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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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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A wire of length 1 m is placed in a uniform magnetic field of 1.5 tesla at an angle
- Two springs with spring constants K₁=1500 N/m and K₂=3000 N/m
- A solid sphere, disc and solid cylinder all of the same mass and made up of same material
- The acceleration due to gravity near the surface of a planet of radius R and density d
- At 0⁰C, 15 gm of ice melts to form water at 0⁰C. The change in entropy is
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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