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Define angular SHM. State the differential equation of angular SHM. Hence derive an expression for the period of angular SHM in terms of (i) the torsion constant (ii) the angular acceleration.

Vidyadip · Accepted Answer

Definition : Angular SHM is defined as the oscillatory motion of a body in which the restoring torque responsible for angular acceleration is directly proportional to the angular displacement and its direction is opposite to that of angular displacement. The differential equation of angular SHM is $I \frac{d^2 \theta}{d t^2}+c \theta=0 \ldots$ (1)
where $I=$ moment of inertia of the
where $I$ = moment of inertia of the oscillating body,
$\frac{d^2 \theta}{d t^2}=$ angular acceleration of the body when its angular displacement is $\theta$, and $c =$ torsion constant of the suspension wire,
$
\therefore \frac{d^2 \theta}{d t^2}+\frac{c}{I} \theta=0
$
Let $\frac{c}{I}=\omega^2$, a constant. Therefore, the angular frequency, $\omega=\sqrt{c / I}$ and the angular acceleration,
$
a=\frac{d^2 \theta}{d t^2}=-\omega^2 \theta
$
The minus sign shows that the $\alpha$ and $\theta$ have opposite directions. The period $T$ of angular SHM is
$
T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{C / I}}=2 \pi \sqrt{\frac{I}{c}}
$
This is the expression for the period in terms of torque constant. Also, from Eq. (2),
$\omega=\sqrt{\left|\frac{\alpha}{\theta}\right|} \quad$
$=\sqrt{\text { angular acceleration per unit angular displacement }}$
$\therefore T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{|\alpha / \theta|}}$
$2 \pi$
$=\frac{2 \pi}{\sqrt{\text { angular acceleration per unit angular displacement }}}$

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