Question 3.2.1: Consider an excitation force of the form given in Figure 3.6......
Consider an excitation force of the form given in Figure 3.6. The force is zero until time t_0, when it jumps to a constant level, F_0. This is called the step function, and when used to excite a single-degree-of-freedom system, it might model some machine operation or an automobile running over a surface that changes level (such as a curb). The step function of unit magnitude is called the Heaviside step function as defined in Example 3.1.4. Calculate the solution of
m \ddot{x}+c \dot{x}+k x=F(t)=\left\{\begin{array}{ll} 0, & t_0>t>0 \\ F_0, & t \geq t_0 \end{array}\right\} (3.14)
with x_0=v_0=0 and F(t) as described in Figure 3.6. Here it is assumed that the values of m, c, and k are such that the system is underdamped (0<\zeta<1).
Applying the convolution integral given by equation (3.13) directly yields
\begin{aligned} x(t) & =\frac{1}{m \omega_d} e^{-\zeta \omega_n t} \int_0^t\left[F(\tau) e^{\zeta \omega_n \tau} \sin \omega_d(t-\tau)\right] d \tau \\ & =\frac{1}{m \omega_d} \int_0^t F(t-\tau) e^{-\zeta \omega_n \tau} \sin \omega_d \tau d \tau \end{aligned} (3.13)
\begin{aligned} x(t) & =\frac{1}{m \omega_d} e^{-\zeta \omega_n t}\left[\int_0^{t_0}(0) e^{\zeta \omega_n \tau} \sin \omega_d(t-\tau) d \tau+\int_{t_0}^t F_0 e^{\zeta \omega_n \tau} \sin \omega_d(t-\tau) d \tau\right] \\ & =\frac{F_0}{m \omega_d} e^{-\zeta \omega_n t} \int_{t_0}^t e^{\zeta \omega_n^\tau \sin \omega_d(t-\tau) d \tau} \end{aligned}
Using a table of integrals to evaluate this expression yields
x(t)=\frac{F_0}{k}-\frac{F_0}{k \sqrt{1-\zeta^2}} e^{-\zeta \omega_n\left(t-t_0\right)} \cos \left[\omega_d\left(t-t_0\right)-\theta\right] \quad t \geq t_0 (3.15)
where
\theta=\tan ^{-1} \frac{\zeta}{\sqrt{1-\zeta^2}} (3.16)
Note that if t_0=0, equation (3.15) becomes just
x(t)=\frac{F_0}{k}-\frac{F_0}{k \sqrt{1-\zeta^2}} e^{-\zeta \omega_k t} \cos \left(\omega_d t-\theta\right) (3.17)
and if there is no damping (\zeta=0), this expression simplifies further to
x(t)=\frac{F_0}{k}\left(1-\cos \omega_n t\right) (3.18)
Examining the damped response given in equation (3.17), it is obvious that for large time the second term of the response dies out and the steady state is just
x_{ ss }(t)=\frac{F_0}{k} (3.19)
In fact, the underdamped step response given by equations (3.15) and (3.17) consists of the constant function F_0 / k minus a decaying oscillation, as illustrated in Figure 3.7.
Often in the design of vibrating systems subject to a step input, the time it takes for the response to reach the largest value, called the time to peak and denoted t_p in Figure 3.7, is used as a measure of the quality of the response. Other quantities used to measure the character of the step response are the overshoot, denoted by O.S. in Figure 3.7, which is the largest value of the response “over” the steady-state value, and the settling time, denoted by t_s in Figure 3.7. The settling time is the time it takes for the response to get and stay within a certain percentage of the steady-state response. For the case of t_0=0, t_p and t_s are given by t_p=\pi / \omega_d and t_s=3.5 / \zeta \omega_n. The peak time is exact (see Problem 3.25), and the settling time is an approximation of when the response stays within 3% of the steady-state value.
