1.

Find the expression for the total energy of a particle executing SHM.

Answer»

For a SHM PE = \(\frac{1}{2}\) kx2

= \(\frac{1}{2}\) mω2 x2 = \(\frac{1}{2}\)mω A2 sin2 ωt

KE= \(\frac{1}{2}\) mv2

v =\(\frac{dx}{dt}\) = – Aω cos ωt

⇒ KE = \(\frac{1}{2}\) mA2ω2 cos2 ωt

Total energy = KE + PE = \(\frac{1}{2}\)2 A2 cos2 ωt + \(\frac{1}{2}\)2 sin2 ωt 

Total energy = \(\frac{1}{2}\)2 A2.



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