Assuming the 2 masses move together (else we will have 2 systems and 2 equations of motions. Hence I assumed that they move together as one body).\[ \left (m_{1}+m_{2}\right ) y^{\prime \prime }+y^{\prime }\mu +ky=f\left ( t\right ) \]
Since \(y\relax (t) =A\sin \left (\omega t\right ) \) hence \[ y\relax (t) =\operatorname {Re}\left (\frac {A}{i}e^{i\omega t}\right ) \] Let \[ f\relax (t) =\operatorname {Re}\left (\frac {\hat {F}}{i}e^{i\left ( \omega t\right ) }\right ) \] Where \(\hat {F}\) is the complex amplitude of the force. Now we substitute all these in the differential equation above.\begin {align*} y^{\prime } & =\operatorname {Re}\left (\omega Ae^{i\omega t}\right ) \\ y^{\prime \prime } & =\operatorname {Re}\left (i\omega ^{2}Ae^{i\omega t}\right ) \end {align*}
\begin {align*} \left (m_{1}+m_{2}\right ) y^{\prime \prime }+y^{\prime }\mu +ky & =\operatorname {Re}\left (\frac {\hat {F}}{i}e^{i\left (\omega t\right ) }\right ) \\ \operatorname {Re}\left (i\omega ^{2}Ae^{i\omega t}\right ) \left (m_{1}+m_{2}\right ) +\operatorname {Re}\left (\omega Ae^{i\omega t}\right ) \mu +k\operatorname {Re}\left (\frac {A}{i}e^{i\omega t}\right ) & =\operatorname {Re}\left (\frac {\hat {F}}{i}e^{i\left (\omega t\right ) }\right ) \\ \operatorname {Re}\left [ \left (i\omega ^{2}\left (m_{1}+m_{2}\right ) +\omega \mu +\frac {1}{i}k\right ) Ae^{i\omega t}\right ] & =\operatorname {Re}\left (\frac {\hat {F}}{i}e^{i\left (\omega t\right ) }\right ) \\ \left (i\omega ^{2}\left (m_{1}+m_{2}\right ) +\omega \mu +\frac {1}{i}k\right ) A & =\frac {\hat {F}}{i} \end {align*}
Hence \[ \hat {F}=\left (-\omega ^{2}\left (m_{1}+m_{2}\right ) +i\omega \mu +k\right ) A \]
\(k=3.2\times 10^{3}\) Nm\(,\mu =40\) Ns/m,\(A=0.02\) meter. When \(\omega =75\)rad/sec the above becomes\begin {align*} \hat {F} & =\left (-75^{2}\left (1.5\right ) +i75\times 40+3.2\times 10^{3}\right ) 0.02\\ & =-104.75+60.0i \end {align*}
Hence \(\operatorname {Re}\left (\hat {F}\right ) =-104.75\)N and the phase is \(\tan ^{-1}\left (-\frac {60.0}{104.75}\right )=2.62\) rad/sec.
When \(\omega =85\)\begin {align*} \hat {F} & =1.5\left (-85^{2}+i85\frac {40}{1.5}+2133.3\right ) 0.02\\ & =-152.75+68.0i \end {align*}
\(\operatorname {Re}\left (\hat {F}\right ) =-152.75\) N and the phase is \(\tan ^{-1}\left (-\frac {68}{152.75}\right )=2.722\) rad/sec.
Force transmitted to floor is given by\[ F_{tr}=cz^{\prime }+kz \]
Let \(f\relax (t) =F\cos \left (\omega t\right ) =\operatorname {Re}\left (Fe^{i\omega t}\right ) =\operatorname {Re}\left (Fe^{i\omega t}\right ) \) where we are given that \(F=20\times 10^{3}\ \)N. \(\omega =2\pi \left (\frac {1800}{60}\right ) =60\pi =188.50\) rad/sec or \(30\) Hz.
