Question

If length of oscillating simple pendulum is made \(\left(\frac{1}{3}\right)\) times at a place keeping amplitude same then its total energy will be

• A. \(2\) times
• B. \(3\) times
• C. \(6\) times
• D. \(4\) times

Hint:

Total energy is half m omega squared A squared, and omega squared equals g over L for a simple pendulum.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Total energy in SHM is \(E=\frac{1}{2}m\omega^{2}A^{2}\).

Step 2: Substitute omega
For a simple pendulum \(\omega^{2}=\frac{g}{L}\), so \(E=\frac{mgA^{2}}{2L}\).
For fixed \(A\), \(E\propto\frac{1}{L}\).

Step 3: Calculation
When \(L\) becomes \(\frac{L}{3}\), \(E'=3E\).

Step 4: Compare the options
The energy becomes 3 times, option (B).

Final Answer:
The total energy becomes 3 times the original. \[ \boxed{\text{(B) }3\text{ times}} \]