Question 2.2.2: A damped spring–mass system with values of c = 100 kg/s, m =......
A damped spring–mass system with values of c = 100 kg/s, m = 100 kg, and k = 910 N/m, is subject to a force of 10 cos (3t) N. The system is also subject to initial conditions: x_0 = 1 mm and v_0 = 20 mm/s. Compute the total response, x(t), of the system.
Following equation (2.26) with the values given here, the system to be solved is
m \ddot{x}+c \dot{x}+k x=F_0 \cos \omega t (2.26)
100 \ddot{x}(t)+100 \dot{x}(t)+910 x(t)=10 \cos 3 t, x_0=0.001 m , v_0=0.02\ m / s
where the units (and thus numerical values) of the initial conditions have been changed to agree with the equation of motion per the previous example. Dividing by the mass yields the vibration properties
\begin{aligned} f_0 & =\frac{F_0}{m}=\frac{10}{100}=0.1 \frac{ m }{ s ^2}, \omega_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{910}{100}}=3.017 \frac{ rad }{ s }, \\ \zeta & =\frac{c}{2 \sqrt{m k}}=\frac{100}{2 \sqrt{100 \cdot 910}}=0.166, \\ \omega_d & =\omega_n \sqrt{1-\zeta^2}=3.017 \sqrt{1-0.166^2}=2.975\ rad / s \end{aligned}
Since ω = 3 rad/s, the system is near resonance. Computing the amplitude and phase for the particular solution from the values given in Window 2.3 yields
Window 2.3
Summary of Phase and Amplitude Relationships for Undamped and Underdamped Single-Degree-of-Freedom Systems for Both the Free Response, F(t) = 0, and for the Forced Response, F(t) = F_0 cos 𝛚t, Cases
The general response for the undamped case has the form
x(t)=A \sin \left(\omega_n t+\phi\right)+X \cos \omega t
where for the free response
\phi=\tan ^{-1} \frac{\omega_n x_0}{v_0}, \quad A=\sqrt{x_0^2+\frac{v_0^2}{\omega_n^2}}, \quad X=0
and for the forced response:
\phi=\tan ^{-1} \frac{\omega_n\left(x_0-X\right)}{v_0}, A=\sqrt{\left(\frac{v_0}{\omega_n}\right)^2+\left(x_0-X\right)^2}, X=\frac{f_0}{\omega_n^2-\omega^2}
The general response in the underdamped case has the form
x(t)=A e^{-\zeta \omega_n t} \sin \left(\omega_d t+\phi\right)+X \cos (\omega t-\theta)
where the free response:
\phi=\tan ^{-1} \frac{x_0 \omega_d}{v_0+\zeta \omega_n x_0}, A=\frac{1}{\omega_d} \sqrt{\left(v_0+\zeta \omega_n x_0\right)^2+\left(x_0 \omega_d\right)^2}, X=0
and for the forced response
\begin{gathered} \theta=\tan ^{-1} \frac{2 \zeta \omega_n \omega}{\omega_n^2-\omega^2}, \quad X=\frac{f_0}{\sqrt{\left(\omega_n^2-\omega^2\right)^2+\left(2 \zeta \omega_n \omega\right)^2}} \\ \phi=\tan ^{-1} \frac{\omega_d\left(x_0-X \cos \theta\right)}{v_0+\left(x_0-X \cos \theta\right) \zeta \omega_n-\omega X \sin \theta} \text { and } A=\frac{x_0-X \cos \theta}{\sin \phi} \end{gathered}
X=\frac{f_0}{\sqrt{\left(\omega_n^2-\omega^2\right)^2+\left(2 \zeta \omega_n \omega\right)^2}}=\frac{0.1}{\sqrt{\left(3.017^2-3^2\right)^2+(2 \cdot 0.166 \cdot 3.017 \cdot 3)^2}}=0.033\ m
\theta=\tan ^{-1} \frac{2 \zeta \omega_n \omega}{\omega_n^2-\omega^2}=\tan ^{-1} \frac{2 \cdot 0.166 \cdot 3.017 \cdot 3}{3.017^2-3^2}=1.537\ rad
Next, compute phase for the transient response
\begin{aligned} \phi & =\tan ^{-1} \frac{\omega_d\left(x_0-X \cos \theta\right)}{v_0+\left(x_0-X \cos \theta\right) \zeta \omega_n-\omega X \sin \theta} \\ & =\frac{2.975(0.001-0.033 \cdot 0.033)}{0.02+(0.001-0.033 \cdot 0.033) 0.166 \cdot 3.017-3 \cdot 0.033 \cdot 0.999}=4.089 \times 10^{-3}\ rad \end{aligned}
The amplitude of the transient is
A=\frac{x_0-X \cos \theta}{\sin \phi}=\frac{0.001-0.033 \cdot 0.0333}{4.089 \times 10^{-3}}=-0.027\ m
The total response is then written from equation (2.37) as
\ddot{x}+2 \zeta \omega_n \dot{x}+\omega_n^2 x=f_0 \cos \omega t (2.37)
x(t) = Ae^{-ζω_nt} \sin (ω_dt + ϕ) + X \cos (ωt – θ)
= -0.027e^{-0.5t}\sin(2.975t + 4.089 \times 10^{-3}) + 0.033 \cos(3t – 1.537)\ m