Question 3.5: The damped mass–spring system shown in Fig. 3.12 has mass m ...
The damped mass–spring system shown in Fig. 3.12 has mass m = 10 kg,
stiffness coefficient k = 1000 N/m, and damping coefficient c = 300 N·s/m.
Determine the displacement of the mass as a function of time.
The natural frequency ω of the system is
ω = \sqrt{k}{m} = \sqrt{\frac{1000}{10}} = 10 rad/s
The critical damping coefficient C_{c} is
C_{c} = 2mω = 2(10)(10) = 200N · s/m
The damping factor ξ is given by
ξ = \frac{c}{C_{c}} = \frac{300}{200} = 1.5
Since ξ > 1, the system is overdamped and the solution is given by
x(t) = A_{1}e^{p_{1}t} + A_{2}e^{p_{2}t}
where p_{1} and p_{2} can be determined using Eq. 3.63 as
p_{1} = −ξω + ω \sqrt{ξ² − 1} = −(1.5)(10) + (10)\sqrt{(1.5)² − 1} = −3.8197
p_{2} = −ξω − ω \sqrt{ξ² − 1} = −(1.5)(10) − (10) \sqrt{(1.5)² − 1} = −26.1803
The solution x(t) is then given by
x(t) = A_{1}e^{p_{1}t} + A_{2}e^{p_{2}t} = A_{1}e^{−3.8197t} + A_{2}e^{−26.1803t}
The constants A_{1} and A_{2} can be determined from the initial conditions.