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{ ( 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 )} $

Answer 1

The Correct Option is C

To solve this problem, we must determine the ratio of the kinetic energy (T) to the potential energy (V) of a simple pendulum at a given displacement \(x\) from the mean position. Given the angular velocity \(\omega\) and amplitude \(a\), we start by finding expressions for the kinetic and potential energy of the pendulum.

1. The total energy (E) of a simple pendulum undergoing simple harmonic motion is given by:
   E = \frac{1}{2}m\omega^2a^2
2. At displacement \(x\) from the mean position, the potential energy (V) is given by:
   V = \frac{1}{2}m\omega^2x^2
3. The kinetic energy (T) can be derived using the concept that total mechanical energy is the sum of kinetic and potential energies:
   T = E - V = \frac{1}{2}m\omega^2a^2 - \frac{1}{2}m\omega^2x^2
   T = \frac{1}{2}m\omega^2(a^2 - x^2)
4. To find the ratio of kinetic energy (T) to potential energy (V), we compute:
   \frac{T}{V} = \frac{\frac{1}{2}m\omega^2(a^2 - x^2)}{\frac{1}{2}m\omega^2x^2}
   On simplifying,
   \frac{T}{V} = \frac{a^2 - x^2}{x^2}

Thus, the correct answer is the expression: \frac{(a^2 - x^2)}{x^2}.

Therefore, the correct option is:

\frac{(a^2 - x^2)}{x^2}