Question

When the bob of a pendulum is at the mean position (minimum displacement) of its motion, its total energy is
(A) All potential
(B) Zero
(C) All kinetic
(D) Partly kinetic partly potential

Answer 1

Hint : To solve this question, we have to use the conservation of the total mechanical energy of the bob of the pendulum. For this we have to consider the potential energy and the kinetic energy of the bob at an arbitrary point of the vertical circle along which it moves.

Formula used: The formulae used to solve this question is given by
$K = \dfrac{1}{2}m{v^2}$ , here $K$ is the kinetic energy of a body of mass $m$ moving with velocity $v$ .
$U = mgh$ , here $U$ is the potential energy of a body of mass $m$ situated at a height $h$ above the reference point.

Complete step by step answer
The bob of a pendulum, while performing oscillations, moves along a part of the vertical circle. Let us consider the bob of a pendulum at an arbitrary point located at a height of $h$ above the mean position, as shown below.

Let the velocity of the bob at this position be $v$ . So the total mechanical energy of the bob at this position is equal to the sum of the kinetic energy and the potential energy. The kinetic energy is given by
$K = \dfrac{1}{2}m{v^2}$ …………………………...(i)
Also, the potential energy is
$U = mgh$ …………………………...(ii)
So the total mechanical energy of the bob is
$E = K + U$
From (i) and (ii)
$E = \dfrac{1}{2}m{v^2} + mgh$…………………………...(iii)
We know that the force of gravity is a conservative force, in the influence of which the bob is moving. Therefore the total mechanical energy of the bob remains conserved at each point of the vertical circle.
As the height is being measured from the mean position, so at the mean position the height $h = 0$ . Substituting this in (iii) we have
$E = \dfrac{1}{2}m{v^2} + mg\left( 0 \right)$
$\Rightarrow E = \dfrac{1}{2}m{v^2}$
As we can see from the above expression of the total energy of the bob that it is all kinetic.
Hence, the correct answer is option C.

Note
We may argue that along with the force of gravity, the bob is also acted upon by the force of tension. But this force acts along the string of the pendulum, which is in the radial direction and therefore is perpendicular to the motion of the bob. Hence, it cannot do any work on the bob to change its mechanical energy. That’s why we have not taken it into consideration.