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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 = \(\frac{1}{2}\)mω2 A2 cos2 ωt + \(\frac{1}{2}\)mω2 sin2 ωt Total energy = \(\frac{1}{2}\) mω2 A2. |
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