91Ó°ÊÓ

The scale on the vertical axis is set as, $u_s=4.0m/s$

The general expression of the wave, $y=y_m\sin \left(kx-\mathrm{Ó\neg }t+\mathrm{Ï•}\right)$ (i)

The displacement of the particle of the wave, perpendicular to the direction of motion, changes continuously. Hence, the slope of the wave also changes with time and position. Using the value of the slope at various points, we can determine the phase angle of the wave at different points.

Using equation (i), we get the slope of the wave as given:

$$\begin{gathered} slope=\frac{dy}{dt} \\ =\mathrm{Ó\neg }{y}_m\cos \left(kx-\mathrm{Ó\neg }t+\mathrm{Ï•}\right) \end{gathered}$$

at x=0 the slope gives the transverse velocity$u_s$

$$\begin{gathered} slope=u_s \\ =\mathrm{Ó\neg }{y}_m\cos \left(k\left(0\right)-\mathrm{Ó\neg }t+\mathrm{Ï•}\right) \\ =\mathrm{Ó\neg }{y}_m\cos \mathrm{Ï•} \end{gathered}$$

From the figure, we can see that at x=0, and t = 0, the transverse velocity is given as-

role="math" localid="1660973415652" $$\begin{gathered} u_s=4.0m/s \\ slope=u_s \\ =\mathrm{Ó\neg }{y}_m\cos \mathrm{Ï•} \\ =-4.0m/s............\left(1\right) \end{gathered}$$

The maximum value of transverse velocity will be-

$\mathrm{Ó\neg }y=5.0m/s\left(âˆ\mu \cos \mathrm{Ï•}=1\right)$

Using this value in equation (1), we get the cosine angle as:

$$\begin{gathered} 5.0\cos \mathrm{Ï•}=-4.0 \\ \cos \mathrm{Ï•}=\frac{-4.0}{5.0} \\ =-0.8 \end{gathered}$$

Since the value of cosine is negative, the angle should lie in quadrant III or IV.

Hence we get,

$$\begin{gathered} \mathrm{ϕ}={\cos }^{-1}\left(-0.8\right) \\ =-0.6435rad \\ ≈-37° \end{gathered}$$

As the cosine value repeats after$2n\mathrm{Ï€}$interval, the valid answer for the angle can also be given as:$\mathrm{Ï•}=-0.6435Â\pm 2n\mathrm{Ï€}$

Hence, the value of phase is, $\mathrm{Ï•}=-0.6435Â\pm 2n\mathrm{Ï€}$.