Question

Prove the law of conservation of energy for a particle performing simple harmonic motion. Hence graphically show the variation of kinetic energy and potential energy w.r.t. instantaneous displacement.

Answer 1

Let the particle be at the mean position initially.

Thus it's displacement can be written as $x=A\sin wt$

Potential energy of the particle performing SHM $P.E=\frac{1}{2}k{x}^2$

Kinetic energy of the particle $K.E=\frac{1}{2}m{v}^2$

Or $K.E=\frac{1}{2}m(w\sqrt{A^2-x^2}{)}^2=\frac{1}{2}m{w}^2(A^2-x^2)$

Using $k=m{w}^2$, we get $K.E=\frac{1}{2}k(A^2-x^2)$

Total energy of the particle $T.E=K.E+P.E$

$\therefore$ $T.E=\frac{1}{2}k(A^2-x^2)+\frac{1}{2}k{x}^2$

$⟹$ $T.E=\frac{1}{2}k{A}^2=$ Constant

Hence total energy of the particle performing SHM is conserved.