91Ó°ÊÓ

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 $\mathrm{λ}=2.0m$and $f=5.0Hz$.

The speed of the wave is given by $v=f\mathrm{λ}$.

Substitute the given values

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

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,

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

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

$D\left(x,t=0\right)=A\sin \left(\frac{2\mathrm{Ï€³\ae }}{\mathrm{λ}}+{\mathrm{Ï•}}_0\right)$

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

The equation for the displacement of the wave is

$D\left(x,t\right)=A\sin \left(\frac{2\mathrm{Ï€³\ae }}{\mathrm{λ}}-2\mathrm{Ï€´Ú³Ù}+{\mathrm{Ï•}}_0\right)$

Substitute the given values

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

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