
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
233.1k+ 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
JEE Main 2026 Session 2 Registration Open, Exam Dates, Syllabus & Eligibility

JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

Trending doubts
Understanding Average and RMS Value in Electrical Circuits

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding Atomic Structure for Beginners

Derive an expression for maximum speed of a car on class 11 physics JEE_Main

Understanding Elastic Collisions in Two Dimensions

JEE Main Syllabus 2026: Download Detailed Subject-wise PDF

Other Pages
NCERT Solutions For Class 11 Physics Chapter 10 Thermal Properties of Matter (2025-26)

NCERT Solutions For Class 11 Physics Chapter 12 Kinetic Theory (2025-26)

Understanding Collisions: Types and Examples for Students

Define thermal expansion for alpha beta and gamma A class 11 physics JEE_Main

Happy New Year Wishes 2026 – 100+ Messages, Quotes, Shayari, Images & Status in All Languages

Valentine Week 2026 List | Valentine Week Days, Dates & Meaning

