2. Benchmark solutions

2.1. Calculation method used for the reference solution

The analytical solution to this problem is presented below.

• Vibration modes and frequencies:

The following system characterizes mass dynamics:

\(\{\begin{array}{}m\ddot{{x}_{1}}+\mathrm{2k}{x}_{1}-k{x}_{2}=0\\ m\ddot{{x}_{2}}+\mathrm{2k}{x}_{2}-k{x}_{1}=0\end{array}\) eq 2.1-1

This is equivalent to the following system:

\(\{\begin{array}{}m(\ddot{{x}_{1}}+\ddot{{x}_{2}})+k({x}_{1}+{x}_{2})=0\\ m(\ddot{{x}_{1}}-\ddot{{x}_{2}})+\mathrm{3k}({x}_{1}-{x}_{2})=0\end{array}\) eq 2.1-2

The 2 natural frequencies of the system are therefore given by:

\({\omega }_{1}=\sqrt{\frac{k}{m}}\) and \({\omega }_{2}=\sqrt{\frac{\mathrm{3k}}{m}}\) eq 2.1-3

and the associated modal deformations are:

\({\Phi }_{1}=(\begin{array}{}1\\ 1\end{array})\) and \({\Phi }_{2}=(\begin{array}{}1\\ -1\end{array})\) eq 2.1-4

The generalized matrices are:

\(\stackrel{ˉ}{M}={\Phi }^{T}M\Phi =(\begin{array}{cc}1& 1\\ 1& -1\end{array})(\begin{array}{cc}m& 0\\ 0& m\end{array})(\begin{array}{cc}1& 1\\ 1& -1\end{array})=(\begin{array}{cc}\mathrm{2m}& 0\\ 0& \mathrm{2m}\end{array})\)

\(\stackrel{ˉ}{K}={\Phi }^{T}K\Phi =(\begin{array}{cc}1& 1\\ 1& -1\end{array})(\begin{array}{cc}\mathrm{2k}& -k\\ -k& \mathrm{2k}\end{array})(\begin{array}{cc}1& 1\\ 1& -1\end{array})=(\begin{array}{cc}\mathrm{2k}& 0\\ 0& \mathrm{6k}\end{array})\) eq 2.1-5

• Transient response:

The sinusoidal force is applied to the first mass: \(F=(\begin{array}{}1\\ 0\end{array})\mathrm{sin}(\omega t)\)

The dynamic system checked is as follows:

\(M\ddot{X}+KX=F\) eq 2.1-6

By projecting on the basis of clean modes, we get:

\({\Phi }^{T}M\Phi \ddot{\eta }+{\Phi }^{T}K\Phi \eta ={\Phi }^{T}F\) eq 2.1-7

Either:

\((\begin{array}{cc}\mathrm{2m}& 0\\ 0& \mathrm{2m}\end{array})(\begin{array}{}\ddot{{\eta }_{1}}\\ \ddot{{\eta }_{2}}\end{array})+(\begin{array}{cc}\mathrm{2k}& 0\\ 0& \mathrm{6k}\end{array})(\begin{array}{}{\eta }_{1}\\ {\eta }_{2}\end{array})=(\begin{array}{cc}1& 1\\ 1& -1\end{array})(\begin{array}{}1\\ 0\end{array})\mathrm{sin}(\omega t)\) eq 2.1-8

We therefore end up with the following decoupled system:

\(\{\begin{array}{}m\ddot{{\eta }_{1}}+k{\eta }_{1}=\frac{1}{2}\mathrm{sin}(\omega t)\\ m\ddot{{\eta }_{2}}+3k{\eta }_{2}=\frac{1}{2}\mathrm{sin}(\omega t)\end{array}\) eq 2.1-9

The solution of this system is given by:

\(\{\begin{array}{}{\eta }_{1}(t)={A}_{1}\mathrm{cos}({\omega }_{1}t)+{B}_{1}\mathrm{sin}({\omega }_{1}t)+\frac{\mathrm{sin}(\omega t)}{\mathrm{2m}({\omega }_{1}^{2}-{\omega }^{2})}\\ {\eta }_{2}(t)={A}_{2}\mathrm{cos}({\omega }_{2}t)+{B}_{2}\mathrm{sin}({\omega }_{2}t)+\frac{\mathrm{sin}(\omega t)}{\mathrm{2m}({\omega }_{2}^{2}-{\omega }^{2})}\end{array}\) eq 2.1-10

Displacements in physical space are obtained by the Ritz formula:

\(X=(\begin{array}{}{x}_{1}\\ {x}_{2}\end{array})=\Phi \eta =(\begin{array}{cc}1& 1\\ 1& -1\end{array})(\begin{array}{}{\eta }_{1}\\ {\eta }_{2}\end{array})=(\begin{array}{}{\eta }_{1}+{\eta }_{2}\\ {\eta }_{1}-{\eta }_{2}\end{array})\) eq 2.1-11

From this we deduce the expressions for \({x}_{1}(t)\) and \({x}_{2}(t)\):

