| ⇦ |
| ⇨ |
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: - The inductive reactance of an inductor of 1/π henry at 50 Hz frequency is
- Which of the following particles will have minimum frequency of revolution
- A man goes 10 m towards north, then 20m towards east, the displacement is
- A body is hangfing from a rigid support by an inextensible string of length
- A balloon with mass m is descending down with an acceleration a
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The inductive reactance of an inductor of 1/π henry at 50 Hz frequency is
- Which of the following particles will have minimum frequency of revolution
- A man goes 10 m towards north, then 20m towards east, the displacement is
- A body is hangfing from a rigid support by an inextensible string of length
- A balloon with mass m is descending down with an acceleration a
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Leave a Reply