Question
The angular velocity and the amplitude of a simple pendulum is $$\omega $$ and $$a$$ respectively. At a displacement $$x$$ from the mean position if its kinetic energy is $$T$$ and potential energy is $$V,$$ then the ratio of $$T$$ to $$V$$ is
A.
$$\frac{{\left( {{a^2} - {x^2}{\omega ^2}} \right)}}{{{x^2}{\omega ^2}}}$$
B.
$$\frac{{{x^2}{\omega ^2}}}{{\left( {{a^2} - {x^2}{\omega ^2}} \right)}}$$
C.
$$\frac{{\left( {{a^2} - {x^2}} \right)}}{{{x^2}}}$$
D.
$$\frac{{{x^2}}}{{\left( {{a^2} - {x^2}} \right)}}$$
Answer :
$$\frac{{\left( {{a^2} - {x^2}} \right)}}{{{x^2}}}$$
Solution :
$$\eqalign{
& P.E.,\,V = \frac{1}{2}m{\omega ^2}{x^2}\,{\text{and}}\,K.E.,\,T = \frac{1}{2}m{\omega ^2}\left( {{a^2} - {x^2}} \right) \cr
& \therefore \frac{T}{V} = \frac{{{a^2} - {x^2}}}{{{x^2}}} \cr} $$