NEET 2026, Physics Q4: Simple Harmonic Motion And Oscillation

NEET2026Physics

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

Solution

At the equilibrium position, all the energy is kinetic, so

$\frac{1}{2} m v_{ma x}^{2} = 0.02$ J

With $m = 20$ g $= 20 \times 1 0^{- 3}$ kg,

$v_{ma x}^{2} = \frac{2 \times 0.02}{20 \times 1 0 ^{- 3}} = 2$

$v_{ma x} = 2 \approx 1.41$ m/s.

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NEET 2026, Physics Q4: Simple Harmonic Motion And Oscillation

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