If a wave$\mathit{y}\left(x,t\right)\mathbf{=}\left(6.0mm\right)\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left(kx+\left(600rad/s\right)t\right)$travels along a string, how much time does any given point on the string take to move between displacements$\mathit{y}\mathbf{=}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{m}$and $\mathit{y}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{m}$?

1. The given equation of the wave,$y\left(x,t\right)=\left(6.0\mathrm{mm}\right)\sin \left(\mathrm{lx}+\left(600\mathrm{rad}/\mathrm{s}\right)\mathrm{t}+\mathrm{\varphi }\right)$
2. Displacements of the string is between, $y=+2.0\mathrm{mm}$and$y=+2.0\mathrm{mm}$

Here, we can apply the standard wave equation for the two given displacements. By rearranging it and subtracting we get the required time.

The given wave function is

$y\left(x,t\right)=\left(6.0\mathrm{mm}\right)\sin \left(\mathrm{lx}+\left(600\mathrm{rad}/\mathrm{s}\right)\mathrm{t}+\mathrm{\varphi }\right)$

Suppose, the point is at y = +2.0 mm at time$t=t_1$. So,

$$\begin{gathered} +2.0\mathrm{mm}=\left(6.0\mathrm{mm}\right)\sin \left(\mathrm{lx}+\left(600\mathrm{rad}/\mathrm{s}\right)\mathrm{t}+\mathrm{\varphi }\right) \\ kx+\left(600\mathrm{rad}/\mathrm{s}\right){\mathrm{t}}_1+\mathrm{\varphi }=\left(\frac{2.0\mathrm{mm}}{6.0\mathrm{mm}}\right)........................................\left(1\right) \end{gathered}$$

In a similar way, let us assume the point is at y =-2.0 mm at time . So,

role="math" localid="1657281223675" $$\begin{gathered} -2.0\mathrm{mm}=\left(6.0\mathrm{mm}\right)sin\left(kx+\left(600\mathrm{rad}/\mathrm{s}\right){\mathrm{t}}_2+\mathrm{\varphi }\right) \\ kx+\left(600\mathrm{rad}/\mathrm{s}\right)t_{\mathit{2}}+\mathrm{\varphi }=\left(\frac{-2.0\mathrm{mm}}{6.0\mathrm{mm}}\right)........................................\left(2\right) \end{gathered}$$

Now, by subtracting equation (1) from equation (2), we get

$$\begin{gathered} 600\times \left(t_1-t_2\right)=0.67614 \\ \left(t_1-t_2\right)=1.1269\times {10}^{-3}\mathrm{s} \\ \approx 1.1\mathrm{ms} \end{gathered}$$

Hence, the value of required time is$1.1\mathrm{ms}$