| ⇦ |
| ⇨ |
For a simple pendulum performing simple harmonic motion, the time period is plotted against its length. The curve will be
Options
(a) straight line
(b) circle
(c) parabola
(d) ellipse
Correct Answer:
parabola
Explanation:
Time period of a simple pendulum performing simple harmonic motion is given by
T = 2π √(l/g) where l is the length of the pendulum
or, T² = 4π²l/g or, T² ∝ l
Graph between T and l is a parabola.
Related Questions: - A proton carrying 1 MeV kinetic energy is moving in a circular path of radius
- An α-particle and a proton travel with the same velocity in a magnetic field
- A uniform rod of length l is placed with one end in contact with the horizontal
- A single slit Fraunhogger diffraction pattern is formed with white light.
- X-ray beam can be deflected by
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A proton carrying 1 MeV kinetic energy is moving in a circular path of radius
- An α-particle and a proton travel with the same velocity in a magnetic field
- A uniform rod of length l is placed with one end in contact with the horizontal
- A single slit Fraunhogger diffraction pattern is formed with white light.
- X-ray beam can be deflected by
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