Question

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

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

Hint:

For a simple pendulum, the potential energy at the 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 A

Step 1: Formula for potential energy in a simple pendulum.
The potential energy at the extreme position (where the amplitude is \( A \)) is given by: \[ PE = mgh \] where \( h \) is the height raised by the pendulum. For small oscillations, \( h = L(1 - \cos \theta) \), where \( \theta \) is the angle corresponding to the extreme position.
Step 2: Using the amplitude.
For maximum displacement \( A \), the potential energy becomes: \[ PE = \frac{mgA^2}{2L} \]
Step 3: Conclusion.
Thus, the correct answer is (A) \( \frac{mgA^2}{2L} \).