Question

Figure 16-32 shows the transverse velocity u versus time t of the point on a string at x = 0 , as a wave passes through it. The scale on the vertical axis is set by ${\mathit{u}}_{\mathbf{s}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathit{m}\mathbf{/}\mathit{s}$ . The wave has the form $\mathit{y}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathit{y}}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{k}\mathbf{x}\mathbf{-}\mathbf{\omega }\mathbf{t}\mathbf{+}\mathbf{\varphi }\mathbf{)}$ . What then is $\mathit{\varphi }$ ? (Caution:A calculator does not always give the proper inverse trig function, so check your answer by substituting it and an assumed value of $\mathit{\omega }$ into $\mathit{y}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}$and then plotting the function.)

Short answer

The phase angle is, $\varphi =-0.6435\pm 2n\mathrm{\pi }$.

Step-by-step solution

The given data

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

The general expression of the wave, $y=y_m\sin \left(kx-\omega t+\varphi \right)$ (i)

Understanding the concept of wave equation

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.

the phase angle

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

$$\begin{gathered} slope=\frac{dy}{dt} \\ =\omega y_m\cos \left(kx-\omega t+\varphi \right) \end{gathered}$$

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

$$\begin{gathered} slope=u_s \\ =\omega y_m\cos \left(k\left(0\right)-\omega t+\varphi \right) \\ =\omega y_m\cos \varphi \end{gathered}$$

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

role="math" localid="1660973415652" $$\begin{gathered} u_s=4.0m/s \\ slope=u_s \\ =\omega y_m\cos \varphi \\ =-4.0m/s............\left(1\right) \end{gathered}$$

The maximum value of transverse velocity will be-

$\omega y=5.0m/s\left(\because \cos \varphi =1\right)$

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

$$\begin{gathered} 5.0\cos \varphi =-4.0 \\ \cos \varphi =\frac{-4.0}{5.0} \\ =-0.8 \end{gathered}$$

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

Hence we get,

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

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

Hence, the value of phase is, $\varphi =-0.6435\pm 2n\mathrm{\pi }$.