Question 2.2.4: Consider a simple spring–mass–damper system with m = 49.2 × ......
Consider a simple spring-mass-damper system with m=49.2 \times 10^{-3}\ kg , c=0.11\ kg / s, and k=8578\ N / m. Calculate the value of the steady-state response if \omega=132\ rad / s for f_0=10\ N / kg. Calculate the change in amplitude if the driving frequency changes to \omega=125\ rad / s.
The frequency and damping ratio are determined from the given values as
\omega_n=\sqrt{\frac{k}{m}}=132 ~rad / s , \quad \zeta=\frac{c}{2 \sqrt{m k}}=0.0085
respectively. From equation (2.39) the magnitude of x_p(t) is
X=\frac{f_0}{\sqrt{\left(\omega_n^2-\omega^2\right)+\left(2 \zeta \omega_n \omega\right)^2}}, \quad \theta=\tan ^{-1} \frac{2 \zeta \omega_n \omega}{\omega_n^2-\omega^2} (2.39)
\begin{aligned} \left|x_p(t)\right| & =X=\frac{f_0}{\sqrt{\left(\omega_n^2-\omega^2\right)+\left(2 \zeta \omega_n \omega\right)^2}} \\ & =\frac{10}{\left\{\left[(132)^2-(132)^2\right]^2+[2(0.0085)(132)(132)]^2\right\}^{1 / 2}} \\ & =\frac{10}{2(0.0085)(132)^2} \\ & =0.034\ m \end{aligned}
If the driving frequency is changed to 125 rad/s, the amplitude becomes
\frac{10}{\left\{\left[(132)^2-(125)^2\right]^2+[2(0.0085)(132)(125)]^2\right\}^{1 / 2}}=0.005\ m
So a slight change in the driving frequency from near resonance at 132 rad/s to 125 rad/s (about 5%) causes an order-of-magnitude change in the amplitude of the steady-state response.