The input file to ANSYS is given to us in plain text file as the following
For a mass-spring system the equation of motion is \[ \ddot{x}+\omega _{n}^{2}x=0 \] Where \(\omega _{n}=\sqrt{\frac{k}{m}}=\sqrt{\frac{10}{10}}=1\) rad/sec. Since \(\omega _{n}=2\pi f_{n}\), hence \(f_{n}=\frac{\omega _{n}}{2\pi }=\frac{1}{2\pi }=0.159\,2\) Hz. Therefore the frequency given by ANSYS is in Hz and not the circular frequency rad/sec.
Unable to verify this result. At first I thought ANSYS uses gravity and the spring is vertically connected, therefore the static displacement would be\[ x_{st}=\frac{W}{k}\] Where \(W\) is the weight attached to end of spring. But this gives \(x_{st}=\frac{mg}{k}=\frac{10g}{10}=g\). And depending on units used (ANSYS do not use units and assumes that the input is using correct units), then value shown which is \(0.316228\) should be numerical value of \(g\). But this would not be valid number using any units. Unable to find out how ANSYS came up with this value.
The free body diagram and the inertial diagram are given below. It is assumed that motion is measured from equilibrium position with the mass already in attached to springs. Hence the weight of the beam do not show up in the FBD.
Taking moments around hinge at point \(o\) and using anti-clockwise as positive gives (assuming small angle \(\theta \))\begin{align*} k_{1}\left ( a\theta \right ) a+k_{2}\left ( b\theta \right ) b-F_{0}\sin \left ( \omega t\right ) L & =-\left ( \frac{1}{3}mL^{2}+ML^{2}\right ) \ddot{\theta }\\ \left ( \frac{1}{3}mL^{2}+ML^{2}\right ) \ddot{\theta }+\theta \left ( k_{1}a^{2}+k_{2}b^{2}\right ) & =F_{0}\sin \left ( \omega t\right ) L \end{align*}
In standard form, the above becomes\[ m_{eq}\ddot{\theta }+k_{eq}\theta =F_{0}\sin \omega t \] Where \begin{align*} \omega _{n}^{2} & =\frac{k_{eq}}{m_{eq}}\\ & =\frac{k_{1}a^{2}+k_{2}b^{2}}{L^{2}\left ( \frac{1}{3}m+M\right ) } \end{align*}
This model is single degree of freedom system, undamped, with forced input. Hence we know its solution is given by\[ \theta \left ( t\right ) =\theta _{h}\left ( t\right ) +\theta _{p}\left ( t\right ) \] Where \(\theta _{p}\left ( t\right ) \) is particular solution and \(\theta _{h}\left ( t\right ) \) is homogenous solution. We know that\[ \theta _{h}\left ( t\right ) =c_{1}\cos \omega _{n}t+c_{2}\sin \omega _{n}t \] And assuming \(\theta _{p}\left ( t\right ) =X\sin \omega t\). Now we need to check if \(\omega \neq \omega _{n}\) so to decide on which solution to pick. Using the numerical values given\begin{align*} k_{eq} & =k_{1}a^{2}+k_{2}b^{2}\\ & =\left ( 5000\right ) \left ( 0.25\right ) ^{2}+\left ( 5000\right ) \left ( 0.5\right ) ^{2}\\ & =1562.5\text{ N/m} \end{align*}
And\begin{align*} M_{eq} & =L^{2}\left ( \frac{1}{3}m+M\right ) \\ & =\left ( 1\right ) ^{2}\left ( \left ( \frac{1}{3}\right ) \left ( 10\right ) +50\right ) \\ & =53.333\text{ kg} \end{align*}
