Question

Derive an expression for K.E and P.E of a particle SHM.

Answer 1

We know that, Potential energy

U = \(\frac{1}{2}\) kx²

For a particle in SHM, k = mω² and x = Asin ωt

⇒ U = \(\frac{1}{2}\) mω² A² sin² ωt

Kinetic energy, k = \(\frac{1}{2}\)mv²

now, v = ωy = ωA cos ωt

⇒ K = \(\frac{1}{2}\) mω² A²cos²ωt.