Question

If the form of a sound wave traveling through air is
$s(x,t)=\left(6.0nm\right)\cos (kx+\left(3000rad/s\right)t+\varphi )$
how much time does any given air molecule along the path take to move between displacements $s=+2.0nm$ and $s=-2.0nm$?

Answer 1

Without loss of generality we take $x=0$, and let $t=0$ be when

$s=0$.

This means the phase is $\varphi =-\pi /2$ and the function is $s=\left(6.0nm\right)\sin \left(\omega t\right)$ at $x=0$.

Noting that $\omega =3000rad/s$,

we note that at $t={\sin }^{-1}\left(1/3\right)/omega=0.1133ms$ the displacement is $s=+2.0nm$.

Doubling that time (so that wqe consider the excursion from $-2.0nm$ to $+2.0nm$)

we conclude that the time required is $2\left(0.1133ms\right)=0.23ms$