Question

A sound wave of the form $s=s_m\cos (kx-\omega t+\varphi )$ travels at 343 m/s through air in a long horizontal tube. At one instant, air molecule A at 2.000 m is at its maximum positive displacement of 6.00 nm and air molecule B at x = 2.070 m is at a positive displacement of 2.00 nm. All the molecules between A and B are at intermediate displacements. What is the frequency of the wave?

Answer 1

The problem says "At one instant..." and we choose that instant (without loss of generality to be $t=0$.

Thus, the displacement of "air molecule A" at that instant is

$S_A=+S_m$

$=S_m\cos (k{x}_A-\omega t+\theta ){|}_{t=0}$

$=S_m\cos (k{x}_A+\theta )$
where

$x_A=2.00m$. Regarding "air molecule B" we have

$S_B=+\frac{1}{3}{S}_m$

$=S_m\cos (k{x}_B-\omega t+\theta ){|}_{t=0}=S_m\cos (k{x}_B+\theta )$

These statements lead to the following conditions:

$k{x}_A+\theta =0$

$k{x}_B+\theta ={\cos }^{-1}\left(1/3\right)=1.231$

where $x_B=2.07m$. Subtracting these equations leads to

$k(x_B-X_A)=1.231\Rightarrow k=17.6rad/m$.

Using the fact that $k=2\pi /lambda$ we find $\lambda =0.357m$. which means

$f=v/\lambda =343/0.357=960Hz$.

Another way to complete this problem (once $k$ is found ) is to use $kv=\omega$ and then the fact that $\omega =2\pi f$.