Example 1 - Find the displacement of a damped single-degree-of-freedom system subjected to the forcing function \(F(t) = F_0e^{-\alpha t}\), where \(\alpha\) is a constant.
Solution - The single degree of freedom system consist of a body of mass \(m\), spring of stifness \(k\) and damper with damping \(c\) as shwon in Figure 1.
Figure 1 Single-degree-of-freedom system consisting of body of mass \(m\), spring with constant \(k\), and damper with damping \(c\).
The differential equation that describes the motion of the system can be written in the following form:
\begin{equation}
m\ddot{x} + c\dot{x} + kx = F(t)
\end{equation}
The resposne of underdamped single-degree-of-freedom system subjecte to arbitrary excitation can be written as:
\begin{equation}
x(t) = \frac{1}{m\omega_d}\int_0^t F(\tau) e^{-\zeta \omega_n (t-\tau)}\sin \omega_d(t-\tau)d\tau
\end{equation}
In this example the arbitrary function that is acting on the system is \(F(\tau) = F_0e^{-\alpha \tau}\), where \(\alpha\) is constant. Inserting the function into previous equation the following response is obtained.
\begin{eqnarray}
x(t) &=& \frac{1}{m\omega_d}\int_0^t F_0 e^{-\alpha \tau} e^{-\zeta\omega_n(t-\tau)}\sin \omega_d (t-\tau) d\tau \nonumber\\
x(t) &=& \frac{F_0}{m\omega_d}e^{-\zeta \omega_n t}\int_0^t e^{-(\alpha - \zeta\omega_n)\tau} \sin \omega_d(t-\tau)d\tau\nonumber\\
x(t) &=& \frac{F_0e^{-\zeta\omega_n t}}{m\omega_d} \int_0^t e^{-(\alpha-\zeta \omega_n)\tau}\left(\sin \omega_d t\cos \omega_d\tau - \cos\omega_d t \sin \omega_d \tau\right)d\tau\nonumber\\
x(t) &=& \frac{F_0e^{-\zeta \omega_n t}}{m\omega_d}\sin\omega_d t\left(\frac{e^{-(\alpha - \zeta\omega_n)\tau}}{(\alpha-\zeta\omega_n)^2 + \omega_d^2}\left(-(\alpha-\zeta\omega_n)\cos\omega_d\tau + \omega_d\sin \omega_d\tau\right)\right)\Bigg|_0^t\nonumber\\
&-&\frac{F_0e^{-\zeta \omega_n t}}{m\omega_d}\cos\omega_d t\left(\frac{e^{-(\alpha - \zeta\omega_n)\tau}}{(\alpha-\zeta\omega_n)^2 + \omega_d^2}\left(-(\alpha-\zeta\omega_n)\sin\omega_d\tau + \omega_d\cos \omega_d\tau\right)\right)\Bigg|_0^t\nonumber\\
&=& \frac{F_0e^{\zeta \omega_n t}}{m\omega_d\left((\alpha-\zeta \omega_n)^2 + \omega_d^2\right)}\left(\omega_d e^{-(\alpha - \zeta \omega_n)t} + (\alpha -\zeta \omega_n)\sin \omega_d t-\omega_d \cos\omega_d t\right)\nonumber\\
&=& \frac{F_0 e^{\alpha t}}{m\omega_d\left((\alpha-\zeta \omega_n)^2 + \omega_d^2\right)} + \frac{F_0e^{-\zeta \omega_n t}}{m\omega_d\left((\alpha-\zeta \omega_n)^2 + \omega_d^2\right)} \sin(\omega_d t - \phi)\nonumber
\end{eqnarray}
It should be noted that \(\phi = \mathrm{arctan} \left(\frac{\omega_d}{\alpha - \zeta\omega_n}\right)\).
Example 2 - Find the transient response of an undamped spring-mass system for \(t>\frac{\pi}{\omega}\) when the mass is subjected to a force: \begin{equation} F(t) = \begin{cases} \frac{F_0}{2}(1-\cos\omega t) & 0\leq t \leq \frac{\pi}{\omega}\\ F_0 & t> \frac{\pi}{\omega} \end{cases} \end{equation} Assumption is that displacement and velocity of the mass are zero at \(t=0\).
