This is a 2 D.O.F. system. The degrees of freedom are \(\theta _{1}\) and \(\theta _{2}\) shown above in the positive sense. The method of power balance is used to obtain the EOM.
The system kinetic energy is \(T=\frac {1}{2}m_{1}\frac {L^{2}}{3}\left ( \theta _{1}^{\prime }\right ) ^{2}+\frac {1}{2}m_{1}\frac {L^{2}}{3}\left ( \theta _{2}^{\prime }\right ) ^{2}\), hence by comparing term to the quadratic form, the mass matrix part of the EOM is obtained\[ \frac {L^{2}}{3}\begin {bmatrix} m_{1} & 0\\ 0 & m_{2}\end {bmatrix}\begin {Bmatrix} \theta _{1}^{\prime \prime }\\ \theta _{2}^{\prime \prime }\end {Bmatrix} \]
To find spring stiffness, the spring deformation is found using stiff spring approximation.\begin {align*} \Delta ^{\prime } & =\left (V_{B}-V_{A}\right ) \cdot e_{B/A}\\ & =\left (L\theta _{2}^{\prime }\mathbf {i-}L\theta _{1}^{\prime }\mathbf {j}\right ) \cdot \left (\cos \beta \mathbf {i}-\sin \beta \mathbf {j}\right ) \end {align*}
Where \(e_{B/A}\,\ \)is unit vector oriented to B from A and \(\tan \beta =\frac {L}{h}\). The above becomes\[ \Delta ^{\prime }=L\theta _{2}^{\prime }\cos \beta \mathbf {+}L\theta _{1}^{\prime }\sin \beta \]
Hence, integrating, squaring and collecting terms gives\begin {align*} \Delta & =L\theta _{2}\cos \beta \mathbf {+}L\theta _{1}\sin \beta \\ \Delta ^{2} & =L^{2}\theta _{2}^{2}\cos ^{2}\beta +L^{2}\theta _{1}^{2}\sin ^{2}\beta +2L^{2}\theta _{1}\theta _{2}\sin \beta \cos \beta \\ & =\theta _{1}^{2}\left (L^{2}\sin ^{2}\beta \right ) +\theta _{2}^{2}\left ( L^{2}\cos ^{2}\beta \right ) +\theta _{1}\theta _{2}\left (2L^{2}\sin \beta \cos \beta \right ) \end {align*}
Using the quadratic form of the power balance method, the spring stiffness matrix part of the EOM is found from \(V_{spring}=\frac {1}{2}k\left ( \Delta ^{2}\right ) \) and by comparing quadratic terms, which leads to\[ V_{spring}=kL^{2}\begin {bmatrix} \sin ^{2}\beta & 2\sin \beta \cos \beta \\ 2\sin \beta \cos \beta & \cos ^{2}\beta \end {bmatrix}\begin {Bmatrix} \theta _{1}\\ \theta _{2}\end {Bmatrix} \]
But \(\sin \beta \cos \beta =\frac {1}{2}\left (\sin 2\beta \right ) \) hence\[ V_{spring}=kL^{2}\begin {bmatrix} \sin ^{2}\beta & \sin 2\beta \\ \sin 2\beta & \cos ^{2}\beta \end {bmatrix}\begin {Bmatrix} \theta _{1}\\ \theta _{2}\end {Bmatrix} \]
