Question

The angular frequency and the amplitude of a simple pendulum are $\omega$ and $a$ respectively. If the pendulum is displaced through a distance $x$ from the mean position then the ratio of its kinetic energy (K) and potential energy (U) will be

• A. $\frac{K}{U}=\frac{a^2-x^2{\omega }^2}{x^2{\omega }^2}$
• B. $\frac{K}{U}=\frac{a^2-x^2}{x^2}$
• C. $\frac{K}{U}=\frac{x^2}{a^2-x^2}$
• D. $\frac{K}{U}=\frac{x^2{\omega }^2}{a^2-{\omega }^2{x}^2}$

Answer 1

Answer: (B)

$\frac{K}{U}=\frac{\frac{1}{2}m{\omega }^2(a^2-{\omega }^2)}{\frac{1}{2}m{\omega }^2{x}^2}$
$=\frac{a^2-x^2}{x^2}$