Let \(z_{ss}=\operatorname {Re}\left (\frac {F}{k}\left \vert D\right \vert e^{i\left (\omega t-\phi \right ) }\right ) \) where \(\phi =\tan ^{-1}\left ( \frac {2\zeta r}{1-r^{2}}\right ) \) and \(\left \vert D\right \vert =\frac {1}{\sqrt {\left (1-r^{2}\right ) ^{2}+\left (2\zeta r\right ) ^{2}}}\). and \(r=\frac {\omega }{\omega _{n}}\) Hence \(z^{\prime }=\operatorname {Re}\left ( i\omega \frac {F}{k}\left \vert D\right \vert e^{i\left (\omega t-\phi \right ) }\right ) =\operatorname {Re}\left (\omega \frac {F}{k}\left \vert D\right \vert e^{i\left (\omega t-\phi +\frac {\pi }{2}\right ) }\right ) \). Therefore\[ F_{tr}=c\operatorname {Re}\left (\omega \frac {F}{k}\left \vert D\right \vert e^{i\left (\omega t-\phi +\frac {\pi }{2}\right ) }\right ) +k\operatorname {Re}\left (\frac {F}{k}\left \vert D\right \vert e^{i\left (\omega t-\phi \right ) }\right ) \]
Where \(c=2\zeta \omega _{n}m\ \) and When \(F_{tr}=2\times 10^{3}\)N . We now solve for \(k\) from\[ 2\times 10^{3}\geq 2\zeta \omega _{n}m\operatorname {Re}\left (\omega \frac {F}{k}\left \vert D\right \vert e^{i\left (\omega t-\phi +\frac {\pi }{2}\right ) }\right ) +k\operatorname {Re}\left (\frac {F}{k}\left \vert D\right \vert e^{i\left (\omega t-\phi \right ) }\right ) \]
Taking the maximum case for RHS where exponential are unity magnitude, hence\begin {align*} 2\times 10^{3} & =2\zeta \omega _{n}m\omega \frac {F}{k}\left \vert D\right \vert +F\left \vert D\right \vert \\ & =\left (2\zeta \omega _{n}m\omega \left (\frac {F}{k}\right ) +F\right ) \left \vert D\right \vert \\ & =\frac {F\left (1+2\zeta \omega _{n}\frac {m}{k}\omega \right ) }{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left (2\zeta r\right ) ^{2}}} \end {align*}
Where \(r=\frac {\omega }{\omega _{n}}=\frac {\omega }{\sqrt {\frac {k}{m}}}\). Hence the above becomes\[ 2\times 10^{3}=\frac {F\left (1+2\zeta \omega _{n}\frac {m}{k}\omega \right ) }{\sqrt {\left (1-\frac {\omega ^{2}}{\frac {k}{m}}\right ) ^{2}+\left ( 2\zeta \frac {\omega }{\sqrt {\frac {k}{m}}}\right ) ^{2}}}\]
In the above everything is known except for \(k\) which we solve for. Plugging the numerical values given. \(\omega =2\pi \left (\frac {1800}{60}\right ) \),\(m=450,F=20\times 10^{3},\zeta =0.03\) hence\[ 2\times 10^{3}=\frac {20\times 10^{3}\left (1+2\left (0.03\right ) \sqrt {\frac {k}{450}}\frac {450}{k}\left (60\pi \right ) \right ) }{\sqrt {\left ( 1-\frac {450\left (60\pi \right ) ^{2}}{k}\right ) ^{2}+\left (2\left ( 0.03\right ) \frac {60\pi }{\sqrt {\frac {k}{450}}}\right ) ^{2}}}\]
Hence \(k=1.2135\times 10^{6}\) N/m. Hence \(\omega _{n}=\sqrt {\frac {k}{m}}=\sqrt {\frac {1.2135\times 10^{6}}{450}}=51.929\) rad/sec or \(8.265\) Hz.
