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stargirl6789 · Nov 14, 2011, 07:35 PM

A 2.15 kg damped harmonic oscillator has an angular frequency of 0.261 rad/s.If the amplitude is 2.9 m at t = 0.0 s and the object is at its maximum displacement at this time, what is the object's displacement at t= 4.01 s if the damping constant is 0.74 kg/s.

ebaines · Nov 17, 2011, 03:10 PM

The basic form of the differential equation of damped oscillation is:

$ \ddot x + \frac c m \dot x + \frac k m x =0 $

The undamped frequency of osciallation (i.e. if c = 0) is:

$ \omega _o = \sqrt {\frac k m } $

The term $ \zeta= \frac c {2 m \omeg_o} $
is the damping ratio for the system. If $\zeta$ < 1 then the system is underdamped and oscillations gradually grow smaller over time. If $\zeta$ > 1 then it's over-damped and no oscillations occur.

If $\zeta$ < 1 then the damped oscillation frequency is given by

$ \omega_1 = \omega_o \sqrt{ 1 - \zeta^2} $

If you plug in the numbers you can find $\omega_1$ , and from that the displacement at t= 4.01s.