| ⇦ | 
| ⇨ |
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 boat of anchor is rocked by waves whose crests are 100 m apart and velocity
- A stone is dropped from a height h. It hits the ground with a certain momentum P. If the same stone is dropped from a height 100
- A proton is moving in a uniform magnetic field B in a circular path of radius a
- The excitation potential of hydrogen atom in the first excited state is
- An insulated container of gas has two champers seperated by an insulating partition.
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A boat of anchor is rocked by waves whose crests are 100 m apart and velocity
- A stone is dropped from a height h. It hits the ground with a certain momentum P. If the same stone is dropped from a height 100
- A proton is moving in a uniform magnetic field B in a circular path of radius a
- The excitation potential of hydrogen atom in the first excited state is
- An insulated container of gas has two champers seperated by an insulating partition.
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