The total displacement is given by\begin {align*} z\relax (t) & =z_{transient}\relax (t) +z_{ss}\relax (t) \\ & =e^{-\zeta \omega _{n}t}\left (A\cos \omega _{d}t+B\sin \omega _{d}t\right ) +\operatorname {Re}\left (\frac {F}{k}\left \vert D\right \vert e^{i\left ( \omega t-\phi \right ) }\right ) \end {align*}
Where \[ z_{transient}\relax (t) =e^{-\zeta \omega _{n}t}\left (A\cos \omega _{d}t+B\sin \omega _{d}t\right ) \]
Assuming at \(t=0\) the system is relaxed hence \(z\relax (0) =0\) and \(z^{\prime }\relax (0) =0\) we can determine \(A,B\) from Eq ??\(.\)
At \(t=0,\)
\begin {align*} z\relax (0) & =0\\ & =A+\operatorname {Re}\left (\frac {F}{k}\left \vert D\right \vert e^{-i\phi }\right ) \end {align*}
Hence \[ A=-\operatorname {Re}\left (\frac {F}{k}\left \vert D\right \vert e^{-i\phi }\right ) \]
and \begin {align*} z^{\prime }\relax (t) & =-\zeta \omega _{n}e^{-\zeta \omega _{n}t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) +e^{-\zeta \omega _{n}t}\left ( -\omega _{d}A\sin \omega _{d}t+\omega _{d}B\cos \omega _{d}t\right ) \\ & +\operatorname {Re}\left (\omega \frac {F}{k}\left \vert D\right \vert e^{i\left (-\phi +\frac {\pi }{2}\right ) }\right ) \end {align*}
Hence at \(t=0\)\begin {align*} z^{\prime }\relax (0) & =0\\ & =-\zeta \omega _{n}A+\omega _{d}B+\operatorname {Re}\left (\omega \frac {F}{k}\left \vert D\right \vert e^{i\left (-\phi +\frac {\pi }{2}\right ) }\right ) \end {align*}
Hence\begin {align*} B & =\frac {\zeta \omega _{n}}{\omega _{d}}A-\frac {1}{\omega _{d}}\operatorname {Re}\left (\omega \frac {F}{k}\left \vert D\right \vert e^{i\left ( -\phi +\frac {\pi }{2}\right ) }\right ) \\ & =-\frac {\zeta \omega _{n}}{\omega _{d}}\operatorname {Re}\left (\frac {F}{k}\left \vert D\right \vert e^{-i\phi }\right ) -\frac {1}{\omega _{d}}\operatorname {Re}\left (\omega \frac {F}{k}\left \vert D\right \vert e^{i\left ( -\phi +\frac {\pi }{2}\right ) }\right ) \end {align*}
Therefore the displacement is\begin {align*} z\relax (t) & =e^{-\zeta \omega _{n}t}\left [ -\operatorname {Re}\left ( \frac {F}{k}\left \vert D\right \vert e^{-i\phi }\right ) \cos \omega _{d}t+\left \{ -\frac {\zeta \omega _{n}}{\omega _{d}}\operatorname {Re}\left (\frac {F}{k}\left \vert D\right \vert e^{-i\phi }\right ) -\frac {1}{\omega _{d}}\operatorname {Re}\left (\omega \frac {F}{k}\left \vert D\right \vert e^{i\left ( -\phi +\frac {\pi }{2}\right ) }\right ) \right \} \sin \omega _{d}t\right ] \\ & +\operatorname {Re}\left (\frac {F}{k}\left \vert D\right \vert e^{i\left ( \omega t-\frac {\pi }{2}-\phi \right ) }\right ) \end {align*}
Hence expressed in \(\sin \) and \(\cos \)\begin {align*} z\relax (t) & =-\left (\frac {F}{k}\left \vert D\right \vert \cos \phi \right ) e^{-\zeta \omega _{n}t}\cos \omega _{d}t+e^{-\zeta \omega _{n}t}\left \{ -\frac {\zeta \omega _{n}}{\omega _{d}}\left (\frac {F}{k}\left \vert D\right \vert \cos \phi \right ) -\frac {1}{\omega _{d}}\left (\omega \frac {F}{k}\left \vert D\right \vert \sin \phi \right ) \right \} \sin \omega _{d}t\\ & +\frac {F}{k}\left \vert D\right \vert \cos \left (\omega t-\phi \right ) \\ & =-\left (\frac {F}{k}\left \vert D\right \vert \cos \phi \right ) e^{-\zeta \omega _{n}t}\cos \omega _{d}t+e^{-\zeta \omega _{n}t}\frac {F\left \vert D\right \vert }{\omega _{d}k}\left (-\zeta \omega _{n}\cos \phi -\omega \sin \phi \right ) \sin \omega _{d}t+\frac {F}{k}\left \vert D\right \vert \cos \left ( \omega t-\phi \right ) \\ & =\frac {F}{k}\left \vert D\right \vert e^{-\zeta \omega _{n}t}\left [ -\cos \phi \cos \omega _{d}t+\frac {1}{\omega _{d}}\left (-\zeta \omega _{n}\cos \phi -\omega \sin \phi \right ) \sin \omega _{d}t\right ] +\frac {F}{k}\left \vert D\right \vert \cos \left (\omega t-\phi \right ) \end {align*}
