| ⇦ | 
| ⇨ |
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 small part of the rim of a fly wheel breaks off while it is rotating at a constant
- A pan is suspended from a steel wire. Weights are slowly put on the pan
- A conducting sphere of radius R is given a charge Q. The electric potential
- The dimension of 1/2ε₀E², where ε₀ is permittivity of free space and E is electric field
- The magnetic susceptibility of a material of a rod is 299. Permeability of vacuum
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A small part of the rim of a fly wheel breaks off while it is rotating at a constant
- A pan is suspended from a steel wire. Weights are slowly put on the pan
- A conducting sphere of radius R is given a charge Q. The electric potential
- The dimension of 1/2ε₀E², where ε₀ is permittivity of free space and E is electric field
- The magnetic susceptibility of a material of a rod is 299. Permeability of vacuum
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