| ⇦ | 
| ⇨ |
A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β. Then, its time period of vibration will be
Options
(a) β²/α
(b) 2πβ/α
(c) β²/α²
(d) α/β
Correct Answer:
2πβ/α
Explanation:
As, we know, in Simple Harmonic Motion
Maximum acceleration of the particle, α = Aω²
Maximum velocity, β = Aω
⇒ ω = α / β
⇒ T = 2π / ω = 2πβ / α [Since, ω = 2π / T].
Related Questions: - In a circuit, L,C and R are connected in series with an alternating voltage source
- The transverse nature of electromagnetic waves is proved by which of the following?
- An ideal gas is expanding such that PT²=constant. The coefficient of volume
- A uniform force of (3i + j) newton acfs on a particle of mass 2kg
- A body is thrown upward from ground covers equal distances
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In a circuit, L,C and R are connected in series with an alternating voltage source
- The transverse nature of electromagnetic waves is proved by which of the following?
- An ideal gas is expanding such that PT²=constant. The coefficient of volume
- A uniform force of (3i + j) newton acfs on a particle of mass 2kg
- A body is thrown upward from ground covers equal distances
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