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Question

(A) For a particle executing simple harmonic motion $(SHM)$ with amplitude $a$ and angular frequency $\omega$,the instantaneous kinetic energy is $E =

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$=$

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(A) For a particle executing simple harmonic motion $(SHM)$ with amplitude $a$ and angular frequency $\omega$,the instantaneous kinetic energy is $E = \frac{1}{2}m\omega^2(a^2 - x^2)$ and potential energy is $U = \frac{1}{2}m\omega^2x^2$.Over one complete period $T$,the average kinetic energy is given by $ = \frac{1}{T} \int_{0}^{T} \frac{1}{2}m\omega^2(a^2 - a^2 \sin^2(\omega t)) dt = \frac{1}{4}m\omega^2a^2$.Similarly,the average potential energy is given by $ = \frac{1}{T} \int_{0}^{T} \frac{1}{2}m\omega^2(a^2 \sin^2(\omega t)) dt = \frac{1}{4}m\omega^2a^2$.Therefore,the correct relation is $ = $.

If $< E >$ and $< U >$ denote the average kinetic and the average potential energies respectively of a mass describing a simple harmonic motion,over one period,then the correct relation is:

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