Question

Two simple pendulums, A of mass \(M_{1}\) and length \(L_{1}\), and B of mass \(M_{2}\) and length \(L_{2}\).
\(M_{1}=M_{2}\) and \(L_{1}=2L_{2}\). If vibrational energies of both are same then which one of the following is correct?

• A. Amplitude of B greater than A
• B. Amplitude of B is smaller than A
• C. Amplitude of both is equal
• D. Amplitude of A is twice that of B

Hint:

The energy of a pendulum in small oscillations is proportional to mass times amplitude squared over length. Equate the energies and compare amplitudes.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For small amplitude \(A\), the energy is \(E=\frac{1}{2}m\omega^{2}A^{2}\) with \(\omega^{2}=\frac{g}{L}\), so \(E=\frac{mgA^{2}}{2L}\).

Step 2: Equate the Energies
With \(M_{1}=M_{2}\) and \(E_{1}=E_{2}\): \(\dfrac{A_{1}^{2}}{L_{1}}=\dfrac{A_{2}^{2}}{L_{2}}\).

Step 3: Use the Lengths
\(L_{1}=2L_{2}\), so \(A_{1}^{2}=2A_{2}^{2}\) and \(A_{1}=\sqrt{2}A_{2}\).

Step 4: Compare the options
\(A_{2}\lt A_{1}\), so the amplitude of B is smaller than that of A.

Final Answer:
The amplitude of B is smaller than that of A. \[ \boxed{\text{(B) }\text{Amplitude of B is smaller than A}} \]