Since \(\zeta =0.03\) then \(\omega _{d}=\omega _{n}\sqrt {1-\zeta ^{2}}=\omega _{n}\sqrt {1-0.03^{2}}=0.99955\left (\omega _{n}\right ) \). Therefore in the above we can just replace \(\omega _{d}\) by \(\omega _{n}\) with very good approximation, hence we now obtain\begin {align*} z\relax (t) & =\frac {F}{k}\left \vert D\right \vert e^{-\zeta \omega _{n}t}\left [ -\cos \phi \cos \omega _{n}t+\frac {1}{\omega _{n}}\left ( -\zeta \omega _{n}\cos \phi -\omega \sin \phi \right ) \sin \omega _{n}t\right ] +\frac {F}{k}\left \vert D\right \vert \cos \left (\omega t-\phi \right ) \\ & =\frac {F}{k}\left \vert D\right \vert e^{-\zeta \omega _{n}t}\left [ -\cos \phi \cos \omega _{n}t-\left (\zeta \cos \phi +\frac {\omega }{\omega _{n}}\sin \phi \right ) \sin \omega _{n}t\right ] +\frac {F}{k}\left \vert D\right \vert \cos \left (\omega t-\phi \right ) \end {align*}
This is the amplitude. In the above \(\left \vert D\right \vert =\frac {1}{\sqrt {\left (1-\left (\frac {\omega }{\omega _{n}}\right ) ^{2}\right ) ^{2}+\left (2\zeta \left (\frac {\omega }{\omega _{n}}\right ) \right ) ^{2}}}\), and \(\phi =\tan ^{-1}\frac {2\zeta \frac {\omega }{\omega _{n}}}{1-\left ( \frac {\omega }{\omega _{n}}\right ) ^{2}}.\) The transient solution usually goes away after 5 or 6 cycles. Hence let us assume that the start up time takes \(6\times \frac {2\pi }{\omega _{n}}=\) \(6\times \frac {2\pi }{\sqrt {\frac {k}{m}}}=6\times \frac {2\pi }{\sqrt {\frac {1.2135\times 10^{6}}{450}}}=0.72597\) seconds. Or \(1\) second at worst.
Therefore we can now plot the amplitude for \(t=0\) to \(t=1\) second in increments of \(0.1\) second, and each time advance, we can increment \(\omega \) from \(0\) to \(60\pi \) in linear fashion, hence each \(0.1\) second we update \(\omega \) by an amount \(6\pi \). After 1 second has passed, the system is assumed to be in steady state, and then we keep \(\omega \) fixed at \(60\pi \) rad/sec. This is a plot showing \(z\relax (t) \) for \(t=0\) to \(2\) seconds given the above method of changing \(\omega \)
To avoid going over \(10mm\), this means we have to avoid the case of \(r=1\) or \(\omega =\omega _{n}\). When I first just incremented \(\omega _{n}\) such that \(r=1\) was not avoided, resonance caused the amplitude to go over \(10mm\) as given in this plot. The transient solution itself stayed just below \(10\)mm but the steady state solution went over \(10\)mm due to resonance
To insure that the amplitude does not go over \(10\)mm, we need to add mass to the generator. Maximum amplitude is given by \(\frac {F}{k}\frac {1}{2\zeta }=\frac {20\times 10^{3}}{1.2135\times 10^{6}}\frac {1}{2\left ( 0.03\right ) }=0.27469\) meter \(\ \)or \(274\)mm
So to insure maximum does not exceed \(10\)mm , solve for new \(k\) from \(0.01=\frac {20\times 10^{3}}{k_{n}}\frac {1}{2\left (0.03\right ) }\), hence \(k_{n}=3.3333\times 10^{7}\). Since \(\omega _{n}=51. 929=\sqrt {\frac {k_{n}}{m_{n}}}\) then new mass is \(m_{n}=\frac {3. 3333\times 10^{7}}{51.929^{2}}=\)\(12361\) kg using these values, the above plot now are redone. This is the result
We see that now the maximum displacement remained below \(10\) mm.
