Question

The kinetic energy and potential energy of a particle executing SHM with amplitude A will be equal when its displacement is

• A. A$\sqrt{2}$
• B. A / 2
• C. A/$\sqrt{2}$
• D. A/$\sqrt{2/3}$

Answer 1

Answer: (C)

A/$\sqrt{2}$

Answer 2

K.E. = $\frac{1}{2}m\omega^{2}(A^{2} - y^{2})$

P.E. = $\frac{1}{2}$mω²y²

$\because$ K.E. = P.E.

∴ $\frac{m\omega^{2}}{2}(A^{2} - y^{2})$ = $\frac{m\omega^{2}y^{2}}{2}$

∴ 2y² = A² or y = A/$\sqrt{2}$