Stiffness due to gravity \(V_{g}\) is now found. Let datum for zero potential energy be at the horizontal level of the top bar, hence \(V_{g}=m_{1}g\frac {L}{2}\sin \theta _{1}-m_{2}g\frac {L}{2}\cos \theta _{2}\). Since the derivatives are evaluated at static equilibrium \(\theta _{1}=0\) and \(\theta _{2}=0,\) the only term that remains is \(m_{2}g\frac {L}{2}\) which is now added to the \(k_{22}\) term of the stiffness matrix. \(FL\) is the generalized force for \(\theta _{2}\) since work done by \(F\) in making virtual \(\delta \theta _{2}\) is \(FL\delta \theta _{2}\). Therefore, the EOM becomes\[ \frac {L^{2}}{3}\begin {bmatrix} m_{1} & 0\\ 0 & m_{2}\end {bmatrix}\begin {Bmatrix} \theta _{1}^{\prime \prime }\\ \theta _{2}^{\prime \prime }\end {Bmatrix} +kL^{2}\begin {bmatrix} \sin ^{2}\beta & \sin 2\beta \\ \sin 2\beta & \cos ^{2}\beta +m_{2}g\frac {L}{2}\end {bmatrix}\begin {Bmatrix} \theta _{1}\\ \theta _{2}\end {Bmatrix} =\begin {Bmatrix} 0\\ FL \end {Bmatrix} \]
To check units of the above EOM, looking at the first EOM from above\[ \frac {L^{2}}{3}m_{1}\theta _{1}^{\prime \prime }+kL^{2}\left (\sin ^{2}\beta \right ) \theta _{1}+kL^{2}\left (\sin 2\beta \right ) \theta _{2}=0 \]
Let \(\theta _{1}=0.\) Hence \(\frac {L^{2}}{3}m_{1}\theta _{1}^{\prime \prime }=-kL^{2}\left (\sin 2\beta \right ) \theta _{2}\). Assume \(\theta _{2}\geq 0\) and the system is now released to move. We should expect the top bar to accelerate down (negative), since the spring is stretched. Looking at the above, we see that \(\theta _{1}^{\prime \prime }\leq 0\). hence this is correct.
Now let \(\theta _{2}=0.\) Hence \(\frac {L^{2}}{3}m_{1}\theta _{1}^{\prime \prime }=-kL^{2}\left (\sin ^{2}\beta \right ) \theta _{1}\). Assume \(\theta _{1}\geq 0\) and the system is now released to move. We should expect the top bar to accelerate down (negative) since the spring was stretched. Looking at the above, we see that \(\theta _{1}^{\prime \prime }\leq 0\). This is correct.
Checking the second EOM\[ \frac {L^{2}}{3}m_{2}\theta _{2}^{\prime \prime }+kL^{2}\left (\sin 2\beta \right ) \theta _{1}+kL^{2}\left (\cos ^{2}\beta \right ) \theta _{2}=FL-m_{2}g\frac {L}{2}\theta _{2}\]
Let \(\theta _{1}=0\) and \(F=0\) then\[ \frac {L^{2}}{3}m_{2}\theta _{2}^{\prime \prime }=-m_{2}g\frac {L}{2}\theta _{2}-L^{2}\left (\cos ^{2}\beta \right ) \theta _{2}\]
Assume \(\theta _{2}\geq 0\) and the system is now released to move. We would expect the right bar to accelerate back (negative) when released to move. From the equation we see that \(\theta _{2}^{\prime \prime }\leq 0\). This is correct.
Now let \(\theta _{2}=0\) and \(F=0\) then\[ \frac {L^{2}}{3}m_{2}\theta _{2}^{\prime \prime }=-kL^{2}\left (\sin 2\beta \right ) \theta _{1}\]
Assume \(\theta _{1}\leq 0\) and the system is now released to move. We would expect the bar to accelerate to the right (positive) since the spring was compressed. From the equation we see that \(\theta _{2}^{\prime \prime }>0\). This is correct.