Let \(\varepsilon =50mm=0.05m\) be the distance of the unbalance mass \(m\). Let \(M=80kg\) be the mass of the motor. The equation of motion is given by\begin {align*} \left (M+m\right ) y^{\prime \prime }+cy^{\prime }+ky & =m\varepsilon \Omega ^{2}\sin \left (\Omega t\right ) \\ y^{\prime \prime }+2\zeta \omega _{n}y^{\prime }+\omega _{n}^{2}y & =\frac {m}{m+M}\varepsilon \Omega ^{2}\operatorname {Re}\left (\frac {1}{i}e^{i\Omega t}\right ) \end {align*}
Where \(\omega _{n}=\sqrt {\frac {k}{m+M}}\) and \(\zeta =\frac {c}{2\left ( M+m\right ) \omega _{n}}\). Let \(y=\operatorname {Re}\left (\frac {Y}{i}e^{i\Omega t}\right ) \). This leads to\[ Y=\frac {m}{m+M}\frac {\varepsilon \Omega ^{2}}{\omega _{n}^{2}-\Omega ^{2}+2i\zeta \omega _{n}\Omega }\]
Since static deflection is \(40mm\), then \begin {align*} \frac {\left (M+m\right ) g}{k} & =0.04\\ k & =\frac {\left (M+m\right ) g}{0.04} \end {align*}
But \(\omega _{n}^{2}=\frac {k}{m+M}=\frac {\left (M+m\right ) g}{0.04\left ( M+m\right ) }=\frac {g}{0.04}\), hence \(\omega _{n}=\sqrt {\frac {9.81}{0.04}}=15.66\) rad/sec or \(2.492\) Hz.
Since at steady state the displacement is \(10\) mm, then \(\Omega =2\pi \frac {145}{60}=15.184\) \(\ \)or \(2.4167\) Hz hence\begin {align*} y & =\operatorname {Re}\left (\frac {Y}{i}e^{i\Omega t}\right ) =\operatorname {Re}\left (\frac {\varepsilon m}{m+M}\frac {r^{2}}{\left ( 1-r^{2}+2i\zeta r\right ) }e^{i\left (\Omega t-\frac {\pi }{2}\right ) }\right ) \\ & =\frac {\varepsilon mr^{2}}{m+M}\frac {1}{\sqrt {\left (\left ( 1-r^{2}\right ) ^{2}+\left (2\zeta r\right ) ^{2}\right ) }}\operatorname {Re}\left (e^{-i\phi }e^{i\left (\Omega t-\frac {\pi }{2}\right ) }\right ) \end {align*}
Where \(\phi =\tan ^{-1}\left (\frac {2\zeta r}{1-r^{2}}\right ) \). \(r=\frac {\Omega }{\omega _{n}}=\frac {15.184}{ 15.66}=\)\(0.9696\) hence the above becomes, at steady state\begin {align} 0.01 & =\frac {\left (0.05\right ) m}{m+80}\frac {0.9696^{2}}{\sqrt {\left ( 1-0.9696^{2}\right ) ^{2}+\left (2\zeta 0.9696\right ) ^{2}}}\operatorname {Re}\left (e^{i\left (15.184\ t-\frac {\pi }{2}-\phi \right ) }\right ) \nonumber \\ & =\frac {\left (0.05\right ) m}{m+80}\frac {0.9696^{2}}{\sqrt {\left ( 1-0.9696^{2}\right ) ^{2}+\left (2\zeta 0.9696\right ) ^{2}}}\sin \left ( 15.184\ t-\phi \right ) \label {eq:3.3} \end {align}
We are now told that at \(\Omega =15.184\) and when \(\Omega t=75^{0}\) then the displacement is zero, hence\[ 0=\frac {\left (0.05\right ) m}{m+80}\frac {0.9696^{2}}{\sqrt {\left ( 1-0.9696^{2}\right ) ^{2}+\left (2\zeta 0.9696\right ) ^{2}}}\sin \left ( 75^{0}-\phi \right ) \]
or\begin {align*} \sin \left (75^{0}-\phi \right ) & =0\\ 75^{0}-\phi & =0\\ \phi & =75^{0} \end {align*}
Since \(\phi =\tan ^{-1}\left (\frac {2\zeta r}{1-r^{2}}\right ) \) then \[ 75\left (\frac {\pi }{180}\right ) =\tan ^{-1}\left (\frac {2\zeta 0.9696}{1-0.9696^{2}}\right ) \]
