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Using the differential equation of linear S.H.M., obtain an expression for acceleration, velocity, and displacement of simple harmonic motion.

Vidyadip · Accepted Answer

i. Expression for acceleration in linear S.H.M:
a. From the differential equation,
$ \frac{ d ^2 x }{ dt ^2}+\omega^2 x =0$
$\frac{ d ^2 x }{ dt ^2}=-\omega^2 x \ldots . . $
b. But, linear acceleration is given by, $ \frac{ d ^2 x }{ dt ^2}= a $ From equations (1) and (2), $ a=-\omega^2 x $
Equation (3) gives acceleration in linear S.H.M.
ii. Expression for velocity in linear S.H.M:
a. From the differential equation of linear S.H.M
$ \frac{ d ^2 x }{ dt ^2}=-\omega^2 x$
$\therefore \frac{ d }{ dt }\left(\frac{ dx }{ dt }\right)=-\omega^2 x$
$\therefore \frac{ dv }{ dt }=-\omega^2 x \ldots \ldots\left(\because \frac{ dx }{ dt }= v \right)$
$\therefore \frac{ dv }{ dx } \cdot \frac{ dx }{ dt }=-\omega^2 x$
$\therefore v \frac{ dv }{ dx }=-\omega^2 x \ldots \ldots .\left(\because \frac{ dx }{ dt }= v \right)$
$\therefore v dv =-\omega^2 xdx \ldots \ldots \ldots . .(4) $
b. Integrating both sides of equation (4),
$ \int v d v=\int-\omega^2 x d x$
$\frac{v^2}{2}=-\frac{\omega^2 x^2}{2}+C $
where, $C$ is the constant of integration.
c. At extreme position, $x= \pm A$ and $v=0$.
Substituting these values in equation (5),
$ 0=-\frac{\omega^2 A ^2}{2}+ C$
$\therefore C =\frac{\omega^2 A ^2}{2} \ldots \ldots \ldots \ldots $
d. Substituting equation (6) in equation (5),
$ \frac{ v ^2}{2}=-\frac{\omega^2 x ^2}{2}+\frac{\omega^2 A ^2}{2}$
$\therefore v ^2=\omega^2 A ^2-\omega^2 x ^2$
$\therefore v ^2=\omega^2\left( A ^2- x ^2\right)$
$\therefore v = \pm \omega \sqrt{ A ^2- x ^2} $
This is the required expression for velocity in linear S.H.M.
iii. Expression for displacement in linear S.H.M:
a. From the differential equation of linear S.H.M, velocity is given by,
$v=\omega \sqrt{ A ^2- x ^2}$
But, in linear motion, $v=\frac{d x}{d t}$
From equation ( 1 ) and ( 2 ),
$ \frac{ dx }{ dt }=\omega \sqrt{ A ^2- x ^2}$
$\therefore \frac{ dx }{\sqrt{ A ^2- x ^2}}=\omega dt . $
b. Integrating both sides of equation (3),
$ \int \frac{d x}{\sqrt{ A ^2- x ^2}}=\int \omega d t$
$\therefore \sin ^{-1}\left(\frac{ x }{ A }\right)=\omega t +\Phi $
where, $\alpha$ is constant of integration which depends upon initial condition (phase angle)
$ \therefore \frac{ x }{ A }=\sin (\omega t +\Phi)$
$\therefore x = A \sin (\omega t +\Phi) $
This is the required expression for displacement of a particle performing linear S.H.M. at time t.

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