A system has mass \(M=20kg\) and \(\omega _{n}=100\) rad/sec. It is observed that steady state response is \(q=20\cos \left (110t-1.5\right ) \) \(mm\), where \(t\) is in seconds. Determine harmonic excitation causing this response for \(\zeta =0\) and \(\zeta =0.4\)
Let the harmonic excitation be \[ F\relax (t) =\operatorname {Re}\left \{ \hat {F}e^{i\omega t}\right \} \]
where \(\hat {F}\) is its complex amplitude. Also let \[ q=\operatorname {Re}\left \{ \hat {Q}e^{i\omega t}\right \} \]
be the steady state response. We are given that \(q=20\times 10^{-3}\cos \left ( 110t-1.5\right ) \), therefore\begin {align*} q & =\operatorname {Re}\left \{ 20\times 10^{-3}e^{i\left (100t-1.5\right ) }\right \} \\ & =\operatorname {Re}\left \{ 20\times 10^{-3}e^{-1.5i}e^{i100t}\right \} \end {align*}
Therefore \[ \hat {Q}=20\times 10^{-3}e^{-1.5i}\]
But the transfer function for second order system is\[ \hat {Q}=\frac {\hat {F}}{k}\frac {1}{\left (1-r^{2}\right ) +2i\zeta r}\]
where \(r=\frac {\omega }{\omega _{n}}\), hence we can now solve for \(\hat {F}\) from the above.\[ \hat {F}=\hat {Q}\left (k\left (\left (1-r^{2}\right ) +2i\zeta r\right ) \right ) \]
But \(k=M\omega _{n}^{2}\) hence\[ \fbox {$\hat {F}=\hat {Q}\left (M\omega _n^2\left (\left (1-r^2\right ) +2i\zeta r\right ) \right ) $}\]
we find\begin {align*} \hat {F} & =20\times 10^{-3}e^{-1.5i}\left (20\times 100^{2}\left (1-\left ( \frac {110}{100}\right ) ^{2}\right ) \right ) \\ & =20\times 10^{-3}e^{-1.5i}\left (-42000.0\right ) \\ & =-42000.0\times 20\times 10^{-3}e^{-1.5i}\\ & =-840.0e^{-1.5i} \end {align*}
Hence\begin {align*} F\relax (t) & =\operatorname {Re}\left \{ \hat {F}e^{i\omega t}\right \} \\ & =\operatorname {Re}\left \{ -840.0e^{-1.5i}e^{i110t}\right \} \\ & =\operatorname {Re}\left \{ -840.0e^{i\left (100t-1.5\right ) }\right \} \end {align*}
Therefore\[ \fbox {$F\relax (t) =-840\cos \left (100t-1.5\right ) $}\]
we find\begin {align*} \hat {F} & =\hat {Q}\left (M\omega _{n}^{2}\left (\left (1-r^{2}\right ) +2i\zeta r\right ) \right ) \\ & =20\times 10^{-3}e^{-1.5i}\left (20\times 100^{2}\left (\left (1-\left ( \frac {110}{100}\right ) ^{2}\right ) +i2\left (0.4\right ) \left (\frac {110}{100}\right ) \right ) \right ) \\ & =20\times 10^{-3}e^{-1.5i}\left (20\times 100^{2}\left (-0.21+0.88i\right ) \right ) \\ & =4000e^{-1.5i}\left (-0.21+0.88i\right ) \\ & =4000e^{-1.5i}\left (\sqrt {\left (0.21\right ) ^{2}+\left (0.88\right ) ^{2}}e^{i\tan ^{-1}\left (\frac {.88}{-0.21}\right ) }\right ) \end {align*}
In[4]:= ArcTan[-0.21, 0.88] Out[4]= 1.80505
Hence\begin {align*} \hat {F} & =4000e^{-1.5i}\left (0.90471e^{i1.80505}\right ) \\ & =3618.\allowbreak 8e^{-1.5i+1.80505i}\\ & =3618.\allowbreak 8e^{0.30505i} \end {align*}
Therefore\begin {align*} F\relax (t) & =\operatorname {Re}\left \{ \hat {F}e^{i\omega t}\right \} \\ & =\operatorname {Re}\left \{ 3618.\allowbreak 8e^{0.30505i}e^{i110t}\right \} \\ & =\operatorname {Re}\left \{ 3618.\allowbreak 8e^{i\left ( 100t+0.30505\right ) }\right \} \end {align*}
Hence\[ \fbox {$F\relax (t) =3618.\allowbreak 8\cos \left (100t+0.30505\right ) $}\]
Let \[ P\relax (t) =\operatorname {Re}\left \{ \hat {F}e^{i\omega t}\right \} \]
where \(\hat {F}\) is the complex amplitude of the excitation. Hence by comparing this to \(P\relax (t) =F\cos \omega t=\operatorname {Re}\left \{ Fe^{i\omega t}\right \} \) we see that \(\hat {F}=F\).