\begin {align*} \det \left (\left [ k\right ] -\omega ^{2}\left [ m\right ] \right ) & =0\\ \det \left ( \begin {bmatrix} k & -k\\ -k & k \end {bmatrix} -\omega ^{2}\begin {bmatrix} 2m & 0\\ 0 & m \end {bmatrix} \right ) & =0\\ \det \left ( \begin {bmatrix} 1 & -1\\ -1 & 1 \end {bmatrix} -\omega ^{2}\frac {m}{k}\begin {bmatrix} 2 & 0\\ 0 & 1 \end {bmatrix} \right ) & =0 \end {align*}
For normalization, let \(t^{\prime }=\omega t\) then \(\frac {dt^{\prime }}{dt}=\omega \) and using \(t^{\prime }\) instead of \(t\) as the independent variable the above becomes\begin {align*} \det \left ( \begin {bmatrix} 1 & -1\\ -1 & 1 \end {bmatrix} -\omega ^{2}\begin {bmatrix} 2 & 0\\ 0 & 1 \end {bmatrix} \right ) & =0\\ \det \left ( \begin {bmatrix} 1-2\omega ^{2} & -1\\ -1 & 1-\omega ^{2}\end {bmatrix} \right ) & =0\\ \left (1-2\omega ^{2}\right ) \left (1-\omega ^{2}\right ) -1 & =0 \end {align*}
The roots are \(\omega =0\) and \(\omega =\sqrt {\frac {3}{2}}\). When \(\omega =0\) it is a rigid body motion, So any \(\varphi \) will do. Let \(\varphi _{1}=\begin {Bmatrix} 1\\ 1 \end {Bmatrix} \). When \(\omega =\sqrt {\frac {3}{2}}\) then\begin {align*} \left ( \begin {bmatrix} 1 & -1\\ -1 & 1 \end {bmatrix} -\frac {3}{2}\begin {bmatrix} 2 & 0\\ 0 & 1 \end {bmatrix} \right ) \begin {Bmatrix} \varphi _{12}\\ \varphi _{22}\end {Bmatrix} & =\begin {Bmatrix} 0\\ 0 \end {Bmatrix} \\\begin {bmatrix} -2 & -1\\ -1 & -\frac {1}{2}\end {bmatrix}\begin {Bmatrix} \varphi _{12}\\ \varphi _{22}\end {Bmatrix} & =\begin {Bmatrix} 0\\ 0 \end {Bmatrix} \end {align*}
let \(\varphi _{12}=1\) then \(-2-\varphi _{22}=0\) or \(\varphi _{22}=-2\) hence \(\varphi _{2}=\begin {Bmatrix} 1\\ -2 \end {Bmatrix} \).\begin {align*} \mu _{1} & =\varphi _{1}^{T}\left [ M\right ] \varphi _{1}=\begin {Bmatrix} 1\\ 1 \end {Bmatrix} ^{T}\begin {bmatrix} 2 & 0\\ 0 & 1 \end {bmatrix}\begin {Bmatrix} 1\\ 1 \end {Bmatrix} =3\\ \mu _{2} & =\varphi _{2}^{T}\left [ M\right ] \varphi _{2}=\begin {Bmatrix} 1\\ -2 \end {Bmatrix} ^{T}\begin {bmatrix} 2 & 0\\ 0 & 1 \end {bmatrix}\begin {Bmatrix} 1\\ -2 \end {Bmatrix} =6 \end {align*}
Hence\begin {align*} \Phi _{1} & =\frac {\varphi _{1}}{\sqrt {\mu _{1}}}=\frac {1}{\sqrt {3}}\begin {Bmatrix} 1\\ 1 \end {Bmatrix} =\begin {Bmatrix} 0.57735\\ 0.57735 \end {Bmatrix} \allowbreak \\ \Phi _{2} & =\frac {\varphi _{2}}{\sqrt {\mu _{2}}}=\frac {1}{\sqrt {6}}\begin {Bmatrix} 1\\ -2 \end {Bmatrix} =\begin {Bmatrix} 0.40825\\ -0.81650 \end {Bmatrix} \end {align*}
\[ \Phi =\begin {bmatrix} \frac {1}{\sqrt {3}} & \frac {1}{\sqrt {6}}\\ \frac {1}{\sqrt {3}} & -\frac {2}{\sqrt {6}}\end {bmatrix} \]