Solution - The differential equation that describes the motion of the system can be written as: \begin{equation} m\ddot{x} + kx = F(t). \end{equation} First case the \(F(t)\) is in range from 0 up to \(\frac{\pi}{\omega}\). \begin{eqnarray} m\ddot{x} + kx &=& \frac{F_0}{2} - \frac{F_0}{2}\cos\omega t\nonumber\\ x(t) &=& A\cos\omega_n t + B\sin \omega_n t +\frac{F_0}{2k} - \frac{F_0}{2k\left(1-\frac{\omega^2}{\omega_n^2}\right)} \cos \omega t\nonumber\\ x(0) &=& 0 \Rightarrow A + \frac{F_0}{2k} - \frac{F_0}{2k\left(1-\frac{\omega^2}{\omega_n^2}\right)} = 0\nonumber\\ A &=& \frac{F_0\left(\frac{\omega}{\omega_n}\right)^2}{2k\left(1-\frac{\omega^2}{\omega_n^2}\right)} \nonumber\\ \dot{x}(0) &=& 0 \Rightarrow B = 0\nonumber\\ x(t) &=& \frac{F_0}{2k\left(1-\left(\frac{\omega}{\omega_n}\right)^2\right)}\left(1-\cos\omega t-\left(\frac{\omega}{\omega_n}\right)^2(1-\cos\omega_n t)\right)\nonumber\\ t &=& \frac{\pi}{\omega} \Rightarrow x\left(\frac{\pi}{\omega}\right) = \frac{F_0}{2k\left(1-\frac{\omega^2}{\omega_n^2}\right)}\left(2- \frac{\omega^2}{\omega_n^2}\left(1-\cos\frac{\omega_n\pi}{\omega}\right)\right)\nonumber\\ \end{eqnarray} In case where \(t>\frac{\pi}{\omega}\). \begin{eqnarray} x(t) &=& \frac{1}{m\omega_n}\int_0^t F(\tau) \sin \omega_n(t-\tau)d\tau\nonumber\\ x(t) &=& \frac{1}{m\omega_n}\int_0^{\frac{\pi}{\omega}}F(\tau)\sin\omega_n (t-\tau )d\tau + \frac{1}{m\omega_n}\int_{\frac{\pi}{\omega}}^t F(\tau) \sin \omega_n(t-\tau)d\tau\nonumber\\ x(t) &=& x(t=\frac{\pi}{\omega}) + \frac{F_0}{m\omega_n}\int_{\frac{\pi}{\omega}}^n \sin \omega_n(t-\tau)d\tau \nonumber\\ x(t) &=& x(t=\frac{\pi}{\omega}) + \frac{F_0}{m\omega_n^2}\left(\cos \omega_n(t-\tau)\right)_{\frac{\pi}{\omega}}^t\nonumber\\ x(t) &=& \frac{F_0}{2k(1-\frac{\omega^2}{\omega_n^2})} \left(2-\frac{\omega^2}{\omega_n^2}\left(1-\cos\frac{\omega_n\pi}{\omega}\right)\right) + \frac{F_0}{k}\left(1-\cos\omega_n\left(t-\frac{\pi}{\omega}\right)\right) \end{eqnarray}
Solution - The single degree of freedom system consist of a body of mass \(m\), spring of stifness \(k\) and damper with damping \(c\) as shwon in Figure 1.
Example 2 - Find the transient response of an undamped spring-mass system for \(t>\frac{\pi}{\omega}\) when the mass is subjected to a force: \begin{equation} F(t) = \begin{cases} \frac{F_0}{2}(1-\cos\omega t) & 0\leq t \leq \frac{\pi}{\omega}\\ F_0 & t> \frac{\pi}{\omega} \end{cases} \end{equation} Assumption is that displacement and velocity of the mass are zero at \(t=0\).
Solution - The differential equation that describes the motion of the system can be written as: \begin{equation} m\ddot{x} + kx = F(t). \end{equation} First case the \(F(t)\) is in range from 0 up to \(\frac{\pi}{\omega}\). \begin{eqnarray} m\ddot{x} + kx &=& \frac{F_0}{2} - \frac{F_0}{2}\cos\omega t\nonumber\\ x(t) &=& A\cos\omega_n t + B\sin \omega_n t +\frac{F_0}{2k} - \frac{F_0}{2k\left(1-\frac{\omega^2}{\omega_n^2}\right)} \cos \omega t\nonumber\\ x(0) &=& 0 \Rightarrow A + \frac{F_0}{2k} - \frac{F_0}{2k\left(1-\frac{\omega^2}{\omega_n^2}\right)} = 0\nonumber\\ A &=& \frac{F_0\left(\frac{\omega}{\omega_n}\right)^2}{2k\left(1-\frac{\omega^2}{\omega_n^2}\right)} \nonumber\\ \dot{x}(0) &=& 0 \Rightarrow B = 0\nonumber\\ x(t) &=& \frac{F_0}{2k\left(1-\left(\frac{\omega}{\omega_n}\right)^2\right)}\left(1-\cos\omega t-\left(\frac{\omega}{\omega_n}\right)^2(1-\cos\omega_n t)\right)\nonumber\\ t &=& \frac{\pi}{\omega} \Rightarrow x\left(\frac{\pi}{\omega}\right) = \frac{F_0}{2k\left(1-\frac{\omega^2}{\omega_n^2}\right)}\left(2- \frac{\omega^2}{\omega_n^2}\left(1-\cos\frac{\omega_n\pi}{\omega}\right)\right)\nonumber\\ \end{eqnarray} In case where \(t>\frac{\pi}{\omega}\). \begin{eqnarray} x(t) &=& \frac{1}{m\omega_n}\int_0^t F(\tau) \sin \omega_n(t-\tau)d\tau\nonumber\\ x(t) &=& \frac{1}{m\omega_n}\int_0^{\frac{\pi}{\omega}}F(\tau)\sin\omega_n (t-\tau )d\tau + \frac{1}{m\omega_n}\int_{\frac{\pi}{\omega}}^t F(\tau) \sin \omega_n(t-\tau)d\tau\nonumber\\ x(t) &=& x(t=\frac{\pi}{\omega}) + \frac{F_0}{m\omega_n}\int_{\frac{\pi}{\omega}}^n \sin \omega_n(t-\tau)d\tau \nonumber\\ x(t) &=& x(t=\frac{\pi}{\omega}) + \frac{F_0}{m\omega_n^2}\left(\cos \omega_n(t-\tau)\right)_{\frac{\pi}{\omega}}^t\nonumber\\ x(t) &=& \frac{F_0}{2k(1-\frac{\omega^2}{\omega_n^2})} \left(2-\frac{\omega^2}{\omega_n^2}\left(1-\cos\frac{\omega_n\pi}{\omega}\right)\right) + \frac{F_0}{k}\left(1-\cos\omega_n\left(t-\frac{\pi}{\omega}\right)\right) \end{eqnarray}