When \(\omega =2\pi 100\) then the response was \(q=\operatorname {Re}\left \{ \hat {Q}e^{i\omega t}\right \} =4\sin \left (2\pi 100t\right ) \) hence \(q=\operatorname {Re}\left \{ \overset {\hat {Q}}{\overbrace {4e^{-i\frac {\pi }{2}}}}e^{i\omega t}\right \} =q=\operatorname {Re}\left \{ 4e^{i\left (\omega t-\frac {\pi }{2}\right ) }\right \} \) therefore \[ \hat {Q}=4e^{-i\frac {\pi }{2}}\]
But, from the transfer function of second order system we know that \[ \hat {Q}=\frac {\hat {F}}{k}\frac {1}{\left (1-r^{2}\right ) +2i\zeta r}\]
Hence\begin {align} 4e^{-i\frac {\pi }{2}} & =\frac {\hat {F}}{k}\frac {1}{\left (1-\left ( \frac {2\pi 100}{\omega _{n}}\right ) ^{2}\right ) +2i\zeta \left (\frac {2\pi 100}{\omega _{n}}\right ) }\nonumber \\ & =\frac {\hat {F}}{k}\frac {1}{\sqrt {\left (1-\left (\frac {2\pi 100}{\omega _{n}}\right ) ^{2}\right ) ^{2}+\left (2\zeta \left (\frac {2\pi 100}{\omega _{n}}\right ) \right ) ^{2}}}e^{-i\tan ^{-1}\left (\frac {2\zeta r}{1-r^{2}}\right ) } \label {eq:2.0} \end {align}
By comparing sides we see that\begin {align} \frac {\pi }{2} & =\frac {2\zeta r}{1-r^{2}}\nonumber \\ & =\frac {2\zeta \frac {2\pi 100}{\omega _{n}}}{1-\left (\frac {2\pi 100}{\omega _{n}}\right ) ^{2}} \label {eq:2.1} \end {align}
When \(\omega =105Hz\) we are told it is half power point, which means the amplitude there is \(0.707\) of the maximum amplitude which occurs when \(r=1\). Hence\begin {align} 0.707\frac {\hat {F}}{k}\frac {1}{\sqrt {\left (1-\relax (1) ^{2}\right ) ^{2}+\left (2\zeta \relax (1) \right ) ^{2}}} & =\frac {\hat {F}}{k}\frac {1}{\sqrt {\left (1-\left (\frac {2\pi 105}{\omega _{n}}\right ) ^{2}\right ) ^{2}+\left (2\zeta \left (\frac {2\pi 105}{\omega _{n}}\right ) \right ) ^{2}}}\nonumber \\ 0.707\frac {1}{2\zeta } & =\frac {1}{\sqrt {\left (1-\left (\frac {2\pi 105}{\omega _{n}}\right ) ^{2}\right ) ^{2}+\left (2\zeta \left (\frac {2\pi 105}{\omega _{n}}\right ) \right ) ^{2}}} \label {eq:2.2} \end {align}
We now have 2 equations ?? and ?? to solve numerically for \(\zeta \) and \(\omega _{n}\) . Solving and keeping the positive solutions results in\begin {align*} \zeta & =0.0309\\ \omega _{n} & =640.8\text { rad/sec}\\ & =101.987\text { Hz} \end {align*}
Hence at \(\omega =105\) hz the phase is\[ \tan ^{-1}\left (\frac {2\zeta r}{1-r^{2}}\right ) =\tan ^{-1}\left ( \frac {2\left (0.0309\right ) \frac {2\pi \left (105\right ) }{640.8}}{1-\left (\frac {2\pi \left (105\right ) }{640.8}\right ) ^{2}}\right ) =133.305^{0}\]
In[35]:= ArcTan[1 - ((2 Pi 105)/640.8)^2, 2 (0.0309) ((2 Pi 105)/640.8)]*180/Pi Out[35]= 133.305
When \(\omega =100\) Hz we found from Eq ?? that\[ 4=\frac {\hat {F}}{k}\frac {1}{\sqrt {\left (1-\left (\frac {2\pi 100}{\omega _{n}}\right ) ^{2}\right ) ^{2}+\left (2\zeta \left (\frac {2\pi 100}{\omega _{n}}\right ) \right ) ^{2}}}\]
