FIGURE P16.46 is a snapshot graph at t = 0 s of a 5.0 Hz wave traveling to the left. a. What is the wave speed? b. What is the phase constant of the wave? c. Write the displacement equation for this wave.

A snapshot graph at t = 0 s of a 5.0 Hz wave traveling to the left is shown below

From the history graph, it can be observed that $\lambda =2.0m$and $f=5.0Hz$.

The speed of the wave is given by $v=f\lambda$.

Substitute the given values

$$\begin{gathered} v=5\times 2 \\ =10m/s \end{gathered}$$

Therefore, the wave speed is$10m/s$.

A phase constant can be obtained by $D\left(0m,0s\right)=0.5mm=\frac{1}{2}A$.

Thus,

$$\begin{gathered} D\left(0m,0s\right)=\frac{1}{2}A=A\sin {\varphi }_0 \\ {\varphi }_0={\sin }^{-1}\left(\frac{1}{2}\right) \\ =\frac{\mathrm{\pi }}{6}rador\frac{5}{6}\mathrm{\pi }\mathrm{rad} \end{gathered}$$

The value of $D\left(0m,0s\right)$alone given the two values of the phase constant. One of the values will cause the displacement to start at $0.5mm$and increases with distance while the other cause the displacement to start at $0.5mm$and decreases with distance.

Therefore the equation of wave for $t=0s$is

$D\left(x,t=0\right)=A\sin \left(\frac{2\mathrm{\pi x}}{\lambda }+{\varphi }_0\right)$

The value ${\varphi }_0=\frac{\mathrm{\pi }}{6}$rad is the phase constant for which displacement increases as$x$increases.

The equation for the displacement of the wave is

$D\left(x,t\right)=A\sin \left(\frac{2\mathrm{\pi x}}{\lambda }-2\mathrm{\pi ft}+{\mathrm{\varphi }}_0\right)$

Substitute the given values

role="math" localid="1649843605976" $D\left(x,t\right)=\left(1.0mm\right)\sin \left[2\mathrm{\pi }\left(\frac{x}{2.0m}-\left(0.5{\mathrm{s}}^{-1}\right)\mathrm{t}\right)+\frac{\mathrm{\pi }}{6}\right]$

Therefore, the displacement equation for this wave is$D\left(x,t\right)=\left(1.0mm\right)\sin \left[2\mathrm{\pi }\left(\frac{x}{2.0m}-\left(0.5{\mathrm{s}}^{-1}\right)\mathrm{t}\right)+\frac{\mathrm{\pi }}{6}\right]$.