Hence\[ \omega _{n}=\sqrt{\frac{k_{eq}}{M_{eq}}}=\sqrt{\frac{1562.5}{53.333}}=5.413\,\text{rad/sec}\] But the forcing frequency is given as \[ \omega =1000\left ( \frac{2\pi }{rev}\right ) \left ( \frac{\min }{60}\right ) =1000\left ( \frac{2\pi }{60}\right ) =104.72\text{ rad/sec}\] Hence \(\omega \neq \omega _{n}\). We also see \(\omega >\omega _{n}\) which means \(r>1\) where \(r=\frac{\omega }{\omega _{n}}\), so we also expect that particular solution displacement maximum displacement to be negative. Now we use the standard solution, which is\[ \theta _{p}\left ( t\right ) =X\sin \omega t \] Where\begin{align*} X & =\frac{F_{0}}{k_{eq}-m_{eq}\omega ^{2}}\\ & =\frac{F_{0}}{m_{eq}}\frac{1}{\frac{k_{eq}}{m_{eq}}-\omega ^{2}}\\ & =\frac{F_{0}}{m_{eq}}\frac{1}{\omega _{n}^{2}-\omega ^{2}}\\ & =\frac{F_{0}}{m_{eq}\omega _{n}^{2}}\frac{1}{1-\left ( \frac{\omega }{\omega _{n}}\right ) ^{2}}\\ & =\frac{F_{0}}{k_{eq}}\frac{1}{1-\left ( \frac{\omega }{\omega _{n}}\right ) ^{2}} \end{align*}
Calling \(\frac{\omega }{\omega _{n}}=r\), which is the standard notation and since \(\frac{F_{0}}{k_{eq}}=x_{st}\) the static deflection, then the above becomes\[ X=\frac{x_{st}}{1-r^{2}}\] We notice again, since \(r>1\) in this problem, then \(X\) is negative. It is out of phase with the forcing function. The particular solution can now be written as\begin{align*} \theta _{p}\left ( t\right ) & =X\sin \omega t\\ & =\frac{x_{st}}{1-r^{2}}\sin \omega t \end{align*}
And the total solution is\begin{equation} \theta \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 \(\theta \left ( 0\right ) =\theta _{0},\dot{\theta }\left ( 0\right ) =\dot{\theta }_{0}\), then (1) at \(t=0\) becomes\[ \theta _{0}=c_{1}\] Hence solution becomes\[ \theta \left ( t\right ) =\theta _{0}\cos \omega _{n}t+c_{2}\sin \omega _{n}t+\frac{x_{st}}{1-r^{2}}\sin \omega t \] Taking derivative\[ \theta ^{\prime }\left ( t\right ) =\omega _{n}\theta _{0}\sin \omega _{n}t+\omega _{n}c_{2}\cos \omega _{n}t+\omega \frac{x_{st}}{1-r^{2}}\cos \omega t \] At \(t=0\) the above becomes\[ \dot{\theta }_{0}=\omega _{n}c_{2}+\omega \frac{x_{st}}{1-r^{2}}\] Hence \begin{align*} c_{2} & =\frac{\dot{\theta }_{0}}{\omega _{n}}-\frac{\omega }{\omega _{n}}\frac{x_{st}}{1-r^{2}}\\ & =\frac{\dot{\theta }_{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} \theta \left ( t\right ) =\overset{\text{homogeneous}}{\overbrace{\theta _{0}\cos \omega _{n}t+\left ( \frac{\dot{\theta }_{0}}{\omega _{n}}-\frac{r}{1-r^{2}}x_{st}\right ) \sin \omega _{n}t}}+\overset{\text{particular}}{\overbrace{\left ( \frac{x_{st}}{1-r^{2}}\right ) \sin \omega t}}\tag{2} \end{equation} The problem now asks for steady state solution. It is not clear to me what is this meant to be, since there is no damping in the system, and hence the full solution remain for all time. Therefore, will show the full solution (using zero initial conditions) and will also show the particular solution.
