Question 4.3: For the single degree of freedom mass–spring system shown in...
For the single degree of freedom mass–spring system shown in Fig. 4.1, m = 10 kg, k = 4000 N/m, F_{0} = 40 N, and ω_{f} = 20 rad/s. The initial conditions are x_{0} = 0.02 m and \dot{x_{0}} = 0. Determine the displacement of the mass after t = 0.5 s and t = 1 s.
The circular frequency of the system is
ω = \sqrt{\frac{k}{m}} = \sqrt{\frac{4000}{10}} = 20 rad/s
In this case, ω = ω_{f} and the system is at resonance. The complete solution is given by Eq. 4.24 as
x(t) = X sin(ωt + Φ) − \frac{ωt X_{0}}{2} cos ωt
where
X_{0} = \frac{F_{0}}{k} = \frac{40}{4000} = 0.01 m
X = \sqrt{x²_{0} + (\frac{\dot{x_{0}}}{ω} + \frac{X_{0}}{2})²}= \sqrt{(0.02)² +(\frac{0.01}{2})²} = 0.0206
Φ = tan^{-1}(\frac{x_{0}}{(\frac{\dot{x_{0}}}{ω} + \frac{X_{0}}{2})})= tan^{-1} (\frac{0.02}{0+ \frac{0.01}{2}})= tan^{−1} 4 = 1.3258 rad
The displacement x(t) can then be written as
x(t) = 0.0206 sin(20t + 1.3258) − 0.1t cos 20t
at t = 0.5 s
x(t = 0.5) = 0.0206 sin(11.3258) − 0.05 cos 10 = 0.02246 m
at t = 1 s
x(t = 1) = 0.0206 sin(21.3258) − (0.1)(1) cos 20 = −0.02809 m