Question

Question: The angle of the pendulum in Figure is given by $\mathit{\theta }\mathbf{=}{\mathit{\theta }}_{\mathbf{m}}\mathit{c}\mathit{o}\mathit{s}\mathbf{[}\left(4.44\text{rad/s}\right)\mathbf{t}\mathbf{+}\mathbf{\Phi }\mathbf{]}$. If at t = 0$\theta =0.040\text{rad}$ , and$\mathit{d}\mathit{\theta }\mathbf{/}\mathit{d}\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{}\mathbf{\text{rad/s}}$ ,

1. what is the phase constant $\mathit{\phi }$,
2. what is the maximum angle${\mathit{\theta }}_{\mathbf{m}}$ ?

(Hint: Don’t confuse the rate $\mathit{d}\mathit{\theta }$at which changes with the $\mathit{\theta }$of the SHM.)

Short answer

Answer

1. The phase constantis 0.845 rad
2. The maximum angle is 0.0602 rad

Step-by-step solution

Identification of given data

1. The angle of the pendulum is
2. At $t=0,\theta =0.040\text{radand}\frac{\text{d}\theta }{\text{dt}}=-0.200\text{rad}/\text{s}$, and$\frac{\text{d}\theta }{\text{dt}}=-0.200\text{rad}/\text{s}$
   

Understanding the concept

The oscillations of the simple pendulum can be defined by the equation of simple harmonic motion. The simple harmonic motion is the motion in which the acceleration of the oscillating object is directly proportional to the displacement. The force caused by the acceleration is called restoring force. This restoring force is always directed towards the mean position.

Compare the given equation with the equation of displacement of the particle in simple harmonic motion.

Formulae:

$$\begin{gathered} x\left(t\right)=x_m\cos \left(\omega t+\theta \right) \\ v\left(t\right)=\omega x_m\sin \left(\omega t+\theta \right) \end{gathered}$$

(a) Determining the phase constant  

The phase constant : $\Phi$

The expression for the displacement of the particle in simple harmonic motion is

$x\left(t\right)=x_m\text{cos}\left(\omega t+\Phi \right)$

Here, x (t) is the displacement, x_m is amplitude, $\omega$angular velocity, t is time, $\Phi$ is phase difference.

For angular displacement, replace x by $\theta$, then

$\theta \left(t\right)={\theta }_m\cos \left(\omega t+\Phi \right)$ …(i)

The expression for velocity of the particle in simple harmonic motion is

$v\left(t\right)=-\omega x_m\sin \left(\omega t+\Phi \right)$

For angular motion, replace x by $\theta$, then

$\frac{d\theta }{dt}=-\omega {\theta }_m\text{sin}\left(\omega t+\Phi \right)$ …(ii)

Divide equation (ii) by equation (i)

$\begin{array}{rcl}\frac{\frac{d\theta }{dt}}{\theta }& =& \frac{-\omega {\theta }_m\text{sin}\left(\omega t+\Phi \right)}{{\theta }_m\text{cos}\left(\omega t+\Phi \right)}\\ & =& \frac{-\omega \text{sin}\left(\omega t+\Phi \right)}{\text{cos}\left(\omega t+\Phi \right)}\\ & =& -\omega \text{tan}\left(\omega \text{t}+\Phi \right)\\ {\left(\frac{\frac{\text{d}\theta }{\text{dt}}}{\theta }\right)}_{t=0}& =& -\omega \text{tan}\left(\omega t+\Phi \right)\\ & =& -\omega \text{tan}\left(\Phi \right)\\ \Phi & =& {\text{tan}}^{-1}\left[-\frac{{\left(\frac{(\text{d}\theta /\text{dt}}{\theta }\right)}_{t=0}}{\omega }\right]\sqrt{a^2+b^2}\\ & =& {\text{tan}}^{-1}\left[-\frac{{\left(\frac{-0.200\text{rad}/\text{s}}{0.040\text{rad}}\right)}_{t=0}}{4.44\text{rad}/\text{s}}\right]\\ & =& 0.845\text{rad}\end{array}$

Therefore, the phase constant is 0.845 rad .

(b) Determining the maximum angle 

For t =0, equation (i) becomes as

$$\begin{gathered} \theta \left(\text{t}\right)={\theta }_m\text{cos}\left(\Phi \right) \\ {\theta }_m=\frac{\theta \left(t\right)}{\text{cos}\left(\Phi \right)} \\ =\frac{0.040\text{rad}}{\text{cos}\left(0.845\right)} \\ =0.0602\text{rad} \end{gathered}$$

The maximum angle is 0.0602 rad