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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].
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Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Ring, hollow ring and solid sphere are rolled down from inclined plane
- A capacitor of 2.5 μF is charged through a resistor of 4 MΩ. In how much time
- When an object is placed 40cm from a diverging lens, its virtual image is formed 20 cm
- In an inductor when current changes from 2 A to 18 A in 0.05 sec, the e.m.f. induced
- A string in musical instrument is 50 cm long and its fundamental frequency is 800 Hz.
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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