# JEE Main & Advanced Physics Simple Harmonic Motion Acceleration in S.H.M.

Acceleration in S.H.M.

Category : JEE Main & Advanced

(1) The acceleration of the particle executing S.H.M. at any instant, is defined as the rate of change of its velocity at that instant. So acceleration

$A=\frac{dv}{dt}=\frac{d}{dt}(a\omega \cos \omega \,t)$$=-{{\omega }^{2}}a\sin \omega \,t$$=-{{\omega }^{2}}y$                                  [As $y=a\sin \omega \,t$]

(2) In S.H.M. as $\left| \,\text{Acceleration}\, \right|\,\,={{\omega }^{2}}y$ is not constant. So equations of translatory motion can not be applied.

(3) In S.H.M. acceleration is maximum at extreme position (at $y=\pm a$). Hence $\left| {{A}_{\max }} \right|={{\omega }^{2}}a$ when $\left| \,\sin \,\omega \,t\, \right|=\text{maximum}=1$ i.e.  at $t=\frac{T}{4}$  or $\omega t=\frac{\pi }{2}$. From equation

(ii) $|{{A}_{\max }}|\,={{\omega }^{2}}a$  when $y=a$. (i) In S.H.M. acceleration is minimum at mean position From equation (i) ${{A}_{\min }}=0$ when $\sin \omega \,t=0$ i.e. at $t=0$ or $t=\frac{T}{2}$ or $\omega \,t=\pi$. From equation

(ii) ${{A}_{\min }}=0$ when $y=0$ (ii) Acceleration is always directed towards the mean position and so is always opposite to displacement

i.e.,  $A\propto -y$ Graph between acceleration (A) and displacement (y) is a straight line as shown

Slope of the line $=-{{\omega }^{2}}$

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