A simple pendulum is oscillating with amplitude ' A ' and angular… - Vidyadip

MCQ

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
$\frac{x^2-A^2}{x^2}$
✓
$\frac{A^2-x^2}{x^2}$
D
$\frac{A-x}{x}$

✓

Answer

Correct option: C.

$\frac{A^2-x^2}{x^2}$

(c) : For a particle executing SHM,
Kinetic energy at displacement $x$ from mean position,
$
K E=\frac{1}{2} m \omega^2\left(A^2-x^2\right)
$
At the same time, its potential energy
$
\begin{aligned}
& P E=\frac{1}{2} m \omega^2 x^2 \\
& \text { Required ratio }=\frac{K E}{P E}=\frac{\frac{1}{2} m \omega^2\left(A^2-x^2\right)}{\frac{1}{2} m \omega^2 x^2}=\frac{A^2-x^2}{x^2}
\end{aligned}
$

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