
If the bob of a simple pendulum starts from x = -A at t=0, then, at x = +A, the values of velocity and acceleration will be
(A) v=max and a=0
(B) v=0 and a=0
(C) v=0 and a=max
(D) v=max and a=max
Answer
229.8k+ views
Hint At mean position velocity is max as the kinetic energy is max and acceleration is angular which is zero because angle made by the pendulum with the vertical is zero.
Complete step-by-step solution
In an oscillatory motion of a simple pendulum, we know that the kinetic energy of the bob is max at the mean position. Kinetic energy is given by
$K = \dfrac{1}{2}m{v^2}$
As kinetic energy is max only due to velocity, at mean position the velocity is max.

The weight of the bob W=mg is resolved into vector components, the cos component is balanced by the tension of the string and sin component is along the circular path of the pendulum. As the motion is part of the circular motion the acceleration is given by
$\alpha = - \dfrac{{mgL}}{I}\sin \theta $
At mean position θ is zero therefore,
$
\sin 0 = 0 \\
\alpha = 0 \\
$
Hence at mean position, velocity is max and acceleration is zero
correct option is A
Note Kinetic energy is zero at extreme position as velocity is zero and potential energy is maximum. Potential energy is minimum at the mean position which is also the lowest position.
If the motion of the pendulum makes a very small angle, then this motion can be approximated to Simple Harmonics Motion. One will need this information to solve SHM and pendulum involving questions.
Complete step-by-step solution
In an oscillatory motion of a simple pendulum, we know that the kinetic energy of the bob is max at the mean position. Kinetic energy is given by
$K = \dfrac{1}{2}m{v^2}$
As kinetic energy is max only due to velocity, at mean position the velocity is max.

The weight of the bob W=mg is resolved into vector components, the cos component is balanced by the tension of the string and sin component is along the circular path of the pendulum. As the motion is part of the circular motion the acceleration is given by
$\alpha = - \dfrac{{mgL}}{I}\sin \theta $
At mean position θ is zero therefore,
$
\sin 0 = 0 \\
\alpha = 0 \\
$
Hence at mean position, velocity is max and acceleration is zero
correct option is A
Note Kinetic energy is zero at extreme position as velocity is zero and potential energy is maximum. Potential energy is minimum at the mean position which is also the lowest position.
If the motion of the pendulum makes a very small angle, then this motion can be approximated to Simple Harmonics Motion. One will need this information to solve SHM and pendulum involving questions.
Recently Updated Pages
States of Matter Chapter For JEE Main Chemistry

Circuit Switching vs Packet Switching: Key Differences Explained

Mass vs Weight: Key Differences Explained for Students

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Trending doubts
JEE Main 2026: Admit Card Out, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

