Question

In an angular simple harmonic motion angular amplitude of oscillation is $\pi$ rad and time period is 0.4 sec then calculate its angular velocity at angular displacement $\pi /2$ rad.
$\begin{array}{rl}& \text{A}\text{. 34}\text{.3 rad}/\text{sec}\\ & \text{B}\text{. 42}\text{.7 rad}/\text{sec}\\ & \text{C}\text{. 22}\text{.3 rad}/\text{sec}\\ & \text{D}\text{. 50}\text{.3 rad}/\text{sec}\end{array}$

Answer 1

Hint: Define angular simple harmonic motion. Obtain the expression for the angular displacement of the particle. From the given quantities, find the time for the angular displacement. Obtain the expression for angular velocity by differentiating the angular displacement. By putting the given values, we can find the answer.

Formula used:
$\theta ={\theta }_A\sin \left(\omega t\right)$
$w={\displaystyle \frac{d\theta }{dt}}$
$\omega ={\displaystyle \frac{2\pi }{T}}$


Complete step by step answer:
 Given in the question that, the angular simple harmonic motion has angular amplitude of $\pi$ rad.
So, the variation of angular amplitude with time for the angular simple harmonic motion can be written as,
$\theta ={\theta }_A\sin \left(\omega t\right)$
Where t is the time and $\omega$ is the angular frequency of the angular simple harmonic motion.
The angular frequency of a motion can be mathematically expressed in terms of the time period of the motion as,
$\omega ={\displaystyle \frac{2\pi }{T}}$
Now, the time period of the oscillation is 0.4 sec.
So, the angular frequency of the oscillation will be,
$\omega ={\displaystyle \frac{2\pi }{0.4}}$
Putting the values for the expression of amplitude, we get that,
$\theta =\pi \sin \left({\displaystyle \frac{2\pi }{0.4}}t\right)$
Now, the angular velocity of the angular simple harmonic motion can be found by differentiating the angular amplitude of the motion with respect to time.
So, we can write,
$w={\displaystyle \frac{d\theta }{dt}}$
Where, w is the angular velocity of the motion.
Putting the value for the angular amplitude on the above equation, we get that,
$\begin{array}{rl}& \Rightarrow w={\displaystyle \frac{d}{dt}}(\pi \sin \left({\displaystyle \frac{2\pi }{0.4}}t\right))\\ & \Rightarrow w=\pi \times {\displaystyle \frac{2\pi }{0.4}}\cos \left({\displaystyle \frac{2\pi }{0.4}}t\right)\\ & \Rightarrow w={\displaystyle \frac{2{\pi }^2}{0.4}}\cos \left({\displaystyle \frac{2\pi }{0.4}}t\right)\end{array}$
Again, we need to find the angular velocity of the particle at an angular displacement of $\pi /2$ rad.
Putting this value on the equation for angular amplitude,
$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{\pi }{2}}=\pi \sin \left({\displaystyle \frac{2\pi }{0.4}}t\right)\\ & \Rightarrow \sin \left({\displaystyle \frac{2\pi }{0.4}}t\right)={\displaystyle \frac{1}{2}}=\sin {\displaystyle \frac{\pi }{6}}\\ & \Rightarrow \left({\displaystyle \frac{2\pi }{0.4}}t\right)={\displaystyle \frac{\pi }{6}}\\ & \therefore t={\displaystyle \frac{0.4}{6\times 2}}={\displaystyle \frac{0.4}{12}}\text{sec}\end{array}$
Putting the value of time on the expression for angular velocity, we get that,
$\begin{array}{rl}& \Rightarrow w={\displaystyle \frac{2{\pi }^2}{0.4}}\cos \left({\displaystyle \frac{2\pi }{0.4}}t\right)\\ & \Rightarrow w={\displaystyle \frac{2{\pi }^2}{0.4}}\cos ({\displaystyle \frac{2\pi }{0.4}}\times {\displaystyle \frac{0.4}{12}})\\ & \Rightarrow w={\displaystyle \frac{2{\pi }^2}{0.4}}\cos {\displaystyle \frac{\pi }{6}}\\ & \Rightarrow w={\displaystyle \frac{2{\pi }^2}{0.4}}\times {\displaystyle \frac{\sqrt{3}}{2}}\\ & \therefore w=42.7\text{rad/sec}\end{array}$
The correct option is (B).

Note:
The angular velocity of a particle can be related with the radial velocity of a particle. Mathematically we can express it as,
$\begin{array}{rl}& v=\omega r\\ & \omega ={\displaystyle \frac{v}{r}}\end{array}$
Where, v is the linear velocity and $\omega$ is the angular velocity.