A simple pendulum performs simple harmonic motion about x = 0 with an amplitude

A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at x = a/2 will be:

Options

(a) πa / T
(b) 3π²a / T
(c) πa√3 / T
(d) πa√3 / 2T

Correct Answer:

πa√3 / T

Explanation:

Speed v = ? √(a² – x²) , x = a / 2
v = ? √(a² – a² / 4) = ? √(3a² / 4) = 2π / T . a√3 / 2
πa√3 / T

admin:

Related Questions

  1. An electron is moving in a circular path under the influence of a transverse magnetic
  2. A particle with charge q is moving along a circle of radius R with unoform speed v.
  3. Which of the following particles will have minimum frequency of revolution
  4. When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid
  5. The excitation potential of hydrogen atom in the first excited state is