Pendulm Damping

Question

A 93.0 cm pendulum is released from a small angle. After 103 oscillations the amplitude is one half of its original value. The damping is proportional to the speed of the pendulum bob.

Find the value of the damping coefficient, α, (in Hz).

Answer 1

Assuming the small angle approximation (oscillations ≤ 15°), the damped pendulum’s position can be modeled with the equation, X(t) = e^(-αt)*(Asin(wt)+Bcos(wt)), where X(t) is the amplitude of the oscillation at time, t, and w is the angular frequency, so w = √(g/L). The period for 1 oscillation is 2pi/w, so the time for 103 oscillations is 103*(2pi/w), and X(t) after 103 oscillations is X(0)/2. Also, X(0) = e^(-α0)*(Asin(w0)+Bcos(w0)) = 1*(A*0 + B*1) = B. So, now substituting in the time of 103 oscillations gives X(103*(2pi/w)) = e^(-α103*(2pi/w))*(Asin(w103*(2pi/w))+Bcos(w103*(2pi/w))) = e^(-α103*(2pi/w))*(Asin(103*(2pi))+Bcos(103*(2pi))). Now, we can use the fact that sin(n*2pi) = 0 and cos(n*2pi) = 1 for integers, n, to simplify it to: X(103*(2pi/w)) = e^(-α103*(2pi/w))*B. And, since we found earlier that B = X(0) and X(q03*(2pi/w)) = X(0)/2, we can substitute those in to get: X(0)/2 = e^(-α103*(2pi/w))*X(0) canceling X(0) and taking ln of both sides then gives ln(1/2) = -α103*(2pi/w). Using the fact that ln(1/2) = -ln2 and isolating α gives: ln2/(103*(2pi/w)) = α, and since w = √(g/L), we finally get: α = ln2/(103*(2pi/√(g/L))) = ln2/(103*(2pi/√(9.8/0.93))) = 0.00348 Hz.