| ⇦ |
| ⇨ |
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 masses of three wires of copper are in the ratio 1:3:5 and lengths
- If the focal length of a concave mirror is 50 cm, then where should the object
- Choose the correct statement
- The kinetic energy of particle moving along a circle of radius R depends
- The half life of radioactive isotope ‘X’ is 50 years. It decays to
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The masses of three wires of copper are in the ratio 1:3:5 and lengths
- If the focal length of a concave mirror is 50 cm, then where should the object
- Choose the correct statement
- The kinetic energy of particle moving along a circle of radius R depends
- The half life of radioactive isotope ‘X’ is 50 years. It decays to
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