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. A point source is kept at a distnace 1000m has an illumination I. To change the illumination
  2. The liquid-crystal phase of a matter is called
  3. The dimensional formula of Plancks’s constant and angular momentum are
  4. A proton of mass m and charge q is moving in a plane with kinetic energy E. If there exists
  5. If the angular momentum of any rotating body increase by 200 % , then the increase