\(\{\begin{array}{}{x}_{1}(t)={A}_{1}\mathrm{cos}({\omega }_{1}t)+{B}_{1}\mathrm{sin}({\omega }_{1}t)+{A}_{2}\mathrm{cos}({\omega }_{2}t)+{B}_{2}\mathrm{sin}({\omega }_{2}t)+\frac{\mathrm{sin}(\omega t)}{\mathrm{2m}}(\frac{1}{{\omega }_{1}^{2}-{\omega }^{2}}+\frac{1}{{\omega }_{2}^{2}-{\omega }^{2}})\\ {x}_{2}(t)={A}_{1}\mathrm{cos}({\omega }_{1}t)+{B}_{1}\mathrm{sin}({\omega }_{1}t)-{A}_{2}\mathrm{cos}({\omega }_{2}t)+{B}_{2}\mathrm{sin}({\omega }_{2}t)+\frac{\mathrm{sin}(\omega t)}{\mathrm{2m}}(\frac{1}{{\omega }_{1}^{2}-{\omega }^{2}}-\frac{1}{{\omega }_{2}^{2}-{\omega }^{2}})\end{array}\)

eq 2.1-12

At the initial moment, the system is at rest, hence the final expressions for \({x}_{1}(t)\) and \({x}_{2}(t)\):

\(\{\begin{array}{}{x}_{1}(t)=\frac{1}{\mathrm{2m}}\left[\frac{\mathrm{sin}(\omega t)-\frac{\omega }{{\omega }_{1}}\mathrm{sin}({\omega }_{1}t)}{{\omega }_{1}^{2}-{\omega }^{2}}+\frac{\mathrm{sin}(\omega t)-\frac{\omega }{{\omega }_{2}}\mathrm{sin}({\omega }_{2}t)}{{\omega }_{2}^{2}-{\omega }^{2}}\right]\\ {x}_{2}(t)=\frac{1}{\mathrm{2m}}\left[\frac{\mathrm{sin}(\omega t)-\frac{\omega }{{\omega }_{1}}\mathrm{sin}({\omega }_{1}t)}{{\omega }_{1}^{2}-{\omega }^{2}}-\frac{\mathrm{sin}(\omega t)-\frac{\omega }{{\omega }_{2}}\mathrm{sin}({\omega }_{2}t)}{{\omega }_{2}^{2}-{\omega }^{2}}\right]\end{array}\) eq 2.1-13

The velocities of the two masses are calculated by deriving the displacements with respect to time:

\(\{\begin{array}{}\dot{{x}_{1}}(t)=\frac{\omega }{\mathrm{2m}}\left[\frac{\mathrm{cos}(\omega t)-\mathrm{cos}({\omega }_{1}t)}{{\omega }_{1}^{2}-{\omega }^{2}}+\frac{\mathrm{cos}(\omega t)-\mathrm{cos}({\omega }_{2}t)}{{\omega }_{2}^{2}-{\omega }^{2}}\right]\\ \dot{{x}_{2}}(t)=\frac{\omega }{\mathrm{2m}}\left[\frac{\mathrm{cos}(\omega t)-\mathrm{cos}({\omega }_{1}t)}{{\omega }_{1}^{2}-{\omega }^{2}}-\frac{\mathrm{cos}(\omega t)-\mathrm{cos}({\omega }_{2}t)}{{\omega }_{2}^{2}-{\omega }^{2}}\right]\end{array}\) eq 2.1-14

The accelerations of the two masses are calculated by deriving the velocities with respect to time:

\(\{\begin{array}{}\ddot{{x}_{1}}(t)=\frac{\omega }{\mathrm{2m}}\left[\frac{\omega \mathrm{sin}(\omega t)-{\omega }_{1}\mathrm{sin}({\omega }_{1}t)}{{\omega }_{1}^{2}-{\omega }^{2}}+\frac{\omega \mathrm{sin}(\omega t)-{\omega }_{2}\mathrm{sin}({\omega }_{2}t)}{{\omega }_{2}^{2}-{\omega }^{2}}\right]\\ \ddot{{x}_{2}}(t)=\frac{\omega }{\mathrm{2m}}\left[\frac{\omega \mathrm{sin}(\omega t)-{\omega }_{1}\mathrm{sin}({\omega }_{1}t)}{{\omega }_{1}^{2}-{\omega }^{2}}-\frac{\omega \mathrm{sin}(\omega t)-{\omega }_{2}\mathrm{sin}({\omega }_{2}t)}{{\omega }_{2}^{2}-{\omega }^{2}}\right]\end{array}\) eq 2.1-15

2.2. Benchmark results

The comparison of the results focuses on the displacements, speeds and accelerations along the axis of the two masses, at five different times.

2.3. Uncertainty about the solution

The reference solution is accurate.

The discrete model perfectly represents the problem posed (the modal base is complete; there is therefore no approximation linked to a possible modal truncation). The number of modes of the modal projection base is equal to the number of measurements, so the solution of the inversion is exact (as opposed to an approximate solution of a generalized inverse problem). If the search for the nodes facing each other is good, the displacements obtained after projection must be in perfect harmony with the experimental values. The speeds and accelerations are determined by deriving the modal contributions identified via a linear approximation scheme in time, which can therefore generate some errors.