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NEET · Physics · STD 11 - 13. 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)

  1. A 0.2 m/s
  2. B 1.41 m/s
  3. C 14.1 m/s
  4. D 2.0 m/s
Verified Solution

Answer & Solution

Correct Answer

(B) 1.41 m/s

Step-by-step Solution

Detailed explanation

(B) 1.41 m/s
Total energy of the simple pendulum is given as E = 0.02 J.
Mass of the bob, m = 20 g = 0.02 kg.
At the equilibrium position, the potential energy is zero and the total energy is purely kinetic.
\(E=\frac{1}{2} m v^2\)
\(0.02=\frac{1}{2} \times 0.02 \times v^2\)
\(v^2=2\)
\(v=\sqrt{2} \approx 1.41 m / s\).
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