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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
- Which of the following rods of same material undergoes maximum elongation
- In atom bomb, the reaction which occurs is
- If A=3i+4j and B=7i+24j the vector having the same magnitude as B and parallel to A is
- Einstein’s photoelectric equation is
- Two identical long conducting wires AOB and COD are placed at right angle
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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