| ⇦ | 
| ⇨ |
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: - The equivalent resistance of two resistors connected in series is 6 Ω
- The energy band gap is maximum in
- A radiactive substance emits 100 beta particles in the first 2s and 50 beta
- Two waves represented by following equations are travelling in same medium
- The apparent weight of the body of mass 10 kg when it is travelling upwards with
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The equivalent resistance of two resistors connected in series is 6 Ω
- The energy band gap is maximum in
- A radiactive substance emits 100 beta particles in the first 2s and 50 beta
- Two waves represented by following equations are travelling in same medium
- The apparent weight of the body of mass 10 kg when it is travelling upwards with
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