Answer
The displacement amplitude is $~~0.260~nm$
Work Step by Step
We can find the displacement amplitude $s_m$:
$p_m = s_m v \rho \omega$
$s_m = \frac{p_m}{v \rho \omega}$
$s_m = \frac{p_m}{v \rho 2\pi f}$
$s_m = \frac{1.13\times 10^{-3}~Pa}{(343~m/s)(1.21~kg/m^3) (2\pi)(1665~Hz)}$
$s_m = 2.60\times 10^{-10}~m$
$s_m = 0.260~nm$
The displacement amplitude is $~~0.260~nm$
