Oscillations Ques 33
33. 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
[1991]
(a) $\frac{(a^{2}-x^{2} \omega^{2})}{x^{2} \omega^{2}}$
(b) $\frac{x^{2} \omega^{2}}{(a^{2}-x^{2} \omega^{2})}$
(c) $\frac{(a^{2}-x^{2})}{x^{2}}$
(d) $\frac{x^{2}}{(a^{2}-x^{2})}$
Show Answer
Answer:
Correct Answer: 33.(c)
Solution:
- (c) P.E., $V=\frac{1}{2} m \omega^{2} x^{2}$
and K.E., $T=\frac{1}{2} m \omega^{2}(a^{2}-x^{2})$
$\therefore \frac{T}{V}=\frac{a^{2}-x^{2}}{x^{2}}$
BETA
