| ⇦ | 
| ⇨ |
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: - Two forces of magnitude 8N and 15N respectively act at a point so as to make the resultant force
- The sun light reaches us as white and not as its components because
- When light of wavelength 300 nm falls on a photoelectric emitter,
- X-ray beam can be deflected by
- The value of workdone for rotating a magnet of magnetic moment M by an angle θ in external
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two forces of magnitude 8N and 15N respectively act at a point so as to make the resultant force
- The sun light reaches us as white and not as its components because
- When light of wavelength 300 nm falls on a photoelectric emitter,
- X-ray beam can be deflected by
- The value of workdone for rotating a magnet of magnetic moment M by an angle θ in external
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