Question

A bob of a simple pendulum, which is initially at rest at its mean position, is now raised to a vertical height of $1.8\text{cm}$ and released from rest. The velocity of the bob at the mean position is

• A. $6\text{m/s}$
• B. $36\text{m/s}$
• C. $0.6\text{m/s}$
• D. $6\text{cm/s}$

Answer 1

The correct option is C $0.6\text{m/s}$

Initial kinetic energy is zero as the body is initially at rest and the final potential energy is zero considering the mean position as a reference point

Using law of conservation of energy
$K_i+U_i=K_f+U_f$
$0+mgh=\frac{1}{2}m{v}^2+0$
$\Rightarrow v^2=2gh$
$\Rightarrow v=\sqrt{2gh}$
$=\sqrt{2\times 10\times 1.8\times {10}^{-2}}=\sqrt{36\times {10}^{-2}}=0.6\text{m/s}$