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)
- A 0.2 m/s
- B 1.41 m/s
- C 14.1 m/s
- D 2.0 m/s
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.
\(E=\frac{1}{2} m v^2\)
\(0.02=\frac{1}{2} \times 0.02 \times v^2\)
\(v^2=2\)
\(v=\sqrt{2} \approx 1.41 m / s\).
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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