Question

Two simple pendulums have (A) mass $M_1$, length $L_1$ and (B) mass $M_2$, length $L_2$. Given $M_1 = M_2$ and $L_1 = 2L_2$. If their total energies are same, then \dots

• A. amplitude of B is greater than amplitude of A.
• B. amplitude of B is smaller than amplitude of A.
• C. amplitude of both will be same.
• D. amplitude of B is twice that of A.

Hint:

Energy in a pendulum scales with $(A^2 / L)$. If you cut the string length in half ($L$ goes down), the pendulum gets "stiffer" and swings faster. To maintain the exact same total energy, its physical swing distance (amplitude) must shrink.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Total energy of a simple pendulum for small oscillations is $E = \frac{1}{2} m \omega^2 A^2$, where $\omega = \sqrt{g/L}$.

Step 2: Formula Application:
$E = \frac{1}{2} M \left( \frac{g}{L} \right) A^2 \implies E \propto \frac{A^2}{L}$ (since $M$ and $g$ are constant).

Step 3: Explanation:
For $E_A = E_B$: $\frac{A_A^2}{L_1} = \frac{A_B^2}{L_2}$. Given $L_1 = 2L_2$, so $\frac{A_A^2}{2L_2} = \frac{A_B^2}{L_2} \implies A_A^2 = 2 A_B^2$. Wait, looking at the math: $A_B^2 = \frac{1}{2} A_A^2$, so $A_B < A_A$. However, let's check the energy formula $E = mgh$. For same energy, $h_1 = h_2$. $h = L(1-\cos\theta) \approx L \frac{\theta^2}{2} = \frac{A^2}{2L}$. If $A^2/2L$ is constant, then $A^2 \propto L$. Since $L_1 > L_2$, $A_A > A_B$. Therefore, amplitude of B is smaller than A.

Step 4: Final Answer:
Amplitude of B is smaller than amplitude of A.