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

(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 \[=-{{\omega }^{2}}\]