Question

Figure ex 15.7 is the position-versus-time graph of a particle in simple harmonic motion.

a. What is the phase constant?

b. What is the velocity at $t=0s?$

c. What is $v_{max}?$

Short answer

a. phase constant ${60}^0$

b. velocity at $t=0s$is $-13.6cm/s$

c. maximum velocity$15.7cm/s$

Step-by-step solution

Given information

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

Explanation (part a)

Suppose the equation of the wave is,

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

Here, $\omega$is the angular frequency of the wave, and $\theta$is the phase constant.

The time period of the wave is,

$T=4s$

Therefore, the angular frequency is,

$\omega =\frac{2\mathrm{\pi }}{T}=\frac{2\mathrm{\pi }}{4}=\frac{\mathrm{\pi }}{2}rad/s$

Let's take what happens at$t=0s$

The equation takes the form,

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

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

Explanation (part b)

Suppose the equation of the wave is,

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

On differentiating the above equation, we get

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

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

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

Explanation (part c)

Suppose the equation of the wave is,$x=a\sin \left(\omega t+\theta \right)$

On differentiating the above equation, we get

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

$\because v_{max}=a\omega$

Plugging the values in the above equation, we get

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