SHMMedium
Question
The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 joule. The speed of the simple pendulum bob at equilibrium position is approximately:
(Consider mass of the $bob = 20\text{ }g$ )
Options
A.0.2 m/s
B.1.41 m/s
C.14.1 m/s
D.2.0 m/s
Solution
Sol. [Key Concept: Pendulum — total energy conserved, v at mean position] At equilibrium position
Total energy = K.E
$$\begin{matrix} & \frac{1}{2}mv^{2} = 0.02 \\ & \frac{1}{2} \times 20 \times v^{2} \times 10^{- 3} = 2 \times 10^{- 2} \\ & v = \sqrt{2} \\ & v = 1.41\text{ }m/s \end{matrix}$$
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