Figure 16-46 shows transverse acceleration${\mathit{a}}_{\mathbf{y}}$versus time tof the point on a string at x=0, as a wave in the form of$\mathbf{y}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathbf{y}}_{\mathbf{m}}\mathbf{sin}\mathbf{(}\mathbf{kx}\mathbf{-}\mathbf{\omega t}\mathbf{+}\mathbf{\varphi }\mathbf{)}$passes through that point. The scale of the vertical axis is set by${\mathbf{a}}_{\mathbf{s}}\mathbf{=}\mathbf{400}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. What 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.

i) Transverse acceleration (a_y) vs t graph at x=0.

ii) The wave equation is $y\left(x,t\right)=y_m\sin \left(kx-\omega t+\varphi \right)$.

iii) The scale of the vertical axis is set by, $a_s=400m/s^2$.

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

The equation of the wave is given as,

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

At x = 0, it becomes,

$y=y_{m}sin(-\omega t+\varphi )$ (1)

At x = 0, acceleration is maximum. Hence,

$$\begin{gathered} a_y=a_m \\ {a}_y=-{\omega }^{2}y \end{gathered}$$ (a)

From equation (1), the acceleration of the wave is given as:

$a_y=-{\omega }^{2}ysin(-\omega t+\varphi )$

This is the equation for the given graph.

The slope of the graph is given as:

$$\begin{gathered} \frac{d{a}_y}{dt}=\frac{d}{dt}\left(-{\omega }^2{y}_m\sin \left(-\omega t+\varphi \right)\right) \\ ={\omega }^2{y}_m\cos \left(-\omega t+\varphi \right) \end{gathered}$$

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

$$\begin{gathered} {\omega }^2{y}_m\cos \left(-\omega \left(0\right)+\varphi \right)<0 \\ {\omega }^3{y}_m\cos \left(\varphi \right)<0 \end{gathered}$$

From this, we can infer that,

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

From the graph, we can write that,
$a_m=400\frac{m}{s^2}and{a}_y=100\frac{m}{s^2}$.

Hence, From equation (a), we get,

$$\begin{gathered} a_y=-400\sin \left(\varphi \right) \\ -100m/s^2=-400\sin \left(\varphi \right) \\ \varphi ={\sin }^{-1}\left(\frac{1}{4}\right) \\ =0.25rad \end{gathered}$$

Since,

$$\begin{gathered} \sin \varphi =\sin \left(\mathrm{\pi }-\mathrm{\varphi }\right) \\ \varphi =2.9rad \end{gathered}$$

Also,

$$\begin{gathered} \varphi =2.9-2\mathrm{\pi } \\ =-3.38 \\ \cong -3.4\mathrm{rad} \end{gathered}$$

Therefore, the value of $\varphi$is 2.9 rad or -3.5 rad.