$FIGUREEX15.8$is the velocity-versus-time graph of a particle in simple harmonic motion.

a. What is the amplitude of the oscillation?

b. What is the phase constant?

c. What is the position at$t=0s?$

$FIGUREEX15.8$is the velocity-versus-time graph of a particle in simple harmonic motion.

. From the figure $v_{max}$ is 60 ms^-1.. A is amplitude and$\omega$is angular frequency.

role="math" localid="1650303983759" $$\begin{gathered} v_{max}=\omega A \\ A=\frac{v_{max}}{\omega } \\ A=\frac{60}{{\displaystyle \frac{\pi }{6}}}=114.65cm. \end{gathered}$$

Suppose the equation of the wave is,

$x=a\cos \left(\omega t+\theta \right)$

The time period of the wave is given by,

localid="1650302892377" $T=12s$

Therefore, angular frequency is

localid="1650302913364" $\omega =\frac{2\mathrm{\pi }}{T}=\frac{2\mathrm{\pi }}{12}=\frac{\mathrm{\pi }}{6}$

Let's take what happens at localid="1650302995667" $t=0s$

The equation takes the form,

On differentiating the above wave equation, we get

$v\left(t\right)=-a\omega \sin (\omega t+\theta )$

localid="1650303164933" $$\begin{gathered} v\left(t\right)=-a\omega \sin (\omega t+\theta )=-v_{max}\sin (\omega t+\theta ) \\ v\left(t\right)=-v_{max}\sin (\omega t+\theta ) \\ 30=-60\sin \left(\frac{\mathrm{\pi }}{6}\right(0)+\theta ) \\ \sin \left(\mathrm{\theta }\right)=-\frac{1}{2} \\ \mathrm{sin\theta }=-\frac{1}{2}=\sin \left(210\right) \\ \mathrm{\theta }={210}^0 \end{gathered}$$

The phase constant is aboutlocalid="1650303186071" ${210}^0$

The velocity equation is given by

$v\left(t\right)=-a\omega \sin (\omega t+\theta )=-v_{max}\sin (\omega t+\theta )$

here, $v_{max}=60cm/s$which is given in the $FIGUREEX.15.8$

Plugging all the values in the above equation we get,

localid="1650303278656" $$\begin{gathered} v\left(0\right)=-(60cm/s)\sin \left(210\right) \\ v\left(0\right)=-(60cm/s)(-1/2) \\ v\left(0\right)=30cm/s \end{gathered}$$

velocity at time$t=0s$is$30cm/s$