Question

Figure 16-44 shows the displacement yversus time tof the point on a string at$\mathit{x}\mathbf{=}\mathbf{0}$, as a wave passes through that point. The scale of the yaxis is set by${\mathbf{y}}_{\mathbf{s}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mm}$. The wave is given by$\mathbf{y}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathbf{y}}_{\mathbf{m}}\mathbf{sin}\mathbf{(}\mathbf{kx}\mathbf{-}\mathbf{\omega t}\mathbf{-}\mathbf{\varphi }\mathbf{)}$. What is$\mathit{\theta }$? (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$\mathbf{y}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{t}\mathbf{)}$) and then plotting the function.)

Short answer

The value of $\mathrm{\varphi }\hspace{0.33em}\mathrm{is}2.8\mathrm{rad}\mathrm{or}-3.5\mathrm{rad}$.

Step-by-step solution

The given data

1. Displacement (y) vs t graph at x=0
2. The wave equation is,$y=y_{m}sin(kx-\omega t+\varphi )$

Understanding the concept of the wave equation

We can find the equation of the wave at x = 0 given in the graph. From the slope of the graph, we can predict the approximate value of the phase constant. Then, from the equation of the wave at t = 0, we can easily get the value of the phase constant.

Formula:

The transverse velocity of the wave (or slope),

$v=\frac{dy}{dt}\left(i\right)$

Calculation of phase constant

The equation of the wave is given as

$y=y_{m}sin(kx-\omega t+\varphi )$

At x = 0, it becomes,

$y=y_{m}sin(-\omega t+\varphi )$

This is the equation for the given graph.

The slope of the graph using equation (i) gives the velocity, which is given as:

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

From the given graph, we can conclude that the slope of the graph at t = 0 is positive.

Hence,

$$\begin{gathered} -\omega \left(y_m\cos \left(-\omega \left(0\right)+\varphi \right)\right)>0 \\ -\omega \left(y_m\cos \left(\varphi \right)\right)>0 \\ -\cos \left(\varphi \right)>0 \end{gathered}$$

This implies that$\varphi$is in between$\frac{\mathrm{\pi }}{2}\mathrm{and}\mathrm{\pi }\hspace{0.33em}\mathrm{or}\hspace{0.33em}\mathrm{\pi }\hspace{0.33em}\mathrm{and}\frac{3\mathrm{\pi }}{2}$.

The equation of the wave at t = 0 is,

$$\begin{gathered} y=y_m\sin \left(-\omega \left(0\right)+\varphi \right) \\ y=y_m\sin \left(\varphi \right) \\ \varphi ={\sin }^{-1}\frac{y}{y_m} \end{gathered}$$

From the given graph, we can figure out that ${\mathrm{y}}_{\mathrm{m}}=6\hspace{0.33em}\mathrm{mm}\mathrm{and}\hspace{0.33em}\mathrm{y}(\mathrm{t}=0)=2\hspace{0.33em}\mathrm{mm}$. Hence,

role="math" localid="1660981403661" $$\begin{gathered} \varphi ={\sin }^{-1}\frac{2}{6} \\ =2.8rad \end{gathered}$$

Since$\mathrm{sin\varphi }=\sin (\mathrm{\pi }-\mathrm{\varphi }),\mathrm{and}\hspace{0.33em}\mathrm{\varphi }=2.8\hspace{0.33em}\mathrm{rad}$

Also,

$$\begin{gathered} \varphi =2.8-2\mathrm{\pi } \\ =-3.48 \\ ~-3.5\mathrm{rad} \end{gathered}$$

Therefore, the value of $\mathrm{\varphi }\hspace{0.33em}\mathrm{is}2.8\mathrm{rad}$or -3.5 rad.