But we found that \(\omega _{n}=640.8\) rad/sec and \(\zeta =0.0309,\) hence \begin {align*} \frac {\hat {F}}{k} & =4\sqrt {\left (1-\left (\frac {2\pi 100}{640.8}\right ) ^{2}\right ) ^{2}+\left (2\left (0.0309\right ) \left (\frac {2\pi 100}{640.8}\right ) \right ) ^{2}}\\ & =0.28733 \end {align*}
At \(\omega =110\) Hz\begin {align*} \left \vert \hat {Q}\right \vert & =\frac {\hat {F}}{k}\frac {1}{\sqrt {\left ( 1-\left (\frac {2\pi 110}{\omega _{n}}\right ) ^{2}\right ) ^{2}+\left ( 2\zeta \left (\frac {2\pi 110}{\omega _{n}}\right ) \right ) ^{2}}}\\ & =0.28733\frac {1}{\sqrt {\left (1-\left (\frac {2\pi 110}{640.8}\right ) ^{2}\right ) ^{2}+\left (2\left (0.0309\right ) \left (\frac {2\pi 110}{640.8}\right ) \right ) ^{2}}}\\ & =\fbox {$1.\allowbreak 6288$} \end {align*}
The phase is\[ \tan ^{-1}\left (\frac {2\zeta r}{1-r^{2}}\right ) =\tan ^{-1}\left ( \frac {2\left (0.0309\right ) \left (\frac {2\pi 110}{640.8}\right ) }{1-\left (\frac {2\pi 110}{640.8}\right ) ^{2}}\right ) =\fbox {$157.798^0$}\]
In[37]:= ArcTan[1 - ((2 Pi 110)/640.8)^2, 2 (0.0309) ((2 Pi 110)/640.8)]*180/Pi Out[37]= 157.798
The two equations are\begin {align*} mx_{1}^{\prime \prime }+2kx_{1}-kx_{2} & =0\\ mx_{2}^{\prime \prime }-kx_{1}+kx_{2} & =f\relax (t) \end {align*}
Since the responses are harmonic and the input is harmonic, then we can write\begin {align*} x_{1}\relax (t) & =\operatorname {Re}\left \{ \hat {X}_{1}e^{i\omega t}\right \} \\ x_{1}\relax (t) & =\operatorname {Re}\left \{ \hat {X}_{2}e^{i\omega t}\right \} \end {align*}
Therefore the two equations can be written in terms of the complex amplitudes as\begin {align} -m\omega ^{2}\hat {X}_{1}+2k\hat {X}_{1}-k\hat {X}_{2} & =0\label {eq:3.1}\\ -m\omega ^{2}\hat {X}_{2}-k\hat {X}_{1}+k\hat {X}_{2} & =F \label {eq:3.2} \end {align}
From Eq ??\[ \frac {\left (-m\omega ^{2}+2k\right ) }{k}\hat {X}_{1}=\hat {X}_{2}\]
Substitute the above into Eq ?? gives\begin {align*} -m\omega ^{2}\frac {\left (-m\omega ^{2}+2k\right ) }{k}\hat {X}_{1}-k\hat {X}_{1}+k\frac {\left (-m\omega ^{2}+2k\right ) }{k}\hat {X}_{1} & =F\\ \left (\frac {\left (-m^{2}\omega ^{4}-m\omega ^{2}2k\right ) }{k}+k-m\omega ^{2}\right ) \hat {X}_{1} & =F\\ \hat {X}_{1} & =kF\frac {1}{\left (-m^{2}\omega ^{4}-m\omega ^{2}2k+k^{2}-km\omega ^{2}\right ) } \end {align*}
Dividing the numerator and denominator of the RHS by \(k^{2}\), and using \(k^{2}=\omega _{n}^{4}m^{2}\) and using \(r=\frac {\omega }{\omega _{n}}\)\begin {align*} \hat {X}_{1} & =\frac {F}{k}\frac {1}{\left (\frac {-m^{2}\omega ^{4}}{\omega _{n}^{4}m^{2}}-\frac {m\omega ^{2}2}{\omega _{n}^{2}m}+1-\frac {m\omega ^{2}}{\omega _{n}^{2}m}\right ) }\\ \hat {X}_{1} & =\frac {F}{k}\frac {1}{\left (-r^{4}-2r^{2}+1-r^{2}\right ) } \end {align*}
