Question

obtain the formula of acceleration of a particle from the formula of displacement of SHM​

Answer 1

Explanation: Click on the image for Solution.

Answer 2

To derive:

The formula of acceleration of a particle from the formula of displacement of SHM ?

Solution:

Displacement function for SHM:

$\therefore \: x = a \sin( \omega t)$

Now, 1st order differentiation will give velocity:

$\implies \: v = \dfrac{dx}{dt}$

$\implies \: v = \dfrac{d \bigg \{a \sin( \omega t) \bigg \} }{dt}$

$\implies \: v = a \omega \cos( \omega t)$

2nd order differentiation will give acceleration:

$\implies \: acc. = \dfrac{dv}{dt} = \dfrac{ {d}^{2} x}{d {t}^{2} }$

$\implies \: acc. = \dfrac{d \bigg \{a \omega \cos( \omega t) \bigg \} }{dt}$

$\implies \: acc. = a { \omega}^{2} \bigg \{ - \sin( \omega t) \bigg \}$

$\implies \: acc. = - { \omega}^{2} \bigg \{ a\sin( \omega t) \bigg \}$

$\implies \: acc. = - { \omega}^{2} \bigg \{ x \bigg \}$

$\implies \: acc. = - { \omega}^{2} x$

So, the required expression is:

$\boxed{ \bf \: acc. = - { \omega}^{2} x}$