⇦ | ![]() | ⇨ |
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:
- Two resistors of resistances 2Ω and 6Ω are connected in parallel.
- The body of mass m hangs at one end of a string of length l, the other end of which
- The parallel beams of monochromatic light of wavelength 4.5 x10⁻⁷ m passes through
- The potential energy of a 1 kg particle free to move along the x-axis is given by
- Two particles of masses m₁ m₂ move with initial velocities u₁ and u₂
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