Hence the transfer function is\[ \hat {X}_1=\frac {F}{k}\frac {1}{\left (1-r^4-3r^2\right ) } \]
Summary of method of solution: There are 2 ways to solve these problem. We will solve it using both methods. The first method is using known standard solution for step input, the solution \(y\relax (t) \) is found for the period of \(0<t<\frac {3\pi }{\omega _{n}}\) using zero initial conditions. Next, the solution \(y\relax (t) \) and \(y^{\prime }\relax (t) \) is evaluated again at \(t=\frac {3\pi }{\omega _{n}}\). These values are now used as the initial conditions for the solution for \(t>\frac {3\pi }{\omega _{n}}\). The solution for \(t>\frac {3\pi }{\omega _{n}}\) will have the same form, but the step input now is \(200N\) instead of \(100\)N\(.\)
The second method as follows: Let \(F\relax (t) =100h\relax (t) +100h\left (t-\frac {3\pi }{\omega _{n}}\right ) \) or \(F\relax (t) =100h\relax (t) +100h\left (\tilde {t}\right ) \) where \(\tilde {t}=t-\) \(\frac {3\pi }{\omega _{n}}\), then assuming the transient solution to \(h\left ( t\right ) \) is \(s\relax (t) \) then the solution to \(F\relax (t) \) is \(100s\relax (t) +100s\left (\tilde {t}\right ) \). The second method is simplet than the first method.
Solution using first method:
The system is \[ my^{\prime \prime }\relax (t) +ky\relax (t) =F\relax (t) \]
When \(F\relax (t) \) is a fixed input, such as a step input of magnitude \(F\) then the response is given by\[ y\relax (t) =\left (y_{0}-\frac {F}{k}\right ) \cos \omega _{n}t+\frac {y_{0}^{\prime }}{\omega _{n}}\sin \omega _{n}t+\frac {F}{k}\]
Where in the above, \(y_{0}\) and \(y_{0}^{\prime }\) are the initial position and initial velocity. For \(0<t<1.5T_{n}\) the solution is\begin {align*} y\relax (t) & =-\frac {F}{k}\cos \omega _{n}t+\frac {F}{k}\\ & =\frac {F}{k}\left (1-\cos \omega _{n}t\right ) \end {align*}
Let \(F=Q_{1}=100\)N\(,\) and since \(k=m\omega _{n}^{2}\) then the above becomes\[ y\relax (t) =\frac {Q_{1}}{m\omega _{n}^{2}}\left (1-\cos \omega _{n}t\right ) \ \ \ \ \ 0<t<\frac {3\pi }{\omega _{n}}\]
Now we need first to evaluate \(y\left (t=\frac {3\pi }{\omega _{n}}\right ) \) and \(y^{\prime }\left (t=\frac {3\pi }{\omega _{n}}\right ) \). From the above\[ y^{\prime }\relax (t) =\frac {Q_{1}}{m\omega _{n}}\sin \omega _{n}t\ \ \ \ \ 0<t<\frac {3\pi }{\omega _{n}}\]
Hence\[ y\left (t=\frac {3\pi }{\omega _{n}}\right ) =\frac {Q_{1}}{m\omega _{n}^{2}}\left (1-\cos \omega _{n}\frac {3\pi }{\omega _{n}}\right ) =\frac {Q_{1}}{m\omega _{n}^{2}}\left (1-\cos 3\pi \right ) =\frac {Q_{1}}{m\omega _{n}^{2}}\left (1-\left (-1\right ) \right ) =\frac {2Q_{1}}{m\omega _{n}^{2}}\]