This is a plot of the full solution, assuming that all initial conditions are zero. Therefore, this is a plot of this solution\[ \theta \left ( t\right ) =-\frac{F_{0}}{k_{eq}}\frac{r}{1-r^{2}}\sin \omega _{n}t+\left ( \frac{F_{0}}{k_{eq}}\frac{1}{1-r^{2}}\right ) \sin \omega t \] Obtained from (2) by setting \(\theta _{0}=0,\dot{\theta }_{0}=0\)\begin{align*} \theta \left ( t\right ) & =-\frac{500}{1562.5}\left ( \frac{3.\,574\,4}{1-\left ( 3.\,574\,4\right ) ^{2}}\right ) \sin \left ( 5.413t\right ) +\left ( \frac{500}{1562.5}\left ( \frac{1}{1-\left ( \allowbreak 3.\,574\,4\right ) ^{2}}\right ) \right ) \sin \left ( 104.72t\right ) \\ & =0.09713\sin \left ( 5.413t\right ) -0.0272\sin \left ( 104.72t\right ) \end{align*}
Here is a plot of the full solution for the first \(1\) second
The particular solution (steady state?) is\begin{align*} \theta _{p}\left ( t\right ) & =\frac{x_{st}}{1-r^{2}}\sin \omega t\\ & =0.0272\sin \left ( 104.72t\right ) \end{align*}
Here is a plot of the particular solution for the first \(0.25\) second
The equation of motion (assuming \(\sin \left ( \omega t\right ) \) for the force) is1 \[ m\ddot{x}+kx=F_{0}\sin \left ( \omega t\right ) \]
Where \(k=5000\) N/m, \(m=10\) kg, \(F_{0}=250\) N. We know the solution to the above is given by (but we here have to assume that \(\omega \neq \omega _{n}\))\[ x\left ( t\right ) =\overset{\text{homogeneous}}{\overbrace{x_{0}\cos \omega _{n}t+\left ( \frac{\dot{x}_{0}}{\omega _{n}}-\frac{r}{1-r^{2}}x_{st}\right ) \sin \omega _{n}t}}+\overset{\text{particular}}{\overbrace{\left ( \frac{x_{st}}{1-r^{2}}\right ) \sin \omega t}}\] Looking now at only the steady state solution (in this case, it is the particular solution) then we see that\[ x_{ss}\left ( t\right ) =\left ( \frac{x_{st}}{1-r^{2}}\right ) \sin \omega t \] Hence maximum is\[ x_{\max }\left ( t\right ) =\frac{x_{st}}{1-r^{2}}\] we are told that \(x_{\max }=0.1\) meter, and . But \(r=\frac{\omega }{\omega _{n}}\) and \(x_{st}=\frac{F_{0}}{k}\). Therefore the above becomes \[ x_{\max }=\frac{F_{0}}{k}\frac{1}{1-\left ( \frac{\omega }{\omega _{n}}\right ) ^{2}}\] In the above equation everything is known except for \(\omega \). Solving for \(\omega \) gives\begin{align*} 1-\left ( \frac{\omega }{\omega _{n}}\right ) ^{2} & =\frac{F_{0}}{kx_{\max }}\\ \left ( \frac{\omega }{\omega _{n}}\right ) ^{2} & =1-\frac{F_{0}}{kx_{\max }}\\ \omega ^{2} & =\left ( 1-\frac{F_{0}}{kx_{\max }}\right ) \omega _{n}^{2} \end{align*}
But \(\omega _{n}=\sqrt{\frac{k}{m}}\), hence\[ \omega =\sqrt{\frac{k}{m}}\sqrt{\left ( 1-\frac{F_{0}}{kx_{\max }}\right ) }\] Substituting numerical values\begin{align*} \omega & =\sqrt{\frac{5000}{10}}\sqrt{\left ( 1-\frac{250}{\left ( 5000\right ) \left ( 0.1\right ) }\right ) }\\ & =22.361\sqrt{0.5}\\ & =15.812\text{ rad/sec} \end{align*}