\[ m\ddot{x}+kx=F_{0}\cos \omega t \] This model is single degree of freedom system, undamped, with forced harmonice input. Its solution is given by\[ x\left ( t\right ) =x_{h}\left ( t\right ) +x_{p}\left ( t\right ) \] Where \(x_{p}\left ( t\right ) \) is particular solution and \(x_{h}\left ( t\right ) \) is homogenous solution. We know that\[ x_{h}\left ( t\right ) =c_{1}\cos \omega _{n}t+c_{2}\sin \omega _{n}t \] And assuming \(x_{p}\left ( t\right ) =X\cos \omega t\) for the case \(\omega \neq \omega _{n}\) Pluggin this into the ODE, we find that \[ X=\frac{x_{st}}{1-r^{2}}\] Where\(\ r=\frac{\omega }{\omega _{n}}\) and \(x_{st}=\frac{F_{0}}{k_{eq}}\) the static deflection. Hence the solution becomes\begin{equation} x\left ( t\right ) =\overset{\text{homogeneous}}{\overbrace{c_{1}\cos \omega _{n}t+c_{2}\sin \omega _{n}t}}+\overset{\text{particular}}{\overbrace{\frac{x_{st}}{1-r^{2}}\cos \omega t}} \tag{1} \end{equation} Assuming initial conditions are \(x\left ( 0\right ) =x_{0},\dot{x}\left ( 0\right ) =\dot{x}_{0}\), then (1) at \(t=0\) becomes\begin{align*} x_{0} & =c_{1}+\frac{x_{st}}{1-r^{2}}\\ c_{1} & =x_{0}-\frac{x_{st}}{1-r^{2}} \end{align*}
Hence solution (1) now becomes\[ x\left ( t\right ) =\left ( x_{0}-\frac{x_{st}}{1-r^{2}}\right ) \cos \omega _{n}t+c_{2}\sin \omega _{n}t+\frac{x_{st}}{1-r^{2}}\cos \omega t \] Taking derivative\[ \dot{x}\left ( t\right ) =-\omega _{n}\left ( x_{0}-\frac{x_{st}}{1-r^{2}}\right ) \sin \omega _{n}t+c_{2}\omega _{n}\cos \omega _{n}t-\omega \frac{x_{st}}{1-r^{2}}\sin \omega t \] At \(t=0\) the above becomes\begin{align*} \dot{x}_{0} & =c_{2}\omega _{n}\\ c_{2} & =\frac{\dot{x}_{0}}{\omega _{n}} \end{align*}
Therefore the solution now becomes (again, this is for \(\omega \neq \omega _{n}\))\begin{equation} x\left ( t\right ) =\left ( x_{0}-\frac{x_{st}}{1-r^{2}}\right ) \cos \omega _{n}t+\frac{\dot{x}_{0}}{\omega _{n}}\sin \omega _{n}t+\frac{x_{st}}{1-r^{2}}\cos \omega t \tag{2} \end{equation}
\[ m\ddot{x}+kx=F_{0}\sin \omega t \] This model is single degree of freedom system, undamped, with forced harmonice input. Its solution is given by\[ x\left ( t\right ) =x_{h}\left ( t\right ) +x_{p}\left ( t\right ) \] Where \(x_{p}\left ( t\right ) \) is particular solution and \(x_{h}\left ( t\right ) \) is homogenous solution. We know that\[ x_{h}\left ( t\right ) =c_{1}\cos \omega _{n}t+c_{2}\sin \omega _{n}t \] And assuming \(x_{p}\left ( t\right ) =X\sin \omega t\) for the case \(\omega \neq \omega _{n}\) Pluggin this into the ODE, we find that \[ X=\frac{x_{st}}{1-r^{2}}\] Where\(\ r=\frac{\omega }{\omega _{n}}\) and \(x_{st}=\frac{F_{0}}{k_{eq}}\) the static deflection. Hence the solution becomes\begin{equation} x\left ( t\right ) =\overset{\text{homogeneous}}{\overbrace{c_{1}\cos \omega _{n}t+c_{2}\sin \omega _{n}t}}+\overset{\text{particular}}{\overbrace{\frac{x_{st}}{1-r^{2}}\sin \omega t}} \tag{1} \end{equation} Assuming initial conditions are \(x\left ( 0\right ) =x_{0},\dot{x}\left ( 0\right ) =\dot{x}_{0}\), then (1) at \(t=0\) becomes\[ x_{0}=c_{1}\] Hence solution (1) now becomes\[ x\left ( t\right ) =x_{0}\cos \omega _{n}t+c_{2}\sin \omega _{n}t+\frac{x_{st}}{1-r^{2}}\sin \omega t \] Taking derivative\[ \dot{x}\left ( t\right ) =-x_{0}\sin \omega _{n}t+c_{2}\omega _{n}\cos \omega _{n}t+\omega \frac{x_{st}}{1-r^{2}}\cos \omega t \] At \(t=0\) the above becomes\begin{align*} \dot{x}_{0} & =c_{2}\omega _{n}+\omega \frac{x_{st}}{1-r^{2}}\\ c_{2} & =\frac{\dot{x}_{0}}{\omega _{n}}-\frac{\omega }{\omega _{n}}\frac{x_{st}}{1-r^{2}}\\ & =\frac{\dot{x}_{0}}{\omega _{n}}-\frac{r}{1-r^{2}}x_{st} \end{align*}
