Answer
The displacement amplitude is $2.77\times 10^{-7}~m$
The displacement amplitude is about 92 times larger than the average distance between molecules in a room.
Work Step by Step
We can find the intensity of the sound:
$\beta = 10~log\frac{I}{I_0}$
$60.0 = 10~log\frac{I}{I_0}$
$6.0 = log\frac{I}{I_0}$
$10^{6.0} = \frac{I}{I_0}$
$I = (10^{6.0})~I_0$
$I = (10^{6.0})~(1.0\times 10^{-12}~W/m^2)$
$I = 1.0\times 10^{-6}~W/m^2$
We can use $343~m/s$ as the speed of sound in air.
We can use $\rho = 1.2~kg/m^3$ as the density of air.
We can find the displacement amplitude:
$s_0 = \sqrt{\frac{I}{2\pi^2 \rho f^2 v}}$
$s_0 = \sqrt{\frac{1.0\times 10^{-6}~W/m^2}{(2\pi^2)(1.2~kg/m^3)(40~Hz)^2(343~m/s)}}$
$s_0 = 2.77\times 10^{-7}~m$
The displacement amplitude is $2.77\times 10^{-7}~m$
We can compare the displacement amplitude to $3~nm$:
$\frac{2.77\times 10^{-7}~m}{3\times 10^{-9}~m} = 92$
The displacement amplitude is about 92 times larger than the average distance between molecules in a room.
