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

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.