⇦ | ![]() | ⇨ |
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.
Related Questions:
- The bandgap in Germanium and silicon, in eV, are respectively
- α- particles, β – particles and γ – rays are all having same energy.
- The range of voltmeter is 10V and its internal resistance is 50Ω.
- A small part of the rim of a fly wheel breaks off while it is rotating at a constant
- If two vectors are equal in magnitude and their resultant is also equal in magnitude
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply