Question

The angular velocity and the amplitude of simple pendulum is $\omega $ and $a$ respectively. At a displacement X from mean position, if the kinetic energy is T and potential energy is V. Then ratio T to V is:

(A) ${X^2}{\omega ^2}/\left( {{a^2} - {X^2}{\omega ^2}} \right)$
(B) ${X^2}/\left( {{a^2} - {X^2}} \right)$
(C) ${a^2} - {X^2}{\omega ^2}/{X^2} - {X^2}{\omega ^2}$
(D) $\left( {{a^2} - {X^2}} \right)/{X^2}$

Answer 1

Hint:
  First start with finding the relation of the potential energy and the kinetic energy in case of simple pendulum then find the ratio of the two and try to find out which of the given options is fit in that relation, you can use the method of elimination and can eliminate the wrong option one by one.








Complete step by step solution:

Potential energy is the energy possessed by the particle when the particle is at rest.
The force acting will be $F = - kx$
Now the work done will be:
$dW = - fdx$
After solving, we get;
Total work done, $W = 1/2{\text{ }}K{\text{ }}{x^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{x^2}$
The total work done here will get stored in the form of potential energy.
So, Potential energy : $P.E. = V = 1/2{\text{ }}k{x^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{x^2}$ (equation 1)
Now, for kinetic energy:
The velocity of the particle at an instant and at a distance x from the mean position is given by:
$v = \pm \omega \sqrt {{a^2} - {x^2}} $
Kinetic energy, $K = \dfrac{1}{2}m{v^2}$
$K = \dfrac{1}{2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)$
as $k = m{\omega ^2}$
Therefore, $K.E = T = \dfrac{1}{2}k\left( {{a^2} - {x^2}} \right)$ (equation 2)
Now, $\dfrac{T}{V} = \dfrac{{{a^2} - {x^2}}}{{{x^2}}}$ (here x=X)

Hence the correct answer is Option(D).













Note:
 First find the force in case of a simple pendulum and then use the value of the force in finding the work done and finally get the potential energy. Then find the kinetic energy and finally get the ratio of the two. Here the angular velocity and the amplitude of the simple pendulum is already given in the question.