Question

Deduce an expression for total energy of particle performing S.H.M

Answer 1

At the mean position, the velocity of the particle in S.H.M. is maximum and displacement is minimum, that is, x=0. Therefore, P.E. =1/2 K x2 = 0 and K.E. = 1/2 k ( a2 – x2) = 1/2 k ( a2 – o2) = 1/2 ka2. Thus, the total energy in simple harmonic motion is purely kinetic.

Answer 2

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