$FIGUREQ15.4$shows a position-versus-time graph for a particle in SHM.

a. What is the phase constant ${\varphi }_0$? Explain.

b. What is the phase of the particle at each of the three numbered points on the graph?

A sinusoidal oscillation with period and amplitude is known as simple harmonic motion.

The particle executing SHM has the following general equation of motion:

$x\left(t\right)=A\cos \left(\omega t+{\varphi }_0\right)$

Here, $A$is the amplitude of the wave, $\omega$is the angular frequency, and ${\varphi }_0$is the phase constant.

Create a figure in SHM that depicts a particle's position-versus-time graph.

(a)

Calculate the phase constant.

At time $t=0$from the above figure, $x\left(t\right)$is equal to $\frac{A}{2}$.

Substitute for $\frac{A}{2}$and for $t$in equation $\left(1\right)$.

$\frac{A}{2}=A\cos \left(0+{\varphi }_0\right)$

$\cos {\varphi }_0=\frac{1}{2}$

Rewrite the above equation for ${\varphi }_0$.

${\varphi }_0={\cos }^{-1}\left(\frac{1}{2}\right)$

$=\frac{\pi }{3}or\frac{-\pi }{3}$

This implies that the phase constant of a wave at $t=0$is either $+\frac{\pi }{3}$or $\frac{-\pi }{3}$.

Determine which direction the particles are travelling.

If the particle is going to the right, the phase constant is negative, and if it is moving to the left, the phase constant is positive.

The phase constant is because the particle is moving to the right.

$\frac{-\pi }{3}$.

(b)

Calculate the phase of the particle at each of the three points on the graph.

At point$1$the displacement of the particle$x\left(t\right)=-\frac{A}{2}$.

Use the following position formula to calculate the phase of the particle $\varphi$.

$x\left(t\right)=A\cos \varphi$

Substitute $-\frac{A}{2}$for $x\left(t\right)$in equation $\left(1\right)$.

$-\frac{A}{2}=A\cos \varphi$

$-\frac{1}{2}=\cos \varphi$

Rewrite the above equation for $\varphi$.

$\varphi ={\cos }^{-1}\left(\frac{-1}{2}\right)$

$\varphi =\frac{2\pi }{3}$

At point $2$the displacement of the particle $x\left(t\right)=A$.

Substitute $A$for $x\left(t\right)$in the equation $\left(2\right)$.

role="math" $$\begin{gathered} A=A\cos \varphi \\ 1=\cos \varphi \end{gathered}$$

Rewrite the above equation for $\varphi$.

$$\begin{gathered} \varphi ={\cos }^{-1}\left(1\right) \\ \varphi =2\pi \end{gathered}$$

At point$3$the displacement of the particle is,

$x\left(t\right)=-\frac{A}{2}$

Substitute $-\frac{A}{2}$for $x\left(t\right)$in equation$\left(2\right)$

width="98">-A2=Acosϕ$-\frac{A}{2}=A\cos \varphi$

$-\frac{1}{2}=\cos \varphi$

Rewrite the above equation for $\varphi$.

$\varphi ={\cos }^{-1}\left(-\frac{1}{2}\right)$

$\varphi =\frac{2\pi }{3}$

$\varphi =\frac{2\pi }{3}$