The EOM is\[\begin {bmatrix} 1 & 0\\ 0 & 1 \end {bmatrix}\begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\end {Bmatrix} +\begin {bmatrix} 0 & 0\\ 0 & \frac {3}{2}\end {bmatrix}\begin {Bmatrix} \eta _{1}\\ \eta _{2}\end {Bmatrix} =\Phi ^{T}\begin {Bmatrix} 0\\ -f\relax (t) \end {Bmatrix} =\begin {bmatrix} \frac {1}{\sqrt {3}} & \frac {1}{\sqrt {6}}\\ \frac {1}{\sqrt {3}} & -\frac {2}{\sqrt {6}}\end {bmatrix} ^{T}\begin {Bmatrix} 0\\ -f\relax (t) \end {Bmatrix} =\begin {Bmatrix} -\frac {1}{3}\sqrt {3}f\relax (t) \\ \frac {1}{3}\sqrt {6}f\relax (t) \end {Bmatrix} \]
initial conditions are \(\begin {Bmatrix} \eta _{1}\relax (0) \\ \eta _{2}\relax (0) \end {Bmatrix} =\begin {Bmatrix} 0\\ 0 \end {Bmatrix} \) and \[\begin {Bmatrix} \eta _{1}^{\prime }\relax (0) \\ \eta _{2}^{\prime }\relax (0) \end {Bmatrix} =\Phi ^{T}\left [ M\right ] \begin {Bmatrix} v_{0}\\ v_{0}\end {Bmatrix} =\begin {bmatrix} \frac {1}{\sqrt {3}} & \frac {1}{\sqrt {6}}\\ \frac {1}{\sqrt {3}} & -\frac {2}{\sqrt {6}}\end {bmatrix} ^{T}\begin {bmatrix} 2 & 0\\ 0 & 1 \end {bmatrix}\begin {Bmatrix} v_{0}\\ v_{0}\end {Bmatrix} =\begin {Bmatrix} \sqrt {3}v_{0}\\ 0 \end {Bmatrix} \]
Therefore, the first ODE is\[ \eta _{1}^{\prime \prime }=-\frac {1}{3}\sqrt {3}F_{0}\left (h\relax (t) -h\left (t-T\right ) \right ) \]
with IC \(\eta _{1}\relax (0) =0\) and \(\eta _{1}^{\prime }\relax (0) =\sqrt {3}v_{0}\). The second ODE is\[ \eta _{2}^{\prime \prime }+\frac {3}{2}\eta _{2}=\frac {1}{3}\sqrt {6}F_{0}\left ( h\relax (t) -h\left (t-T\right ) \right ) \]
with IC \(\eta _{1}\relax (0) =0\) and \(\eta _{1}^{\prime }\relax (0) =0\)
\begin {align*} x_{1}\relax (t) & =\Phi _{11}\eta _{1}\relax (t) +\Phi _{12}\eta _{2}\relax (t) \\ & =\frac {1}{\sqrt {3}}\eta _{1}\relax (t) +\frac {1}{\sqrt {6}}\eta _{2}\relax (t) \end {align*}
Therefore, \(x_{1}\relax (t) \) solution has contribution from \(\eta _{1}\relax (t) \) and \(\eta _{2}\relax (t) \). But \(\eta _{1}\relax (t) \) is linear with positive slope of \(\nu _{0}\) and \(\eta _{2}\relax (t) \) is a sinusoidal, with no damping. So adding both together, here is a sketch of possible solution
\begin {align*} \Phi ^{T}\left [ K\right ] \Phi ^{T} & =\begin {bmatrix} \omega _{1}^{2} & 0\\ 0 & \omega _{2}^{2}\end {bmatrix} \\\begin {bmatrix} 0.85 & 1.1\\ 0.65 & -0.5 \end {bmatrix} ^{T}\begin {bmatrix} 100 & -100\\ -100 & 200 \end {bmatrix}\begin {bmatrix} 0.85 & 1.1\\ 0.65 & -0.5 \end {bmatrix} & =\begin {bmatrix} \omega _{1}^{2} & 0\\ 0 & \omega _{2}^{2}\end {bmatrix} \\\begin {bmatrix} 46.25 & -0.5\\ -0.5 & 281.0 \end {bmatrix} & =\begin {bmatrix} \omega _{1}^{2} & 0\\ 0 & \omega _{2}^{2}\end {bmatrix} \end {align*}
Hence \(\omega _{1}^{2}=46.25\) or \(\omega _{1}=6.8\) rad/sec
