Question

A simple pendulum is oscillating with amplitude A and angular frequency $\omega$. At displacement $x$ from mean position, the ratio of kinetic energy to potential energy is

• A. $\frac{x^2}{A^2-x^2}$
• B. $\frac{x^2-A^2}{x^2}$
• C. $\frac{A^2-x^2}{x^2}$
• D. $\frac{A-x}{x}$

Answer 1

The correct option is C $\frac{A^2-x^2}{x^2}$

Velocity of the particle executing $SHM$ is: $v=\omega \sqrt{A^2-x^2}$, where, $x$ is the displacement of the particle from the mean position.

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

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

Potential energy of the particle, $P.E=\frac{1}{2}k{x}^2$, where $k=m{\omega }^2$

$\therefore$ $P.E=\frac{1}{2}\left(m{\omega }^2\right)x^2$ ...........(2)

From (1) and (2), we get $\Rightarrow \frac{K.E}{P.E}=\frac{A^2-x^2}{x^2}$