91Ó°ÊÓ

$FIGUREEX15.7$is the position-versus-time graph of a particle in simple harmonic motion.

Suppose the equation of the wave is,

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

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 at$t=0s$

The equation takes the form,

$$\begin{gathered} 5cm=(10cm)\sin \left(\frac{\mathrm{π}}{2}\left(0s\right)+\mathrm{θ}\right) \\ \sin \left(\mathrm{θ}\right)=\frac{1}{2} \\ \sin \mathrm{θ}=\frac{1}{2}=\sin {30}^0 \\ \mathrm{θ}={30}^0 \end{gathered}$$

Therefore, the phase constant or the initial phase is${30}^0$

Suppose the equation of the wave is,

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

On differentiating the above equation, we get

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

Plugging the values in the above equation, we get velocity at $t=0s$

$$\begin{gathered} v\left(0\right)=\left(10cm\right)\frac{\mathrm{π}}{2}\cos (0+{30}^0) \\ v\left(0\right)=10×\frac{\mathrm{π}}{2}×\frac{\sqrt{3}}{2}=13.6cm/s \end{gathered}$$

Suppose the equation of the wave is,$x=a\sin \left(\mathrm{Ó\neg }t+\mathrm{θ}\right)$

On differentiating the above equation, we get

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

$âˆ\mu v_{max}=a\mathrm{Ó\neg }$

Plugging the values in the above equation, we get

$v_{max}=\left(10cm\right)\frac{\mathrm{Ï€}}{2}=15.7cm/s$