Question 3.1.1: Consider a spring–mass–damper system with m = 100 kg, c = 20......

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

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)

$x(t)=\frac{\hat{F} e^{-\zeta \omega_c t}}{m \omega_d} \sin \omega_d t=0.022 e^{-0.1 t} \sin (4.471\ t)$