Question 3.1.1: Consider a spring–mass–damper system with m = 100 kg, c = 20......
Consider a spring–mass–damper system with m = 100 kg, c = 20 kg/s, and k = 2000 N/m with an impulse force applied to it of 1000 N for 0.01 s. Compute the resulting response.
A 1000\ N force acting over 0.01\ s provides (area under the curve) a value of \hat{F}=F \Delta t=1000 \cdot 0.01=10\ N \cdot s. Using the values given, the equation of motion is
100 \ddot{x}(t)+20 \dot{x}(t)+2000 x(t)=10δ(t)
Thus the natural frequency, damping ratio, and damped natural frequency are
\begin{aligned} & \omega_n=\sqrt{\frac{2000}{100}}=4.427\ rad / s , \zeta=\frac{20}{2 \sqrt{100 \cdot 2000}}=0.022, \\ & \omega_d=4.472 \sqrt{1-0.022^2}=4.471\ rad / s \end{aligned}
Using equation (3.6), the response becomes
x(t)=\frac{\hat{F} e^{-\zeta \omega_n t}}{m \omega_d} \sin \omega_d t (3.6)
x(t)=\frac{\hat{F} e^{-\zeta \omega_c t}}{m \omega_d} \sin \omega_d t=0.022 e^{-0.1 t} \sin (4.471\ t)