Question

A simple pendulum with a length of 2.23 m and a mass of 6.74 kg is given an initial speed of 2.06 m/s at its equi- librium position. Assume it undergoes simple harmonic motion. Determine (a) its period, (b) its total energy, and (c) its maximum angular displacement.

Answer 1

Answer:

a)   T = 2.997 s

b)   K = 14.3 J

c)   φ = 0.444 rad

Explanation:

a) Determine its period

  The pendulum simple’s period is:

 

  T = 2π$\sqrt{\frac{l}{g} }$

        Where l: Pendulum’s length

                    g = 9.8 m/s2

  T = 2π$\sqrt{\frac{2.23}{9.8} }$

  T = 2.997 s

b) Total energy

  Initially his total energy is kinetic

  K = $\frac{mv^{2} }{2}$

  K = $\frac{(6.74)(2.06)^{2} }{2}$

  K = 14.3 J

c) Maximum angular displacement

  φ = $cos^{-1}(1-\frac{E}{mgl} )$

  φ = $cos^{-1}(1-\frac{14.3}{(6.74)(9.8)(2.23)} )$

  φ = 0.444 rad