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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
- An α-particle and a proton are accelerated through same potential difference
- The pair of quantities having same dimensions is
- The de-Broglie wavelength of neutron in thermal equilibrium at temperature T is
- In the reaction ₁²H+₁³H+₂⁴He+₀¹n, if the binding energies of ₁²H,₁³H and ₂⁴He
- Photons of 5.5 eV energy fall on the surface of the metal emitting photoelectrons
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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