Question

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 $\mathit{y}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{m}$?

Short answer

Any given point on the string takes 1.1 ms to move between displacements$y=+2.0\mathrm{mm}$ and$y=+2.0\mathrm{mm}$

Step-by-step solution

The given data

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$y=+2.0\mathrm{mm}$

Understanding the concept of wave equation

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

Calculation for the time taken by the point of string

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$1.1\mathrm{ms}$