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Derive expressions for the period of SHM in terms of (1) angular frequency (2) force constant (3) acceleration.

Vidyadip · Accepted Answer

The general expression for the displacement $( x )$ of a particle performing SHM is $x = A \sin$ ( $\omega t$ $+\alpha)$
(1) Let T be the period of the SHM and $x_1$ the displacement after a further time interval T. Then
$ x_1=A \sin [\omega(t+T)+\alpha]$
$=A \sin (\omega t+\omega T+\alpha)$
$=A \sin (\omega t+\alpha+\omega T) $
Since $T \neq 0$, for $x_1$ to be equal to $x$, we must have $(\omega T)_{\min }=2 \pi$.
Hence, the period $(T)$ of SHM is $T=2 \pi / \omega$
This is the expression for the period in terms of the constant co, the angular frequency.
(2) If $m$ is the mass of the particle and $k$ is the force constant, $\omega=\sqrt{ k / m}$.
$\therefore T =\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{k / m}}=2 \pi \sqrt{\frac{m}{k}}$
(3) The acceleration of a particle performing SHM has a magnitude $a=\omega^2 x$
$ \therefore \omega=\sqrt{a / x}$
$$
$\therefore T=\frac{2 \pi}{\omega}=\frac{\sqrt{\text { acceleration per unit displacement }}}{\sqrt{\text { acceleration per unit displacement }}} $

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