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 $$U,$$ then the ratio of $$T$$ to $$U$$ is
A.
$$\left( {\frac{{{a^2} - {x^2}{\omega ^2}}}{{{x^2}{\omega ^2}}}} \right)$$
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 :
Consider a particle of mass $$m,$$ executing linear $$SHM$$ with amplitude $$a$$ and constant angular frequency $$\omega .$$ Suppose $$t$$ second after starting from the mean position, the displacement of the particle is $$x,$$ which is given by
$$x = a\sin \omega t$$
So, potential energy of particle is
$$U = \frac{1}{2}m{\omega ^2}{x^2}\,......\left( {\text{i}} \right)$$
and kinetic energy of particle is
$$T = \frac{1}{2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)\,......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii)
$$\frac{T}{U} = \frac{{{a^2} - {x^2}}}{{{x^2}}}$$