 2.83 \times {10^{ - 4}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\end{align}\)

Therefore the intensity of the wave is \(2.83 \times {10^{ - 4}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\)