Angular Simple Harmonic Motion - Practice Questions & MCQ

The general equation of linear SHM is given by $x=A \sin (\omega t+\alpha)$
Similarly, The general equation of angular SHM is given by $\theta=\theta_0 \sin (\omega t+\phi)$ where $\theta$ and $\theta_0$ are angular displacement and angular amplitude of the bob respectively, as shown in the below figure

If I=length of the bob then we can write $\theta=\frac{x}{l}$ and $\theta_0=\frac{A}{l}$.
Similarly, The angular velocity if the bob which is in angular SHM is given by

$
\begin{aligned}
& \dot{\theta}=\frac{d \theta}{d t}=\theta_0 \omega \operatorname{Cos}(\omega t+\phi) \\
& \text { or } \dot{\theta}=\omega \sqrt{\theta_0^2-\theta^2}
\end{aligned}
$

Similarly, The angular acceleration if the bob which is in angular SHM is given by

$
\begin{aligned}
& \alpha=\frac{d^2 \theta}{d t}=-\theta_0 \omega^2 \operatorname{Sin}(\omega t+\phi) \\
& \text { or } \alpha=-\omega^2 \theta
\end{aligned}
$

And Thus restoring torque on the body is given as

$
\tau_R=-I \alpha=-I \omega^2 \theta
$

Thus we can state that in angular SHM, the angular acceleration of the body and the restoring torque on the body are directly
proportional to the angular displacement of body from its mean position and are directed toward the mean position.

Similarly, a basic differential equation for angular SHM can be written as

$
\frac{d^2 \theta}{d t^2}+\omega^2 \theta=0
$