The following two waves are sent in opposite directions on a horizontal string so as to create a standing wave in a vertical plane:

$$\begin{gathered} {\mathit{y}}_{\mathbf{1}}\left(x,t\right)\mathbf{=}\left(6.00mm\right)\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\left(4.00\pi x-400\pi t\right) \\ {\mathit{y}}_{\mathbf{2}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mm}\mathbf{)}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\pi x}\mathbf{+}\mathbf{400}\mathbf{\pi t}\mathbf{)} \end{gathered}$$

within X meters andin seconds. An antinode is located at point A. In the time interval that point takes to move from maximum upward displacement to maximum downward displacement, how far does each wave move along the string?

Two waves that create a standing wave are given:

$$\begin{gathered} {\mathrm{y}}_1(\mathrm{x},\mathrm{t})=(6.00\mathrm{mm})\sin (4.00\mathrm{\pi x}-400\mathrm{\pi t}) \\ {\mathrm{y}}_2(\mathrm{x},\mathrm{t})=(6.00\mathrm{mm})\sin (4.00\mathrm{\pi x}+400\mathrm{\pi t}) \end{gathered}$$

We can find the time taken by the wave to move from maximum upward displacement to maximum downward displacement in terms of time period. Then, using the relation between T and angular speed we can write an expression for time in terms of angular speed. Also, we can write the velocity in terms of angular speed. Then, from the velocity and time, we can easily calculate the distance travelled by the wave along the string in the time interval that the point takes to move from maximum upward displacement to maximum downward displacement.

Formulae:

The time period of oscillation, $T=\frac{2\mathrm{\pi }}{\omega }...........\left(1\right)$

The velocity of the wave, $v=\frac{\omega }{k}...........\left(2\right)$

The displacement change of the wave, $∆x=vt......\left(3\right)$

Two waves that create the standing wave are

$$\begin{gathered} {\mathrm{y}}_1(\mathrm{x},\mathrm{t})=(6.00\mathrm{mm})\sin (4.00\mathrm{\pi x}-400\mathrm{\pi t}) \\ {\mathrm{y}}_2(\mathrm{x},\mathrm{t})=(6.00\mathrm{mm})\sin (4.00\mathrm{\pi x}+400\mathrm{\pi t}) \end{gathered}$$

Therefore, according to the superposition principle, the equation of the resultant wave is

$$\begin{gathered} y\text{'}=\left[2y_m\sin \left(kx\right)\cos \left(\omega t\right)\right] \\ =12\mathrm{mm}\sin \left(4.00\mathrm{\pi x}\right)\cos \left(400\mathrm{\pi t}\right).........\left(4\right) \end{gathered}$$

The time taken by the wave to move from maximum upward displacement to maximum downward displacement is T/2.

$$\begin{gathered} t=\frac{T}{2} \\ =\frac{2\pi }{2\omega }\left(\mathrm{from}\mathrm{equation}\left(1\right)\right) \\ =\frac{\pi }{\omega } \end{gathered}$$

Substituting the value of time and velocity from equation (2) in equation (3), we get the displacement as:

$$\begin{gathered} ∆x=\left(\frac{\omega }{k}\right)\left(\frac{\pi }{\omega }\right) \\ =\left(\frac{\pi }{k}\right) \\ =\left(\frac{\pi }{4.00\pi }\right)(\mathrm{from}\mathrm{equation}(4),\mathrm{we}\mathrm{get}\mathrm{k}=4.00\mathrm{\pi }) \end{gathered}$$

Therefore, the distance through which each wave moves along the string in the time interval that the point takes to move from maximum upward displacement to maximum downward displacement is 0.25 m