Therefore the solution now becomes (again, this is for \(\omega \neq \omega _{n}\))\begin{equation} x\left ( t\right ) =x_{0}\cos \omega _{n}t+\left ( \frac{\dot{x}_{0}}{\omega _{n}}-\frac{r}{1-r^{2}}x_{st}\right ) \sin \omega _{n}t+\frac{x_{st}}{1-r^{2}}\sin \omega t \tag{2} \end{equation}
Notice the difference in the solution. Here is summary
| ODE | solution |
| \(m\ddot{x}+kx=F_{0}\cos \omega t\) | \(x\left ( t\right ) =\left ( x_{0}-\frac{x_{st}}{1-r^{2}}\right ) \cos \omega _{n}t+\frac{\dot{x}_{0}}{\omega _{n}}\sin \omega _{n}t+\overset{x_{p}}{\overbrace{\frac{x_{st}}{1-r^{2}}\cos \omega t}}\) |
| \(m\ddot{x}+kx=F_{0}\sin \omega t\) | \(x\left ( t\right ) =x_{0}\cos \omega _{n}t+\left ( \frac{\dot{x}_{0}}{\omega _{n}}-\frac{r}{1-r^{2}}x_{st}\right ) \sin \omega _{n}t+\overset{x_{p}}{\overbrace{\frac{x_{st}}{1-r^{2}}\sin \omega t}}\) |
| ODE | particular solution only |
| \(m\ddot{x}+c\dot{x}+kx=\frac{a_{0}}{2}\) | \(x_{p}\left ( t\right ) =\frac{a_{0}}{2}\frac{1}{k}\) |
| \(m\ddot{x}+c\dot{x}+kx=a_{n}\cos \left ( n\omega t\right ) \) | \(x_{p}\left ( t\right ) =\frac{a_{n}}{k}\frac{1}{\sqrt{\left ( 1-\left ( nr\right ) ^{2}\right ) ^{2}+\left ( 2\zeta nr\right ) ^{2}}}\cos \left ( n\omega t-\phi _{n}\right ) \) |
| \(m\ddot{x}+c\dot{x}+kx=b_{n}\sin \left ( n\omega t\right ) \) | \(x_{p}\left ( t\right ) =\frac{b_{n}}{k}\frac{1}{\sqrt{\left ( 1-\left ( nr\right ) ^{2}\right ) ^{2}+\left ( 2\zeta nr\right ) ^{2}}}\sin \left ( n\omega t-\phi _{n}\right ) \) |
Where \begin{align*} r & =\frac{\omega }{\omega _{n}}\\ \phi _{n} & =\tan ^{-1}\left ( \frac{2\zeta nr}{1-\left ( nr\right ) ^{2}}\right ) \end{align*}
For undamped system \(m\ddot{x}+kx=\delta \left ( t\right ) \) the response (solution) is (notes calls these \(g\left ( t\right ) \)) \[ g\left ( t\right ) =\frac{1}{m\omega _{n}}\sin \left ( \omega _{n}t\right ) \] And for an underdamped \(m\ddot{x}+c\dot{x}+kx=\delta \left ( t\right ) \) the response is
\[ g\left ( t\right ) =\frac{1}{m\omega _{d}}e^{-\zeta \omega _{n}t}\sin \left ( \omega _{d}t\right ) \]
For arbitray forcing function \(F\left ( t\right ) \) which can be of any forum, the response of the system to \(F\left ( t\right ) \), assuming the system was at rest is
\[ x_{conv}\left ( t\right ) =\int _{0}^{t}F\left ( \tau \right ) g\left ( t-\tau \right ) d\tau \]
DLF Dynamic locad factor. \(DLF=\frac{x\left ( t\right ) }{x_{st}}\). But we really only care for the maximum DLF. When the input is constant (step input), the \(DLF_{\max }=2\).
Response spectrum Plots the DLF\(_{\max }\) on the \(y\) axis vs \(\frac{t}{T}\) where \(T\) is the period of the system on the \(x\) axis. This is done for typical inputs such as unit step, triangle, half sine, etc...