Question

A particle is executing simple harmonic motion with amplitude of 10 cm. At what distance from the mean position the potential and kinetic energies of the particle will be equal? (Assume that the minimum potential energy is zero)

• A. cm.
• B. cm.
• C. cm.
• D. cm.

Answer 1

The correct option is B cm.

The potential energy in SHM is U $=\frac{1}{2}m{\omega }^2{x}^2$
The kinetic energy in SHM is K $=\frac{1}{2}m{\omega }^2(A^2-x^2)$
Equating $\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\omega }^2(A^2-x^2)\Rightarrow x^2=A^2$
$\therefore x=\frac{A}{\sqrt{2}}=\frac{10}{\sqrt{2}}=5\sqrt{2}cm.$