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)

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Correct answer: a) 1.41 m/s. Total energy (1/2)KA² = 0.02 J (1/2)mω²A² = 0.02 J (1/2)m${V}_{max}^{2}$ = 0.02 J ${V}_{max}^{2}$ = 0.04 / (20 × 10 ⁻³ ) = 2 V max = 1.41 m/s

The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 joule. The speed of the simple pendulum …

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