Question: If the form of a sound wave traveling through air is$\mathbf{s}\left(x,t\right)\mathbf{=}\left(6.0nm\right)\mathbf{\backslash }\mathit{c}\mathit{o}\mathit{s}\left(\mathrm{kx}+\left(3000\frac{rad}{s}\right)t+f\right)$.How much time does any given air molecule along the path take to move between displacement s = + 2.0 nmand s = - 2.0 nm?

• The form of sound wave,$s\left(x,t\right)=\left(6.0\text{nm}\right)\cos \left(kx+3000\frac{\text{rad}}{\text{s}}t+\phi \right)$.
• The molecule move between s = + 2.0 nmand s = - 2.0 nm.

By using the values of displacements in the given form of the sound wave, two equations can be found. By subtracting and solving them, find the time taken by the air molecule to move between displacements s = + 2.0 nmand s = - 2.0 nm..

$1.90=kx+3000\frac{\text{rad}}{\text{s}}{t}_2+\phi$The form of the sound wave,$s\left(x,t\right)=\left(6.0\text{nm}\right)\cos \left(kx+3000\frac{\text{rad}}{\text{s}}t+\phi \right)$.

At$s=+2.0\text{nm,}$

$$\begin{gathered} +2.0\text{nm}=\left(6.0\text{nm}\right)\cos \left(kx+3000\frac{\text{rad}}{\text{s}}{t}_1+\phi \right) \\ \frac{\text{2.0nm}}{\text{6.0nm}}=\cos \left(kx+3000\frac{\text{rad}}{\text{s}}{t}_1+\phi \right) \\ \left(\frac{2.0}{6.0}\right)=kx+3000\frac{\text{rad}}{\text{s}}{t}_1+\phi \\ \left(\frac{1}{3}\right)=kx+3000\frac{\text{rad}}{\text{s}}{t}_1+\phi \\ \left(0.33\right)=kx+3000\frac{\text{rad}}{\text{s}}{t}_1+\phi \end{gathered}$$

$1.23=kx+3000\frac{\text{rad}}{\text{s}}{t}_1+\phi$……. (i)

At,$s=-2.0\text{nm,}$

role="math" localid="1661754951618" $$\begin{gathered} -2.0\text{nm}=\left(6.0\text{nm}\right)\cos \left(kx+3000\frac{\text{rad}}{\text{s}}{t}_2+\phi \right) \\ \frac{\text{- 2.0nm}}{\text{6.0nm}}=\cos \left(kx+3000\frac{\text{rad}}{\text{s}}{t}_2+\phi \right) \\ \left(\frac{-2.0}{6.0}\right)=kx+3000\frac{\text{rad}}{\text{s}}{t}_2+\phi \\ \left(-\frac{1}{3}\right)=kx+3000\frac{\text{rad}}{\text{s}}{t}_2+\phi \\ \left(-0.33\right)=kx+3000\frac{\text{rad}}{\text{s}}{t}_2+\phi \end{gathered}$$

$1.90=kx+3000\frac{\text{rad}}{\text{s}}{t}_2+\phi$……. (ii)

Subtracting equation 1 from 2,

$$\begin{gathered} 1.90-1.23=3000\frac{\text{rad}}{\text{s}}{t}_2-3000\frac{\text{rad}}{\text{s}}{t}_1 \\ 0.67=3000(t_2-t_1) \\ {t}_2-t_1=\frac{0.67}{3000} \\ {t}_2-t_1=0.00022 \end{gathered}$$

$\therefore t_2-t_1=0.23\text{ms}$

Therefore the time taken by the air molecule to move between displacements to s = + 2.0 nm and s = - 2.0 nm is 0.23 ms .