| ⇦ | 
| ⇨ |
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 ideal heat engine works between temperature T₁=500 K and T₂=375 K.
- Which of the following statement about scalar quantity is true?
- Zener diode is used
- Radiofrequency choke uses core of
- If a nucleus ᴢXᴬ emits 9α and 5β-particles, then the ratio of total protons and neutrons
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An ideal heat engine works between temperature T₁=500 K and T₂=375 K.
- Which of the following statement about scalar quantity is true?
- Zener diode is used
- Radiofrequency choke uses core of
- If a nucleus ᴢXᴬ emits 9α and 5β-particles, then the ratio of total protons and neutrons
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