Question

In a science museum, a 110 kg brass pendulum bob swings at the end of a 12.3 m -long wire. The pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.4 m to the side and releasing it. Because of its compact shape and smooth surface, the pendulum's damping constant is only 0.010kg/s.

a.) At exactly 12:00 noon, how many oscillations will the pendulum have completed?

b.) What is its amplitude at noon?

Answer 1

given, in a science museum
m = 110 kg
l = 12.3 m
Ti = 8 am every morning
damping constat, b = 0.01 kg/s
and xo = 1.4 m

a. for the equation of motion mx" + bx' + kx = 0, for a damped harmonic oscilator
   angular frequency is givne by
   w = sqrt(wn^2 - gamma^2)
   where gamma = b/2m
   and wn = sqrt(g/l) for pendulum
   w = sqrt(g/l - b^2/4m^2)
   w = 0.89306 rad/s
   hence
   f = w/2*pi = 0.14213534207 Hz

   at exactly 12 noon, total time spent = 4 hours = 4*360 seconds
   so numebr of oscillations = 4*360/0.142135342 = 10131.188900 osscilations

b. at noon amplitude is given by
   x = xo*exp(-b*t/2m)
because the pendulum would follow exponential decay of the amplitude
hence
at noon
x = 1.4*exp(-0.01*4*3600/2*110) = 0.72754951679 m