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The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where E is the total energy)
Options
(a) E/8
(b) E/4
(c) E/2
(d) 2E/3
Correct Answer:
E/4
Explanation:
P.E. = (1/2) mω²y² ⇒ At y = a/2 ⇒ (1/2) (mω²a²/4)
Total energy (E) = P.E. at extreme position = (1/2) mω²a²
P.E. = (1/4).[(1/2) mω²a²] = E/4
Related Questions: - A 50Hz AC signal is applied in a circuit of inductance of (1/π)H and resistance 2100Ω
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Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A 50Hz AC signal is applied in a circuit of inductance of (1/π)H and resistance 2100Ω
- A circular disc of radius R is removed from a bigger circular disc of radius 2R
- A cricket ball of mass 250 g collides with a bat with velocity 10 m/s
- Two particles A and B having equal charges +6C, after being accelerated
- The component of vector A=2i+3j along the vector i+j is
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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