Using the first natural frequency, since this has the longest time constant \(\tau =\frac {1}{\zeta _{1}\omega _{1}}\)and solving for the number of periods using logarithmic decrement method\begin {equation} \frac {1}{N}\ln \left (\frac {y_{1}}{y_{N}}\right ) =2\pi \zeta _{1} \tag {1} \end {equation}
\(\zeta _{1}\) is not known but can be found by evaluating \(\Phi ^{T}\left [ C\right ] \Phi ^{T}\) \[ \Phi ^{T}\left [ K\right ] \Phi ^{T}=\begin {bmatrix} 0.85 & 1.1\\ 0.65 & -0.5 \end {bmatrix} ^{T}\begin {bmatrix} 0.04 & 0\\ 0 & 0.05 \end {bmatrix}\begin {bmatrix} 0.85 & 1.1\\ 0.65 & -0.5 \end {bmatrix} =\begin {bmatrix} 0.05 & 0.021\\ 0.021\, & 0.061 \end {bmatrix} \]
and assuming small damping approximation, then \(2\zeta _{1}\omega _{1}=0.05\). Hence \(\zeta _{1}=\frac {0.05}{2\omega _{1}}=\frac {0.05}{2\left (6.8\right ) }=0.0038\). Now that the critical damping ratio for the first mode is found, we can use the method of logarithmic decrement to find how many periods it takes to attenuate by \(99\%\)
Let \(\frac {y_{1}}{y_{N}}=\frac {1}{0.01}=100\) then Eq (1) becomes \begin {align*} \frac {1}{N}\ln \left (100\right ) & =2\pi \left (0.0038\right ) \\ N & =\frac {\left (4.605\right ) }{2\pi \left (0.0038\right ) }=192.87\\ & =193 \end {align*}
Where \(N\) is the number or periods needed. But \(T=\frac {2\pi }{\omega _{1}}\),hence the time needed is\[ t=NT=192T=192\frac {2\pi }{\omega _{1}}=192\frac {2\pi }{6.8}=177.41\text { sec}\]
So it takes \(178\) seconds for the first modal (decoupled) solution to attenuate in amplitude by \(99\%\). Since this is the dominant time constant, we expect the physical solution to attenuate in approximately the same amount of time as well.
The EOM is, in modal coordinates\[\begin {bmatrix} 1 & 0\\ 0 & 1 \end {bmatrix}\begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\end {Bmatrix} +\Phi ^{T}\begin {bmatrix} 0.04 & 0\\ 0 & 0.05 \end {bmatrix} \Phi \begin {Bmatrix} \eta _{1}^{\prime }\\ \eta _{2}^{\prime }\end {Bmatrix} +\begin {bmatrix} \omega _{1}^{2} & 0\\ 0 & \omega _{2}^{2}\end {bmatrix}\begin {Bmatrix} \eta _{1}\\ \eta _{2}\end {Bmatrix} =\Phi ^{T}\begin {Bmatrix} \operatorname {Re}\left (Ae^{i\varpi t}\right ) \\ 0 \end {Bmatrix} \]
But \[ \Phi ^{T}\begin {bmatrix} 0.04 & 0\\ 0 & 0.05 \end {bmatrix} \Phi =\begin {bmatrix} 0.85 & 1.1\\ 0.65 & -0.5 \end {bmatrix} ^{T}\begin {bmatrix} 0.04 & 0\\ 0 & 0.05 \end {bmatrix}\begin {bmatrix} 0.85 & 1.1\\ 0.65 & -0.5 \end {bmatrix} =\begin {bmatrix} 0.05 & 0.021\\ 0.021 & 0.061 \end {bmatrix} \]
Hence EOM in modal coordinates become\[\begin {bmatrix} 1 & 0\\ 0 & 1 \end {bmatrix}\begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\end {Bmatrix} +\begin {bmatrix} 0.05 & 0.021\\ 0.021 & 0.061 \end {bmatrix}\begin {Bmatrix} \eta _{1}^{\prime }\\ \eta _{2}^{\prime }\end {Bmatrix} +\begin {bmatrix} 6.8^{2} & 0\\ 0 & 16.9^{2}\end {bmatrix}\begin {Bmatrix} \eta _{1}\\ \eta _{2}\end {Bmatrix} =\begin {bmatrix} 0.85 & 1.1\\ 0.65 & -0.5 \end {bmatrix} ^{T}\begin {Bmatrix} \operatorname {Re}\left (Ae^{i\varpi t}\right ) \\ 0 \end {Bmatrix} \]