and\[ y^{\prime }\left (t=\frac {3\pi }{\omega _{n}}\right ) =\frac {Q_{1}}{m\omega _{n}}\sin \left (\omega _{n}\frac {3\pi }{\omega _{n}}\right ) =\frac {Q_{1}}{m\omega _{n}}\sin \left (3\pi \right ) =0 \]
Now let \(\tilde {t}=t-\frac {3\pi }{\omega _{n}}\)Hence the solution for \(\tilde {t}>0\) is\begin {align*} y\left (\tilde {t}\right ) & =\left (y\left (\tilde {t}=0\right ) -\frac {Q_{2}}{m\omega _{n}^{2}}\right ) \cos \omega _{n}\tilde {t}+\frac {y^{\prime }\left (\tilde {t}=0\right ) }{\omega _{n}}\sin \omega _{n}\tilde {t}+\frac {Q_{2}}{k}\\ & =\left (\frac {2Q_{1}}{m\omega _{n}^{2}}-\frac {Q_{2}}{m\omega _{n}^{2}}\right ) \cos \omega _{n}\tilde {t}+\frac {Q_{2}}{k} \end {align*}
Therefore, we have obtain the complete solution, which is
| time | solution |
| \(0<t<\frac {3\pi }{\omega _{n}}\) | \(\frac {Q_{1}}{m\omega _{n}^{2}}\left ( 1-\cos \omega _{n}t\right ) =\frac {100}{5\left (50\right ) ^{2}}\left ( 1-\cos 50t\right ) =\) \(0.008\left (1-\cos 50t\right ) \) |
| \(\tilde {t}=t-\frac {3\pi }{\omega _{n}}\) | \(\left (\frac {2Q_{1}}{m\omega _{n}^{2}}-\frac {Q_{2}}{k}\right ) \cos \omega _{n}\tilde {t}+\frac {Q_{2}}{k}=\left ( \frac {2\left (100\right ) }{5\left (50\right ) ^{2}}-\frac {200}{5\left ( 50\right ) ^{2}}\right ) \cos \omega _{n}\tilde {t}+\frac {200}{5\left ( 50\right ) ^{2}}=\) \(0.016\) |
This is a plot of the solution. Then a numerical ODE solver is used to verify the result
Now a numerical ODE solver was used to verify. Here is the result
We can see the solutions match very well.
Solution using second method:
Let \(F\relax (t) =100h\relax (t) +100h\left (t-\frac {3\pi }{\omega _{n}}\right ) \) then assuming the transient solution to \(h\left ( t\right ) \) is \(s\relax (t) \) then the solution to \(F\relax (t) \) is \(100s\relax (t) +100s\left (t-\frac {3\pi }{\omega _{n}}\right ) h\left (t-\frac {3\pi }{\omega _{n}}\right ) \). From appendix \(B\), the solution to \(h\relax (t) \) is given by \[ s\relax (t) =\frac {1}{m\omega _{n}^{2}}\left (1-\cos \omega _{n}t\right ) \] hence the solution to \(F\relax (t) =100h\relax (t) +100h\left ( t-\frac {3\pi }{\omega _{n}}\right ) \) is\begin {align*} y\relax (t) & =100s\relax (t) +100s\left (t-\frac {3\pi }{\omega _{n}}\right ) h\left (t-\frac {3\pi }{\omega _{n}}\right ) \\ & =\frac {100}{m\omega _{n}^{2}}\left (1-\cos \omega _{n}t\right ) +\frac {100}{m\omega _{n}^{2}}\left (1-\cos \omega _{n}\left (t-\frac {3\pi }{\omega _{n}}\right ) \right ) h\left (t-\frac {3\pi }{\omega _{n}}\right ) \end {align*}
To verify, this is a plot of the above solution. We see it is the same as the first analytical solution, and it is the same solution as the one using numerical ODE solver as well.
