| ⇦ | 
| ⇨ |
The particle executing simple harmonic motion has a kinetic energy K₀ cos²ωt. The maximum values of the potential energy and the total energy are respectively
Options
(a) K₀/2 and K₀
(b) K₀ and 2 K₀
(c) K₀ and K₀
(d) 0 and 2 K₀
Correct Answer:
K₀ and K₀
Explanation:
Kinetic energy + Potential energy = total energy
When kinetic energy is maximum, potential energy is zero and vice versa.
.·. Maximum potential energy = total energy
0 + K₀ = K₀
(K.E + P.E = Total energy) .
Related Questions: - Two particles P and Q start from origin and execute simple harmonic motion along X-axis
- ₉₂U²³⁴ has 92 protons and 238 nucleons. It decays by emitting an alpha particle
- Find out the angle of projection, if range is 4 times of maximum height
- In a tangent galvanometer, a current of 0.1 A produces a deflection of 30⁰.
- A boat taken 2h to travel 8km and back in still water.If the velocity of water 4 km/h
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two particles P and Q start from origin and execute simple harmonic motion along X-axis
- ₉₂U²³⁴ has 92 protons and 238 nucleons. It decays by emitting an alpha particle
- Find out the angle of projection, if range is 4 times of maximum height
- In a tangent galvanometer, a current of 0.1 A produces a deflection of 30⁰.
- A boat taken 2h to travel 8km and back in still water.If the velocity of water 4 km/h
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