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}}} $