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

Energy in S.H.M.

Category : JEE Main & Advanced

(1) Potential energy : This is an account of the displacement of the particle from its mean position.

(i) The restoring force $F=-ky$ against which work has to be done. Hence potential energy U is given by

$U=\int{dU}=-\int{dW}=-\int_{\,0}^{\,x}{Fdx}=\int_{\,0}^{\,y}{ky\,dy}=\frac{1}{2}k{{y}^{2}}+{{U}_{0}}$

where ${{U}_{0}}=$ Potential energy at equilibrium position.

If   ${{U}_{0}}=0$ then $U=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}$                                                        [As${{\omega }^{2}}=k/m$]

(ii) Also $U=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}{{\sin }^{2}}\omega \,t$$=\frac{1}{4}m{{\omega }^{2}}{{a}^{2}}(1-\cos 2\omega \,t)$

[As $y=a\sin \omega \,t$]

Hence potential energy varies periodically with double the frequency of S.H.M.

(iii) Potential energy maximum and equal to total energy at extreme positions

${{U}_{\max }}=\frac{1}{2}k{{a}^{2}}=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}$ when $y=\pm a$; $\omega \,t=\pi /2$; $t=\frac{T}{4}$

(iv) Potential energy is minimum at mean position

${{U}_{\min }}=0$         when    $y=0$;  $\omega \,t=0$;  $t=0$

(2) Kinetic energy : This is because of the velocity of the particle

Kinetic Energy  $K=\frac{1}{2}m{{v}^{2}}$$=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})$                                [As $v=\omega \sqrt{{{a}^{2}}-{{y}^{2}}}$]

(i) Also $K=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}{{\cos }^{2}}\omega \,t$$=\frac{1}{4}m{{\omega }^{2}}{{a}^{2}}(1+\cos 2\omega t)$

[As $v=a\omega \,\cos \omega \,t$]

Hence kinetic energy varies periodically with double the frequency of S.H.M.

(ii) Kinetic energy is maximum at mean position and equal to total energy at mean position.

${{K}_{\max }}=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}$ when $y=0$; $t=0$; $\omega \,t=0$

(iii) Kinetic energy is minimum at extreme position. ${{K}_{\min }}=0$ when $y=a$;  $t=T/4$, $\omega \,t=\pi /2$

(3) Total mechanical energy : Total mechanical energy always remains constant and it is equal to sum of potential energy and kinetic energy i.e. $E=U+K$

$E=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})+\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}$$=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}$

Total energy is not a position function.

(4) Energy position graph (i) At $y=0;\,\,U=0$ and $K=E$

(ii) At $y=\pm a;\,\,U=E$ and $K=0$

(iii) At $y=\pm \frac{a}{2}$; $U=\frac{E}{4}$and $K=\frac{3E}{4}$

(iv) At $y=\pm \frac{a}{\sqrt{2}}$; $U=K=\frac{E}{2}$

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