Two simple pendulums, A and B, have lengths $L$ and $4L$ respectively. If the bob of pendulum A is displaced by a small angle $\theta$ and released, what angle should the bob of pendulum B be displaced by to have the same maximum velocity as pendulum A?

## CORE INFORMATION\n✓ **Answer**: C\n## CONCEPT FRAMEWORK\n **Primary Concept**: Conservation of energy in simple pendulum\n **Related Topics**: Simple harmonic motion, potential and kinetic energy\n **Prerequisites**: Understanding of simple pendulum motion, potential energy, kinetic energy, and small angle approximation.\n## SOLUTION PATHWAY\n**Given Information**:\n Pendulum A: Length = $L$, initial angle = $\theta$\n Pendulum B: Length = $4L$, initial angle = $\theta_B$ (to be determined)\n Maximum velocity is the same for both pendulums.\n [Important conditions]: Small angle approximation is valid. \n**Method & Steps**:\n 1. Find the maximum velocity of pendulum A using energy conservation.\n Reasoning: At the lowest point, all potential energy is converted into kinetic energy.\n Formula/Rule: $mgh = \frac{1}{2}mv_{max}^2$, where $h = L(1-\cos\theta) \approx \frac{L\theta^2}{2}$ for small $\theta$.\n $mgL\frac{\theta^2}{2} = \frac{1}{2}mv_{max,A}^2$\n $v_{max,A}^2 = gL\theta^2$\n $v_{max,A} = \theta\sqrt{gL}$\n 2. Find the maximum velocity of pendulum B using energy conservation.\n Reasoning: Similar to step 1, but with length $4L$ and angle $\theta_B$.\n Formula/Rule: Same as step 1.\n $mg(4L)\frac{\theta_B^2}{2} = \frac{1}{2}mv_{max,B}^2$\n $v_{max,B}^2 = 4gL\theta_B^2$\n $v_{max,B} = 2\theta_B\sqrt{gL}$\n 3. Equate the maximum velocities and solve for $\theta_B$.\n Reasoning: We are given that $v_{max,A} = v_{max,B}$.\n Working: $\theta\sqrt{gL} = 2\theta_B\sqrt{gL}$\n $\theta = 2\theta_B$\n $\theta_B = \frac{\theta}{2}$\n Result: The angle for pendulum B should be $\frac{\theta}{2}$.\n Verification: Substitute $\theta_B = \frac{\theta}{2}$ into the expression for $v_{max,B}$: $v_{max,B} = 2(\frac{\theta}{2})\sqrt{gL} = \theta\sqrt{gL}$, which is equal to $v_{max,A}$.\n ## OPTION ANALYSIS\nA] :x: Incorrect. If the angles were equal, the maximum velocity would not be the same.\nB] :x: Incorrect. This would lead to a much larger maximum velocity for pendulum B.\n[C] ✓: Correct. This ensures the maximum velocities are equal due to the difference in lengths.\nD] :x: Incorrect. This would lead to a much larger maximum velocity for pendulum B.\n ## LEARNING AIDS\n **Common Mistakes**:\n 1.Forgetting to account for the length difference in the potential energy calculation: Remember that the potential energy depends on both the angle and the length of the pendulum.\n 2.Incorrectly applying the small angle approximation: The approximation $\cos\theta \approx 1 - \frac{\theta^2}{2}$ is crucial for simplifying the energy equation.\n **Quick Tips**:\n 1.Focus on the energy conservation principle: This is the key to solving this problem efficiently.\n 2.Remember the small angle approximation: This simplifies the calculations significantly.

The correct Option is "$\theta/2$"

NEET Physics Oscillations Simple Pendulum MCQs

Prepare for NEET Physics Oscillations (Simple Pendulum) with MCQs & PYQs on NEET.GUIDE. Access free practice, previous year questions, and expert solutions to analyze period and motion of a simple pendulum.