| ⇦ | 
| ⇨ |
The velocity of a particle performing simple harmonic motion, when it passes through its mean position is
Options
(a) Infinity
(b) Zero
(c) Minimum
(d) Maximum
Correct Answer:
Maximum
Explanation:
ν = ω √(a² – y²), At mean position y = 0
ν = ωa = This is the maximum velocity.
Related Questions: - A straight conductor of length 0.4 m is moved with a speed of 7 ms⁻¹perpendicular
- One mole of an ideal diatomic gas undergoes a transition from A to B along a path
- A person has a minimum distance of distinct vision as 50 cm. The power of lenses
- If two bodies are projected at 30⁰ and 60⁰ respectively with the same velocity, then
- The ratio of the weight of a man in a stationary lift and when it is moving downwards
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A straight conductor of length 0.4 m is moved with a speed of 7 ms⁻¹perpendicular
- One mole of an ideal diatomic gas undergoes a transition from A to B along a path
- A person has a minimum distance of distinct vision as 50 cm. The power of lenses
- If two bodies are projected at 30⁰ and 60⁰ respectively with the same velocity, then
- The ratio of the weight of a man in a stationary lift and when it is moving downwards
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
At mean position x=0
Therefore v=w(√a^2-0^2)
Therefore v=w(√a^2)=aw
Vmax=aw