Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

Asked by Abhisek | 1 year ago |  169

##### Solution :-

Let m be the mass of the particle executing simple harmonic motion. The displacement of the particle at an instant t is given by

x = A sin ωt

Velocity of the particle, v= $$\dfrac{dx}{dt}$$= Aωcos ωt

Instantaneous Kinetic Energy, K = ($$\dfrac{1}{2}$$) mv2

= ($$\dfrac{1}{2}$$) m (Aωcos ωt)2

= ($$\dfrac{1}{2}$$) m (A2ω2cos2ωt)

Average value of kinetic energy over one complete cycle

..

Answered by Pragya Singh | 1 year ago

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