Question

A 2.15 kg lightly damped harmonic oscillator has an angular oscillation frequency of 0.261 rad/s. If the maximum displacement of 2.0 m occurs when t = 0.00 s, and the damping constant b is 0.74 kg/s what is the objectʹs displacement when t = 4.01 s?

Answer 1

A lightly damped harmonic oscillator has a differential equation of the form:\[\frac{d^{2}x}{dt^{2}}+2\gamma\frac{dx}{dt}+\omega _{0}^{2}x=0\]where gamma is the damping factor and omega_0 is the natural angular frequency.


For this ODE, the solution is of the following form:\[x(t)=Ae^{-bt}\cos(\omega_{d}t+\phi)\]where A is the amplitude, gamma is the damping factor, omega_d is the damped angular frequency, and phi is the phase angle.


We have a lot of information above given:
m = 2.15 kg
omega_d = 0.261 rad/s
A = 2.0 m
b = 0.74 kg/s

gamma = b/2m = 0.74/4.30 = 0.172 Hz


Let's put this information in:\[x(t) = (2.0)e^{-0.172t}\cos(0.261t)\]

Now we can solve for the displacement at a time of t = 4.01 s:\[x(4.01)=(2.0)e^{-0.172(4.01)}\cos(0.261(4.01))=0.502 \]

So in t=4.01s, the displacement is 0.502 m.