The input can be written as \(F_{0}h\relax (t) -F_{0}h\left ( t-T\right ) +F_{0}e^{-\beta \left (t-T\right ) }h\left (t-T\right ) \) or, by letting \(\tilde {t}=t-T\), the input becomes \[ F_{0}h\relax (t) -F_{0}h\left (\tilde {t}\right ) +F_{0}e^{-\beta \tilde {t}}h\left (\tilde {t}\right ) \]
If the response to \(h\relax (t) \) is \(s\relax (t) \) and the response to \(e^{-\beta \tilde {t}}\) is \(s_{1}\left (\tilde {t}\right ) \) then the response to the above becomes\[ F_{0}s\relax (t) -F_{0}s\left (\tilde {t}\right ) h\left (\tilde {t}\right ) +F_{0}s_{1}\left (\tilde {t}\right ) h\left (\tilde {t}\right ) \]
From appendix B, we see that \[ s\relax (t) =\frac {1}{m\omega _{n}^{2}}\left (1-\cos \omega _{n}t\right ) \]
and\[ s_{1}\left (\tilde {t}\right ) =\frac {1}{m\left (\omega _{n}^{2}+\beta ^{2}\right ) }\left (e^{-\beta \tilde {t}}-\left (\cos \left (\omega _{n}\tilde {t}\right ) +\frac {-\beta }{\omega _{n}}\sin \omega _{n}\tilde {t}\right ) \right ) h\left (\tilde {t}\right ) \]
Therefore the the final response is\begin {align*} y\relax (t) & =F_{0}h\relax (t) -F_{0}h\left (\tilde {t}\right ) +F_{0}e^{-\beta \tilde {t}}h\left (\tilde {t}\right ) \\ & =F_{0}\frac {1}{m\omega _{n}^{2}}\left (1-\cos \omega _{n}t\right ) h\left ( t\right ) -F_{0}\frac {1}{m\omega _{n}^{2}}\left (1-\cos \omega _{n}\tilde {t}\right ) h\left (\tilde {t}\right ) +\\ & F_{0}\frac {1}{m\left (\omega _{n}^{2}+\beta ^{2}\right ) }\left ( e^{-\beta \tilde {t}}-\left (\cos \left (\omega _{n}\tilde {t}\right ) +\frac {-\beta }{\omega _{n}}\sin \omega _{n}\tilde {t}\right ) \right ) h\left ( \tilde {t}\right ) \\ & \\ & =F_{0}\frac {1}{m\omega _{n}^{2}}\left (1-\cos \omega _{n}t\right ) -F_{0}\frac {1}{m\omega _{n}^{2}}\left (1-\cos \left (\omega _{n}\left ( t-T\right ) \right ) \right ) h\left (t-T\right ) +\\ & F_{0}\frac {1}{m\left (\omega _{n}^{2}+\beta ^{2}\right ) }\left ( e^{-\beta \left (t-T\right ) }-\left (\cos \left (\omega _{n}\left ( t-T\right ) \right ) +\frac {-\beta }{\omega _{n}}\sin \omega _{n}\left ( t-T\right ) \right ) \right ) h\left (t-T\right ) \end {align*}
To plot this, we need to choose values for parameters. Let \(F_{0}=100,\omega _{n}=50rad/\sec ,m=5kg,\beta =1,T=1\), then a plot of the above is below, followed by solution from numerical ODE solver.
Plot of the analytical solution
To verify, this is the result from numerical ODE solver
We can see that the solutions agree.