The bob of a simple pendulum is released when the string is horizontally straight. If E is kinetic energy of bob when string makes an angle 60° with vertical, then its kinetic energy at mean position is:
A) $\sqrt{2}$​E
B) $\sqrt{3}$​E
C) $2$E
D) $2\sqrt{2}$​E

Hint: When a pendulum undergoes oscillatory motion, energy is being converted from potential energy and kinetic energy

Complete step by step solution:
K.E.gain = P.E.loss
Work Energy Theorem: K.E is equal to work done by gravity
at $60^\circ$ with vertical,
$\Rightarrow$ Potential energy at horizontal position is the work done by gravity
$\Rightarrow$ ${W.D.}_g$ ​= $\dfrac{{{\rm{mgR}}}}{{\rm{2}}}$ = E
$\Rightarrow $ Mean position of the pendulum is it's initial position.
$\therefore$ At, $90^\circ$
${W.D.}​_g$ = mgR = 2E

Additional information:
A simple pendulum consists of a relatively large object hung by a string from a stable support. It usually hangs vertically in its equilibrium position. The huge object is affectionately known as the Bob of the Pendulum. When Bob is displaced from the imbalance and then released, it begins to vibrate back and forth about his steady balance position. The motion is regular and repetitive, an example of periodic motion.

Note: An object that is vibrating is acted upon by a restoring force. The retrieval force causes the equilibrium object to move slowly as it moves away from the equilibrium position and increases in speed as it approaches the equilibrium position. It is the recovery energy that is responsible for the vibration.