Question 4.1: For the undamped single degree of freedom mass–spring system...
For the undamped single degree of freedom mass–spring system shown in Fig. 4.1, let m = 5 kg, k = 2500 N/m, F_{0} = 20 N, and the force frequency ω_{f} = 18 rad/s. The mass has the initial conditions x_{0} = 0.01 m and \dot{x_{0}} = 1 m/s. Determine the displacement of the mass at t = 1 s.
The circular frequency of the system is
ω = \sqrt{\frac{k}{m}} = \sqrt{\frac{2500}{5}} = 22.361 rad/s
The frequency ratio r is
r = \frac{ω_{f}}{ω} = \frac{18}{22.361} = 0.80499
The steady state solution is
x_{p} = X_{0}β \sin ω_{f} t
where
X_{0} = \frac{F_{0}}{k} = \frac{20}{2500} = 0.008m
β = \frac{1}{1 − r²} =\frac{1}{1 − (0.80499)²} = 2.841
The complete solution, which is the sum of the complementary function and the particular integral, is
x = X sin(ωt + Φ) + X_{0}β \sin ω_{f} t
Using the initial conditions
x_{0} = 0.01 = X sin Φ
\dot{x_{0}} = 1 = ωX cos Φ + ω_{f}X_{0}β = 22.361 X cos Φ + (18)(0.008)(2.841)
or
X sin Φ = 0.01, X cos Φ = 0.0264
from which tan Φ = 0.3784, or
Φ = 20.728°,
and the amplitude X is
X = 0.02825
Therefore, the displacement x is
x = 0.02825 sin(22.361t + 20.728°) + 0.022728 sin 18t
The displacement at t = 1 s is
x(t = 1) = 0.02825 sin(22.361 + 0.3617) + 0.022728 sin 18
= −0.0359m