Question 3.1.4: Consider the system (mass normalized) x(t) + 2x(t) + 4x(t) =......

By inspection, the natural frequency is \omega_n=2\ rad / s. Examining the velocity coefficient yields

2=2 \zeta \omega_n \text { or } \zeta=0.5

Thus, the system is underdamped and the response given in Window 3.1 applies. Computing the damped natural frequency yields

\omega_d=\omega_n \sqrt{1-\zeta^2}=2 \sqrt{1-\left(\frac{1}{2}\right)^2}=\sqrt{3}

First, compute the response for the time interval 0 \leq t \leq 4\ s. In this interval, only the first impulse is active. The corresponding impulse solution is, by equation (3.6),

x(t)=\frac{\hat{F} e^{-\zeta \omega_n t}}{m \omega_d} \sin \omega_d t       (3.6)

x_I(t)=\frac{\hat{F}}{m \omega_d} e^{-\zeta \omega_n t} \sin \omega_d t=\frac{1}{\sqrt{3}} e^{-t} \sin \sqrt{3} t, \quad 0 \leq t<4

The total solution for the first time interval is then equal to the sum of the homogeneous and impulse solutions. The homogeneous solution is

x_h(t)=e^{-t}\left(A \sin \omega_d t+B \cos \omega_d t\right), \quad 0 \leq t<4

where A and B are the constants of integration to be determined by the initial conditions and the subscript h denotes the solution due to the initial conditions. Differentiating the displacement yields the velocity:

\begin{aligned} \dot{x}_h(t)= & -e^{-t}(A \sin \sqrt{3} t+B \cos \sqrt{3} t) \\ & +e^{-t}(\sqrt{3} A \cos \sqrt{3} t-\sqrt{3} B \sin \sqrt{3} t) \end{aligned}

Setting t=0 in these last two expressions and using the initial conditions yields the following two equations:

\begin{aligned} & x_h(0)=1=B \\ & \dot{x}_h(0)=-1=-B+\sqrt{3} A \end{aligned}

Solving for A and B yields A=0 and B=1, so that x_h(t)=e^{-t} \cos \sqrt{3} t. Next, compute the response due to the impulse at t=0, which is equivalent to solving the initial value problem for x_l(0)=0 and \dot{x}_l(0)=1. Following the same procedure to compute the constants of integration for the impulse yields

B=1 \text { and } A=\frac{1}{\sqrt{3}} \text { so that } x_I(t)=\frac{e^{-t}}{\sqrt{3}} \sin \sqrt{3} t

Adding the homogenous response and the impulse response yields

x_1(t)=e^{-t}\left(\cos \sqrt{3} t+\frac{1}{\sqrt{3}} \sin \sqrt{3} t\right), \quad 0 \leq t<4

Next, compute the response of the system to the second impulse, which starts at t=4\ s. Using equation (3.9) with \tau=4\ s, the response to the second impulse is

h(t-\tau)=\frac{1}{m \omega_d} e^{-\zeta \omega_n(t-\tau)} \sin \omega_d(t-\tau) \quad t>\tau       (3.9)

x_2(t)=\frac{\hat{F}}{m \omega_d} e^{-\zeta \omega_n(t-\tau)} \sin \omega_d(t-\tau)=-\frac{1}{\sqrt{3}} e^{-t+4} \sin \sqrt{3}(t-4), \quad t>4

The Heaviside step function defined by

\Phi(t-\tau)=\left\{\begin{array}{ll} 0, & t<\tau \\ 1, & t \geq \tau \end{array}\right\}

is perfect for writing functions that “turn on” after some time has evolved. Heaviside functions are also denoted by H(t-\tau). Using superposition, the total solution is x = x_1 + x_2, and the Heaviside function is used to indicate that x_2 “starts” after τ = 4. The solution can be written as

x(t)=e^{-t}\left(\cos \sqrt{3} t+\frac{1}{\sqrt{3}} \sin \sqrt{3} t\right)-\left[\frac{e^{-(t-4)}}{\sqrt{3}} \sin \sqrt{3}(t-4)\right] \Phi(t-4)\ mm

This is plotted in Figure 3.4. Note the sharp change in the response as the second impact is applied. This is in contrast to the double hit in the previous example. In the previous example, the second impulse occurs in the same “direction” as the current response. However, in this example, the second impact occurs out of phase with the response of the first impact and causes an abrupt change in direction.