and using small damping approximation\[\begin {bmatrix} 1 & 0\\ 0 & 1 \end {bmatrix}\begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\end {Bmatrix} +\begin {bmatrix} 0.05 & 0\\ 0 & 0.061 \end {bmatrix}\begin {Bmatrix} \eta _{1}^{\prime }\\ \eta _{2}^{\prime }\end {Bmatrix} +\begin {bmatrix} 6.8^{2} & 0\\ 0 & 16.9^{2}\end {bmatrix}\begin {Bmatrix} \eta _{1}\\ \eta _{2}\end {Bmatrix} =\begin {Bmatrix} 0.85\operatorname {Re}\left (Ae^{it\varpi }\right ) \\ 1.1\operatorname {Re}\left (Ae^{it\varpi }\right ) \end {Bmatrix} \]
Hence the 2 EOM’s are\begin {align*} \eta _{1}^{\prime \prime }+0.05\eta _{1}^{\prime }+46.24\eta _{1} & =\operatorname {Re}\left (0.85Ae^{i\varpi t}\right ) \\ \eta _{2}^{\prime \prime }+0.061\eta _{2}^{\prime }+285.61\eta _{2} & =\operatorname {Re}\left (1.1Ae^{i\varpi t}\right ) \end {align*}
Let \(\eta _{1}=\operatorname {Re}\left (X_{1}e^{it\varpi }\right ) \) then \(X_{1}=\frac {0.85A}{-\varpi ^{2}+i0.05\varpi +46.24}\)and \(\eta _{2}=\operatorname {Re}\left (X_{2}e^{it\varpi }\right ) \)then \(X_{2}=\frac {1.1A}{-\varpi ^{2}+i0.0609\varpi +285.\,61}\) then\begin {align*} \mathbf {x} & =\Phi _{1}\eta _{1}+\Phi _{2}\eta _{2}\\\begin {Bmatrix} x_{1}\relax (t) \\ x_{2}\relax (t) \end {Bmatrix} & =\begin {Bmatrix} 0.85\\ 0.65 \end {Bmatrix} \operatorname {Re}\left (X_{1}e^{i\varpi t}\right ) +\begin {Bmatrix} 1.1\\ -0.5 \end {Bmatrix} \operatorname {Re}\left (X_{2}e^{i\varpi t}\right ) \end {align*}
Hence \begin {align*} x_{1}\relax (t) & =0.85\operatorname {Re}\left (X_{1}e^{i\varpi t}\right ) +1.1\operatorname {Re}\left (X_{2}e^{i\varpi t}\right ) \\ x_{2}\relax (t) & =0.65\operatorname {Re}\left (X_{1}e^{i\varpi t}\right ) -0.5\operatorname {Re}\left (X_{2}e^{i\varpi t}\right ) \end {align*}
hence\begin {align*} x_{1}\relax (t) & =0.85\operatorname {Re}\left (\frac {0.85A}{-\varpi ^{2}+i0.05\varpi +46.24}e^{i\varpi t}\right ) +1.1\operatorname {Re}\left (\frac {1.1A}{-\varpi ^{2}+i0.061\varpi +285.61}e^{i\varpi t}\right ) \\ x_{2}\relax (t) & =0.65\operatorname {Re}\left (\frac {0.85A}{-\varpi ^{2}+i0.05\varpi +46.24}e^{i\varpi t}\right ) -0.5\operatorname {Re}\left (\frac {1.1A}{-\varpi ^{2}+i0.061\varpi +285.61}e^{i\varpi t}\right ) \end {align*}
These can be combined to
\begin {align*} x_{1}\relax (t) & =\operatorname {Re}\left (\left [ \frac {0.85^{2}A}{-\varpi ^{2}+i0.05\varpi +46.24}+\frac {1.1^{2}A}{-\varpi ^{2}+i0.0609\varpi +285.61}\right ] e^{i\varpi t}\right ) \\ x_{2}\relax (t) & =\operatorname {Re}\left (\left [ \frac {\left ( 0.65\right ) \left (0.85\right ) A}{-\varpi ^{2}+i0.05\varpi +46.24}-\frac {\left (0.5\right ) \left (1.1\right ) A}{-\varpi ^{2}+i0.061\varpi +285.61}\right ] e^{i\varpi t}\right ) \end {align*}
