| ⇦ | 
| ⇨ |
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: - A rod of weight W is supported by two parallel knife edge A and B
- The string of a pendulum is horizontal.The mass of bob attached to it is m
- On bombarding U²³⁵ by slow neutron, 200 MeV energy is released. If the power output
- A dynamo converts
- 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
- A rod of weight W is supported by two parallel knife edge A and B
- The string of a pendulum is horizontal.The mass of bob attached to it is m
- On bombarding U²³⁵ by slow neutron, 200 MeV energy is released. If the power output
- A dynamo converts
- 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
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