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 MCQs

Prepare for NEET Physics Oscillations with MCQs & PYQs on NEET.GUIDE. Access free practice, previous year questions, and detailed solutions to grasp oscillatory motion thoroughly.

📚 Subject:Physics

📑 Chapter:Oscillations

🎯Topic:Velocity-Acceleration

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :2 mins

🔄Ncert Concepts :Kinematics, Equations of Motion, Relative Motion

A particle is projected vertically upwards with an initial speed 'u'. At a height 'h' above the ground, its speed is 'v'. What is the maximum height reached by the particle?

u²/2g

v²/2g

v²/2g + h

Question Tags

📚 Subject:Physics

📑 Chapter:Oscillations

🎯Topic:Velocity-Acceleration

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :2 mins

🔄Ncert Concepts :Kinematics, Equations of Motion, Projectile Motion

A simple pendulum of length $L$ oscillates with a time period $T$ . If the bob is replaced with another bob of mass four times the original, and the amplitude is halved, the new time period will be:

$2T$

$T/2$

$T$

$4T$

Question Tags

📚 Subject:Physics

📑 Chapter:Oscillations

🎯Topic:Simple Pendulum

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :1 mins

🔄Ncert Concepts :Simple Pendulum, Time Period, Oscillations

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?

$heta$

$2heta$

$heta/2$

$4heta$

Question Tags

📚 Subject:Physics

📑 Chapter:Oscillations

🎯Topic:Simple Pendulum

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :2 mins

🔄Ncert Concepts :Simple Pendulum, Energy Conservation, Oscillations

A simple pendulum with a bob of mass $m$ and length $L$ is oscillating in a lift accelerating upwards with acceleration $a$ . The effective acceleration due to gravity experienced by the pendulum is:

$g$

$g - a$

$g + a$

$\sqrt{g^2 + a^2}$

Question Tags

📚 Subject:Physics

📑 Chapter:Oscillations

🎯Topic:Simple Pendulum

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :1 mins

🔄Ncert Concepts :Simple Pendulum, Accelerated Frames, Effective Gravity

10.

A simple pendulum of length $L$ has a time period $T$ . If the length is increased by $\Delta L$ such that $\Delta L << L$ , the change in time period $\Delta T$ is approximately: