A simple pendulum consists of a 650cm long rod with a 320g m

A simple pendulum consists of a 65.0-cm long rod with a 320-g mass at the end of it. The motion is damped due to friction at the pivot. The rod is set in oscillation by displacing it 15° from its equilibrium position and releasing it . After 10.0 s, the amplitude has reduced to 5.00°. Find the ‘damping constant’ b, and the time for the original amplitude to be reduced to 2.00°. [7.03e–002 kg/s, 1.83e+001 s]

Solution

For damped oscillation, the pendulum oscillates following the equation:

= ₀ exp (-b/2m t) cos (t+)

(assuming small angles ; note that = 15° = 0.262 rad, while sin 15.0° = 0.259 rad)

At t=0 the amplitude is           (0)= ₀ exp (-b/2m t) = ₀ = 15°

At t=10s amplitude is              (10)= ₀ exp (-b/2m t) = 5°

15°exp (-b/2m t) = 5°

b/2m = -1/t ln (5°/15°) = 0.11s^-1   .... (1)

or b= 7.03x10^-2 kg s^-1

Going back to eqn (1), for (t) = 2°

0.11 = -1/t ln (2°/15°)

Calculating, t= 18.3 seconds