| ⇦ | 
| ⇨ |
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: - What is the energy released by fission of 1 g of U²³⁵?
- One mole of an ideal diatomic gas undergoes a transition from A to B along a path
- The most important characteristic of electron in the production of X-rays is
- In Wheatstone bridge, three resistors P,Q and R are conncected in three arms
- A silver wire of radius 0.1 cm carries a current of 2A. If the charge density in silver
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- What is the energy released by fission of 1 g of U²³⁵?
- One mole of an ideal diatomic gas undergoes a transition from A to B along a path
- The most important characteristic of electron in the production of X-rays is
- In Wheatstone bridge, three resistors P,Q and R are conncected in three arms
- A silver wire of radius 0.1 cm carries a current of 2A. If the charge density in silver
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