| ⇦ |
| ⇨ |
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: - Light of wavelength λ is incident on slit of width d. The resulting diffraction
- The most important characteristic of electron in the production of X-rays is
- The radius of orbit of a planet is two times that of the earth. The time period
- A body is moving with velocity 30 m/s towards east. After 10 seconds its velocity
- A susceptibility of a certain magnetic material is 400. What is the class
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Light of wavelength λ is incident on slit of width d. The resulting diffraction
- The most important characteristic of electron in the production of X-rays is
- The radius of orbit of a planet is two times that of the earth. The time period
- A body is moving with velocity 30 m/s towards east. After 10 seconds its velocity
- A susceptibility of a certain magnetic material is 400. What is the class
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