91Ó°ÊÓ

Given:

$x\left(0\right)=5cm$left of equilibrium position

$v\left(0\right)=36.3cm/s$to the right side of the equilibrium position

$T=1.5s$

Equation of wave is given,

$x\left(t\right)=A\cos \left(\frac{2\mathrm{Ï€}t}{T}+{\mathrm{Ï•}}_o\right)$

$v\left(t\right)=-\frac{2\mathrm{Ï€}A}{T}\sin \left(\frac{2\mathrm{Ï€}t}{T}+{\mathrm{Ï•}}_o\right)$

We have to search out the Amplitude by divide the above equations,

$\frac{v\left(t\right)}{x\left(t\right)}=-\frac{2\mathrm{Ï€}}{T}\tan \left(\frac{2\mathrm{Ï€}t}{T}+{\mathrm{Ï•}}_o\right)$

Substitute the values and at $t=0$,

$$\begin{gathered} \frac{36.3}{5}=-\frac{2\mathrm{π}}{1.5}\tan \left({\mathrm{ϕ}}_o\right) \\ ⇒\left|{\mathrm{ϕ}}_o\right|=\arctan \left(\frac{1.5×36.3}{10\mathrm{π}}\right) \\ =1.047\mathrm{rad} \\ =\frac{\mathrm{π}}{3}\mathrm{rad}≈60° \end{gathered}$$

The absolute angle is $60°$, and also the displacement is $5$, thus the direction is from $-A$to $0$at $270°$, where $x=0$, that the angle is within the third quadratic and is given from $+x$by,

$$\begin{gathered} {\mathrm{Ï•}}_o=\mathrm{Ï€}+\frac{\mathrm{Ï€}}{3}=\frac{4\mathrm{Ï€}}{3}\mathrm{rad} \\ =-\frac{2\mathrm{Ï€}}{3}\mathrm{rad} \end{gathered}$$

Then, $\mathrm{Ï•}=\mathrm{Ï•}+2\mathrm{Ï€}=\mathrm{Ï•}-2\mathrm{Ï€}$, we write $\mathrm{Ï•}$within the variety of $-\frac{2\mathrm{Ï€}}{3}$

By,

$x\left(t\right)=A\cos \left(\frac{2\mathrm{Ï€}t}{T}-\frac{2\mathrm{Ï€}}{3}\right)$

To achieve the best possible positive amplitude, $t=0s$

$A=\left|\frac{5}{\cos (2\mathrm{Ï€}/3)}\right|=10\mathrm{cm}$

so, the equation is

$$\begin{gathered} x\left(t\right)=10\cos \left(\mathrm{Ï•}\right) \\ =10\cos \left(\frac{2\mathrm{Ï€}t}{1.5}-\frac{2\mathrm{Ï€}}{3}\right) \end{gathered}$$

The angle $\mathrm{Ï•}$is,

$\mathrm{Ï•}\left(t\right)=\frac{2\mathrm{Ï€}t}{1.5}-\frac{2\mathrm{Ï€}}{3}$

Substitute,

$t=0s$, $\mathrm{Ï•}\left(0\right)=-\frac{2\mathrm{Ï€}}{3}$

$t=0.5s$, $\mathrm{Ï•}(0.5)=0$

$t=1s$, $\mathrm{Ï•}\left(1\right)=\frac{2\mathrm{Ï€}}{3}$

$t=1.5s$,$\mathrm{Ï•}(1.5)=\frac{4\mathrm{Ï€}}{3}$