A body of mass 'm' performs linear S.H.M. giv | vedclass

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(C) The given equation is $x = P \sin \omega t + Q \sin \left(\omega t + \frac{\pi}{2}\right)$.Since $\sin \left(\omega t + \frac{\pi}{2}\right) = \co

Vedclass · Accepted Answer

$\frac{1}{2} m \omega^2 (P^2 + Q^2)$

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(C) The given equation is $x = P \sin \omega t + Q \sin \left(\omega t + \frac{\pi}{2}\right)$.Since $\sin \left(\omega t + \frac{\pi}{2}\right) = \cos \omega t$,the equation becomes $x = P \sin \omega t + Q \cos \omega t$.This represents the superposition of two simple harmonic motions with amplitudes $P$ and $Q$ that have a phase difference of $\frac{\pi}{2}$.The resultant amplitude $R$ is given by $R = \sqrt{P^2 + Q^2}$.The total energy $E$ of a particle performing $S$.$H$.$M$. is given by $E = \frac{1}{2} m \omega^2 R^2$.Substituting the value of $R^2 = P^2 + Q^2$,we get $E = \frac{1}{2} m \omega^2 (P^2 + Q^2)$.

$A$ body of mass '$m$' performs linear $S$.$H$.$M$. given by the equation $x = P \sin \omega t + Q \sin \left(\omega t + \frac{\pi}{2}\right)$. The total energy of the particle at any instant is

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