The faintest sound that the human ear can detect at frequency $1\ \mathrm{kHz}$ corresponds to an intensity…
The faintest sound that the human ear can detect at frequency $1\ \mathrm{kHz}$ corresponds to an intensity of about $10^{-12}\ \mathrm{Wm^{-2}}$. Determine the pressure amplitude and the maximum displacement associated with this sound, assuming the density of the air = $1.3\ \mathrm{kg\ m^{-3}}$ and velocity of sound in air = $332\ \mathrm{ms^{-1}}$.
Solution
Sol. Intensity of sound wave, I = $p^2/2\rho v$
$\Rightarrow p = \sqrt{I\times 2\rho v} = \sqrt{10^{-12}\times 2\times 1.3\times 332}$
$= 2.94\times 10^{-5}\ \mathrm{Nm^{-2}}$
Now,
$p = \rho v \omega A$
$\Rightarrow A = \dfrac{p}{\rho v \omega} = \dfrac{2.94\times 10^{-5}}{1.3\times 332\times 2\pi\times 10^3}$
$= 1.1\times 10^{-11}\ \mathrm{m}$
Answer: A = $1.1\times 10^{-11}$ m