An air-track glider attached to a spring oscillates with a period of $1.5s$. At $t=0s$the glider is $5.00cm$left of the equilibrium position and moving to the right at $36.3cm/s$.

a. What is the phase constant?

b. What is the phase at $t=0s$, $0.5s$, $1.0s$, and $1.5s$?

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,

$x\left(t\right)=A\cos \left(\frac{2\pi t}{T}+{\varphi }_o\right)$

$v\left(t\right)=-\frac{2\pi A}{T}\sin \left(\frac{2\pi t}{T}+{\varphi }_o\right)$

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

$\frac{v\left(t\right)}{x\left(t\right)}=-\frac{2\pi }{T}\tan \left(\frac{2\pi t}{T}+{\varphi }_o\right)$

Substitute the values and at $t=0$,

$$\begin{gathered} \frac{36.3}{5}=-\frac{2\pi }{1.5}\tan \left({\varphi }_o\right) \\ \Rightarrow \left|{\varphi }_o\right|=\arctan \left(\frac{1.5\times 36.3}{10\pi }\right) \\ =1.047\mathrm{rad} \\ =\frac{\pi }{3}\mathrm{rad}\approx 60° \end{gathered}$$

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,

$$\begin{gathered} {\varphi }_o=\pi +\frac{\pi }{3}=\frac{4\pi }{3}\mathrm{rad} \\ =-\frac{2\pi }{3}\mathrm{rad} \end{gathered}$$

Then, $\varphi =\varphi +2\pi =\varphi -2\pi$, we write $\varphi$within the variety of $-\frac{2\pi }{3}$

By,

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

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

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

so, the equation is

$$\begin{gathered} x\left(t\right)=10\cos \left(\varphi \right) \\ =10\cos \left(\frac{2\pi t}{1.5}-\frac{2\pi }{3}\right) \end{gathered}$$

The angle $\varphi$is,

$\varphi \left(t\right)=\frac{2\pi t}{1.5}-\frac{2\pi }{3}$

Substitute,

$t=0s$, $\varphi \left(0\right)=-\frac{2\pi }{3}$

$t=0.5s$, $\varphi (0.5)=0$

$t=1s$, $\varphi \left(1\right)=\frac{2\pi }{3}$

$t=1.5s$,$\varphi (1.5)=\frac{4\pi }{3}$