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.

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.