Use the convolution integral to find the response of the single-DOF system represented by Eq. (3.2), assuming that the damping is zero, when a step force of magnitude \overline{F} is applied at t = 0, if the initial displacement, z, and the initial velocity, \dot{z}, are both zero at t = 0.
\ddot{z} + 2\gamma \omega _{n}\dot{z} + \omega ^{2}_{n}z = F/m (3.2)
From Eq. (C) of Example 3.2, the unit impulse response of the system is
h(t)= \frac{1}{m \omega _{n}}\sin \omega _{n}t (A)
The applied force, F(\tau), is
\begin{aligned}&\begin{array}{lll}F(\tau ) = 0 &&& \tau < 0 \\F(\tau ) = \overline{F} &&& 0 < \tau <\infty &&&&&& (B) \end{array} \end{aligned}From Eq. (3.21) and Eq. (A) (bearing in mind that \sin \left(-\theta \right) = – \sin \theta \mathrm{, and} \cos \left(-\theta \right) = \cos \theta ):
z(t) = \int_{0}^{t}{F(\tau)} \cdot h(t-\tau ) \;d\tau = \frac{\overline{F} }{m\omega _{n}} \int_{0}^{t}{\sin \omega _{n}(t -\tau )} \;d\tau = \frac{\overline{F} }{m\omega _{n}}\int_{0}^{t}{\sin (-\omega _{n}\tau + \omega _{n}t)} \;d\tau = \frac{\overline{F} }{m\omega _{n}}\int_{0}^{t}{-\sin (\omega _{n}\tau – \omega _{n}t)} \;d\tau = \frac{\overline{F} }{m\omega _{n}^{2}} {{}_{0}^{t}}\left[\cos (\omega _{n}\tau – \omega _{n}t)\right]\begin{aligned}&\begin{array}{lll} z(t) = \frac{\overline{F} }{m\omega _{n}^{2}} \left(1 – \cos \omega _{n}t\ \right) = \frac{\overline{F} }{k} \left(1 – \cos \omega _{n}t\ \right) \\ (\mathrm{since} k = m\omega _{n}^{2}). \end{array} (C) \end{aligned}