The transient response is given in appendix B as \[ x\relax (t) =\frac {F_{0}}{k}\left (1-\cos \omega _{n}t\right ) h\left ( t\right ) \]
Hence maximum amplitude of the response is \(u_{\max }=\frac {2F_{0}}{k}.\) Compare this to static deflection which is \(u_{static}=\frac {F_{0}}{k}\) then we can say that dynamic load is twice as large as the static load. Therefor using \(2F_{0}\) in place of \(F\) in the expression for stress gives the result needed\begin {align*} \sigma _{y} & =\frac {-Mc}{I}\\ & =\frac {-\left (2F_{0}\right ) \frac {L}{4}\left (\frac {3}{4}\right ) }{I/c}\\ & =-\frac {3}{8}L\frac {c}{I}F_{0} \end {align*}
Therefore \[ F_{0}=-\frac {8I}{3Lc}\sigma _{y}\]
At \(t=0\) then \(x\relax (t) =\operatorname {Re}\left (e^{i\frac {2\pi }{3}}\right ) \) which is \(-\cos \left (60^{0}\right ) =-\frac {1}{2}\). Using \(\omega =2\pi \) rad/sec then \(x\relax (t) \) can be traced. Here is a plot
I=sqrt(-1);
w=2*pi;
x=@(t) real(exp(I*2*pi/3)*exp(I*w*t))
t=0:.01:1;
plot(t,x(t))
grid
xlabel('time (sec)'); ylabel('x(t)');
Damped resonances are seen at \(\omega =8.5,14\) and \(23\) rad/sec. This is where \(r=\frac {\varpi }{\omega _{i}}\) is close to unity, where \(\varpi \) is the forcing frequency and \(\omega _{i}\) is the natural frequency. Since this is a 3 dof system, it will have 3 natural frequencies.
The response of each dof will take contributions from each mode of vibration. Each mode vibrates at different natural frequency. From the plot above it is seen that the response of \(x_{1}\relax (t) \) has the largest response when the forcing frequency is close to the \(\omega _{2}=14\) rad/sec.
The new force now has the following set of discrete harmonics in it: (\(n=0\) is not counted, DC). \(\frac {100}{1}e^{3t},\frac {99}{2}e^{6t},\frac {98}{3}e^{9t},\frac {97}{4}e^{12t},\frac {96}{5}e^{15t},\frac {95}{6}e^{18t},\frac {94}{7}e^{21t},\frac {98}{8}e^{24t},\cdots \) or
\[ f\relax (t) =100e^{3t},49.5e^{6t},32.7e^{9t},24.3e^{12t},19.2e^{15t},15.8e^{18t},13.4e^{21t},12.3e^{24t}\]
So the input force has only discrete frequencies. Since linear sum, each \(f_{i}\relax (t) \) will cause the response \(\left \vert X\right \vert \) at that specific forcing frequency as shown in the plot. Looking the plot it can be seen that when forcing frequency is \(9\) rad/sec, this will cause the largest \(\left \vert X\right \vert \) among all these set of discrete frequencies. Hence the dominant harmonic is \(9\) rad/sec and will have amplitude around \(2.4\) from looking at the plot.