Answer
$s_m=3.7\times10^{-8}m$
Work Step by Step
Amplitude of the air oscillations $'s_m'$ can be determined from the following equation:
$s_m^2$=$\frac{2I}{\rho{(\omega)^2}}$
where $'I'$ is the intensity of sound, $\rho=1.21\frac{kg}{m^3}$ is the density of air and $v=343\frac{m}{s}$ is the speed of the air.
First of all, let us find angular frequency
$\omega=2\pi$$f=2(3.14)(3000)=1.88\times10^3$
Now putting the values in the very first equation, we get
$s_m^2$=$\frac{2(1.0\times10^{-6})}{1.21\times343{(1.88\times10^3)^2}}$
$s_m^2=1.36\times10^{-15}$
taking square root on both sides, we get
$s_m=3.7\times10^{-8}m$
