2026 OscillationsTotal mechanical energy and speed at equilibrium in SHM Medium Question 3 of 63 in Oscillations

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 g)

A 1.41 m/s
B 14.1 m/s
C 0.2 m/s
D 2.0 m/s

The answer is A.

Solution

Total energy $= \frac{1}{2}KA^2 = \frac{1}{2}m\omega^2A^2 = 0.02$ J At equilibrium, all energy is kinetic: $\frac{1}{2}mv_{max}^2 = 0.02$ J $v_{max}^2 = \frac{0.04}{20 \times 10^{-3}} = 2$ $v_{max} = 1.41$ m/s

Oscillations in past papers

63 questions from this chapter have appeared across 26 exam years.