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A particle executes simple harmonic oscillations with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
Options
(a) T/8
(b) T/12
(c) T/2
(d) T/4
Correct Answer:
T/12
Explanation:
Displacement from the mean position y = a sin(2π / T) t
According to the problem y = a/2
a/2 = a sin (2π / T) t
⇒ π / 6 = (2π / T) t ⇒ t = T / 12
This is the minimum time taken by the particle to travel half of the amplitude from the equilibrium position.
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Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If capillary experiment is performed in vaccum, then a liquid under the condition
- An atom bomb weighing 1 kg explodes releasing 9×10¹³ J of energy.
- A current of 2 A is made to flow through a coil which has only one turn. The magnetic
- Steam at 100⁰C is passed into 20 g of water at 10°C. When water acquires a temperature
- The particle executing simple harmonic motion has a kinetic energy K₀cos²ωt.
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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