91Ó°ÊÓ

From the given $FIGUREEX15.6$amplitude, frequency and phase constant.

The amplitude is the length of the peak from the time axis. From , it is clear that the amplitude is$a=10cm$

The frequency is simply the number of waves per unit time. From $FIGUREEX15.6$, in $8s$the number of waves completed is$2.$

Therefore, frequency is,

localid="1650282912538" $f=\frac{numberofwaves}{time}=\frac{1}{4s}=0.25Hz$

Suppose the equation of the wave is,

$x=a\cos \left(\mathrm{Ó\neg }t+\mathrm{θ}\right)$

Here, $\mathrm{Ó\neg }$is the angular frequency of the wave, and $\mathrm{θ}$is the phase constant.

The time period of the wave is,

$T=4s$

Therefore, the angular frequency is,

$\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}=\frac{2\mathrm{Ï€}}{4}=\frac{\mathrm{Ï€}}{2}rad/s$

Let's take what happens atlocalid="1650282979393" $t=0s$

The equation takes the form,

localid="1650283019843" $$\begin{gathered} -5cm=(10cm)\cos \left(\frac{\mathrm{Ï€}}{2}\left(0s\right)+\mathrm{θ}\right) \\ \cos \left(\mathrm{θ}\right)=-\frac{1}{2} \\ \mathrm{³¦´Ç²õθ}=-\frac{1}{2}=\cos {120}^0 \\ \mathrm{θ}={120}^0 \end{gathered}$$

Therefore, the initial phase constant or the initial phase is${120}^0$