Hence\begin {align*} \tan ^{-1}\left (\frac {2\zeta 0.9696}{1-0.9696^{2}}\right ) & =1. 3090\\ \frac {2\zeta 0.9696}{1-0.9696^{2}} & =\tan \left (1.3090\right ) \\ \frac {2\zeta 0.9696}{1-0.9696^{2}} & =3.7321 \end {align*}
Hence \(\zeta =0.11523\)
From Eq ??\[ 0.01=\frac {\left (0.05\right ) m}{m+80}\frac {0.9696^{2}}{\sqrt {\left ( 1-0.9696^{2}\right ) ^{2}+\left (2\zeta 0.9696\right ) ^{2}}}\sin \left ( 15.184\ t-\phi \right ) \]
The maximum amplitude is when\[ 0.01=\frac {\left (0.05\right ) m}{m+80}\frac {0.9696^{2}}{\sqrt {\left ( 1-0.9696^{2}\right ) ^{2}+\left (2\zeta 0.9696\right ) ^{2}}}\]
But \(\zeta =0.11523\), hence we now solve for \(m\)\[ 0.01=\frac {\left (0.05\right ) m}{m+80}\frac {0.9696^{2}}{\sqrt {\left ( 1-0.9696^{2}\right ) ^{2}+\left (2\left (0.11523\right ) 0.9696\right ) ^{2}}}\] Hence \[ \fbox {$m=4.1\ kg$}\]
Hence \(\varepsilon m=\left (0.05\right ) \left (4.1\right ) =\)\(0.20\) kg meter
since
\[ y=\frac {\varepsilon mr^{2}}{m+M}\frac {1}{\sqrt {\left (\left (1-r^{2}\right ) ^{2}+\left (2\zeta r\right ) ^{2}\right ) }}\operatorname {Re}\left ( e^{i\left (\Omega t-\frac {\pi }{2}-\phi \right ) }\right ) \]
As \(\Omega \) becomes much larger than \(\omega _{n}\) then \(\left (1-r^{2}\right ) ^{2}\rightarrow r^{4}\) . Now dividing numerator and denominator by \(r^{2}\) gives\begin {align*} y & =\frac {\varepsilon m}{m+M}\frac {1}{\sqrt {\frac {\left (r^{4}+\left ( 2\zeta r\right ) ^{2}\right ) }{r^{4}}}}\sin \left (\Omega t-\phi \right ) \\ & =\frac {\varepsilon m}{m+M}\frac {1}{\sqrt {\left (1+\frac {4\zeta ^{2}}{r^{2}}\right ) }}\sin \left (\Omega t-\phi \right ) \end {align*}
as \(r\) becomes large then \(\frac {4\zeta ^{2}}{r^{2}}\rightarrow 0\) hence\[ y\simeq \frac {\varepsilon m}{m+M}\sin \left (\Omega t-\phi \right ) \]
The smallest possible amplitude is\begin {align*} \left \vert y\right \vert & =\frac {0.20}{4.1+80}\\ & =2.3781\times 10^{-3}\text { meter} \end {align*}
or\[ \left \vert y\right \vert =2.38\text { mm}\]
(note: total mass of system includes the small unbalanced masses) Since static deflection is \(8.5mm\), then \begin {align*} \frac {Mg}{k} & =0.0085\\ k & =\frac {Mg}{0.0085} \end {align*}
But \(\omega _{n}^{2}=\frac {k}{M}=\frac {Mg}{0.0085M}=\frac {g}{0.0085}\), hence \(\omega _{n}=\sqrt {\frac {9.81}{0.0085}}=33.972\) rad/sec or \(5.4068\) Hz
The equation of motion is (angle \(\Omega \) is now measured from horizontal, anti-clock wise positive)\[ My^{\prime \prime }+cy^{\prime }=2m\varepsilon \Omega ^{2}\sin \left (\Omega t\right ) =\operatorname {Re}\left (\frac {1}{i}2m\varepsilon \Omega ^{2}e^{i\left (\Omega t\right ) }\right ) \]
