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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