The Question is from neet-physics oscillations-waves , and it was created by the author P.NARASIMHA

Question

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

P.NARASIMHA · Accepted Answer

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

P.NARASIMHA · Answer

$	heta$

P.NARASIMHA · Answer

$2	heta$

P.NARASIMHA · Answer

The correct Option is "$	heta/2$"

P.NARASIMHA · Answer

$4	heta$

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