Let \(y\relax (t) =\operatorname {Re}\left (\frac {1}{i}Ye^{i\Omega t}\right ) \) hence \(y^{\prime }\relax (t) =\operatorname {Re}\left ( Y\Omega e^{i\Omega t}\right ) \),\(y^{\prime \prime }\relax (t) =\operatorname {Re}\left (iY\Omega ^{2}e^{i\Omega t}\right ) \), hence the above becomes\begin {align*} \operatorname {Re}\left (iY\Omega ^{2}e^{i\Omega t}\right ) +\frac {c}{M}\operatorname {Re}\left (Y\Omega e^{i\Omega t}\right ) & =\operatorname {Re}\left (\frac {1}{i}\frac {2m\varepsilon \Omega ^{2}}{M}e^{i\Omega t}\right ) \\ \operatorname {Re}\left (\left (i\Omega ^{2}+\frac {c\Omega }{M}\right ) Ye^{i\Omega t}\right ) & =\operatorname {Re}\left (\frac {1}{i}\frac {2m\varepsilon \Omega ^{2}}{M}e^{i\Omega t}\right ) \\ \left [ i\Omega ^{2}+\frac {c\Omega }{M}\right ] Y & =\frac {1}{i}\frac {2m\varepsilon \Omega ^{2}}{M}\\ Y & =\frac {1}{i}\frac {\frac {2m\varepsilon \Omega ^{2}}{M}}{i\Omega ^{2}+\frac {c\Omega }{M}}\\ & =\frac {2m\varepsilon \Omega ^{2}}{ic\Omega -M\Omega ^{2}} \end {align*}
Hence\begin {align*} y_{ss}\relax (t) & =\operatorname {Re}\left (\frac {1}{i}Ye^{i\left ( \Omega t\right ) }\right ) \\ & =\operatorname {Re}\left (\frac {1}{i}\frac {2m\varepsilon \Omega ^{2}}{ic\Omega -M\Omega ^{2}}e^{i\left (\Omega t\right ) }\right ) \end {align*}
Now we are told when \(\Omega t=\frac {\pi }{2}\) (upright position) then \(y=0\)(since it passes static equilibrium). At this moment \(\Omega =2\pi \frac {900}{60}=\)\(94.248\) rad/sec\(,\) At this moment the centripetal forces equal the damping force downwards (since the mass was moving upwards). Hence \[ \fbox {$m\varepsilon \Omega ^2=cy^\prime \relax (t) $}\]
But from above we found that \begin {align*} y^{\prime }\relax (t) & =\operatorname {Re}\left (\frac {2m\varepsilon \Omega ^{2}}{ic\Omega -M\Omega ^{2}}\Omega e^{i\Omega t}\right ) \\ & =\operatorname {Re}\left (\frac {\left (0.5\right ) 94.248^{2}}{ic\left (94.248\right ) -200\left (94.248\right ) ^{2}}94.248e^{i\frac {\pi }{2}}\right ) \\ & =\operatorname {Re}\left (\frac {8.3718\times 10^{5}}{94.248ic-1.7765\times 10^{6}}e^{i\frac {\pi }{2}}\right ) \end {align*}
Hence\begin {align*} m\varepsilon \Omega ^{2} & =c\left \vert y^{\prime }\relax (t) \right \vert \\ \left (0.5\right ) 94.248^{2} & =c\frac {8. 3718\times 10^{5}}{\sqrt {\left (94.248c\right ) ^{2}+\left ( 1.7765\times 10^{6}\right ) ^{2}}}\\ 4441.3 & =8.3718\times 10^{5}\frac {c}{\sqrt {8882.7c^{2}+3.1560\times 10^{12}}} \end {align*}
Solving numerically for \(c\) gives\[ c=1.0882\times 10^{4}\text { N second per meter}\]
When \(\Omega =\left (2\pi \frac {1000}{60}\right ) =104. 72\) rad/sec or \(16.667\) Hz. From\begin {align*} y & =\operatorname {Re}\left (\frac {1}{i}\frac {2m\varepsilon \Omega ^{2}}{ic\Omega -M\Omega ^{2}}e^{i\left (\Omega t\right ) }\right ) \\ & =\operatorname {Re}\left (\frac {2\left (0.5\right ) \left ( 104.72\right ) ^{2}}{i\left (1.0882\times 10^{4}\right ) 104.72-200\left (104.72\right ) ^{2}}e^{i\left ( 104.72t-\frac {\pi }{2}\right ) }\right ) \\ & =\operatorname {Re}\left (\frac {10966.}{i1.1396\times 10^{6}-2.1933\times 10^{6}}e^{i\left (104.72t-\frac {\pi }{2}\right ) }\right ) \\ \left \vert y\right \vert & =4.4\text { mm} \end {align*}