Question

A simple pendulum of length \( L \) has mass \( M \) and it oscillates freely with amplitude \( A \). At extreme position its potential energy is \( (g = \text{acceleration due to gravity}) \)

• A. \( \frac{MgA}{2L} \)
• B. \( \frac{2 MgA}{L} \)
• C. \( \frac{MgA^2}{L} \)
• D. \( \frac{MgA^2}{2L} \)

Hint:

The potential energy of a pendulum at its extreme position is proportional to the square of the amplitude and inversely proportional to the length of the pendulum.

Answer 1

The Correct Option is D

Step 1: Understand the potential energy of a pendulum.
The potential energy \( U \) of the pendulum at its extreme position is given by: \[ U = M g h \] where \( h \) is the height the pendulum rises to from its equilibrium position. Step 2: Use the geometry of the pendulum.
For small angles of oscillation, the height \( h \) at the extreme position can be approximated by: \[ h = L - L \cos(\theta) \] where \( \theta \) is the angular displacement of the pendulum. Using the small angle approximation \( \theta \approx \frac{A}{L} \), we get: \[ h \approx \frac{A^2}{2L} \] Step 3: Calculate the potential energy.
Thus, the potential energy at the extreme position is: \[ U = M g \frac{A^2}{2L} \] Step 4: Conclusion.
Thus, the potential energy at the extreme position is \( \frac{MgA^2}{2L} \), which corresponds to option (D).