NEET2026Physics-Simple Harmonic Motion
NEET 2026 Physics Energy in SHM MCQ Question
Type: MCQ-numerical-Medium-Class 11
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
0.2 m/s
B
1.41 m/s
C
14.1 m/s
D
2.0 m/s
Correct Answer
Option B
Detailed Explanation
At equilibrium position, the total energy is equal to the kinetic energy (K.E). Using the formula K.E = 1/2 mv², we can solve for v: 1/2 * 20 * 10⁻³ * v² = 0.02. Solving gives v = √2, which is approximately 1.41 m/s.
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