| ⇦ | 
| ⇨ |
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: - An electron revolves in a circle at the rate of 10¹⁹ rounds per second.
- If an electron and a proton have the same de-Broglie wavelength, then the kinetic
- The refractive index of water, glass and diamond are 1.33, 1.50, 2.40 respectively
- A train of mass 2x 10⁵ kg has a constant speed of 20 m/s up a hill inclined
- A uniform wire of resistance 9Ω is joined end-to-end to form a circle. Then
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An electron revolves in a circle at the rate of 10¹⁹ rounds per second.
- If an electron and a proton have the same de-Broglie wavelength, then the kinetic
- The refractive index of water, glass and diamond are 1.33, 1.50, 2.40 respectively
- A train of mass 2x 10⁵ kg has a constant speed of 20 m/s up a hill inclined
- A uniform wire of resistance 9Ω is joined end-to-end to form a circle. Then
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