Using Rayleigh method, we need to find \(T_{\max }\) and \(U_{\max }\) where \(T\) is the kinetic energy of the system and \(U\) is the potential energy and then solve for \(\omega _{n}\) by setting \(T_{\max }=U_{\max }\).
Kinetic energy is\[ T=\frac{1}{2}m\dot{x}^{2}+\frac{1}{2}J_{o}\dot{\theta }^{2}\] But \(x=r_{1}\theta \), therefore \(\dot{x}=r_{1}\dot{\theta }\) and the above becomes\begin{equation} T=\frac{1}{2}m\left ( r_{1}\dot{\theta }\right ) ^{2}+\frac{1}{2}J_{o}\dot{\theta }^{2}\tag{1} \end{equation} And potential energy only comes from the spring, since we assume \(x\) is measured from static equilibrium. Hence\begin{align} U & =\frac{1}{2}kx^{2}\nonumber \\ & =\frac{1}{2}k\left ( r_{2}\theta \right ) ^{2}\tag{2} \end{align}
To get \(\omega _{n}\) into (1) and (2), we now assume that motion is harmonic, hence \(\theta =\theta _{\max }\sin \left ( \omega _{n}t\right ) \), Therefore \(\dot{\theta }=\theta _{\max }\omega _{n}\cos \left ( \omega _{n}t\right ) \) and rewriting (1,2) using these expressions results in\begin{align*} T & =\frac{1}{2}m\left ( r_{1}\theta _{\max }\omega _{n}\cos \left ( \omega _{n}t\right ) \right ) ^{2}+\frac{1}{2}J_{o}\left ( \theta _{\max }\omega _{n}\cos \left ( \omega _{n}t\right ) \right ) ^{2}\\ U & =\frac{1}{2}k\left ( r_{2}\left ( \theta _{\max }\sin \left ( \omega _{n}t\right ) \right ) \right ) ^{2} \end{align*}
Hence, maximum is when \(\theta =\theta _{\max }\) and \(\dot{\theta }=\theta _{\max }\omega _{n}\) and the above becomes\begin{align*} T_{\max } & =\frac{1}{2}mr_{1}^{2}\theta _{\max }^{2}\omega _{n}^{2}+\frac{1}{2}J_{o}\theta _{\max }^{2}\omega _{n}^{2}\\ U_{\max } & =\frac{1}{2}kr_{2}^{2}\theta _{\max }^{2} \end{align*}
Now \begin{align*} T_{\max } & =U_{\max }\\ \frac{1}{2}mr_{1}^{2}\theta _{\max }^{2}\omega _{n}^{2}+\frac{1}{2}J_{o}\theta _{\max }^{2}\omega _{n}^{2} & =\frac{1}{2}kr_{2}^{2}\theta _{\max }^{2}\\ mr_{1}^{2}\omega _{n}^{2}+J_{o}\omega _{n}^{2} & =kr_{2}^{2} \end{align*}
Hence\begin{align*} \omega _{n}^{2} & =\frac{kr_{2}^{2}}{mr_{1}^{2}+J_{o}}\\ \omega _{n} & =\sqrt{\frac{kr_{2}^{2}}{mr_{1}^{2}+J_{o}}} \end{align*}
The equation of motion is given by\[ \frac{d}{dt}\left ( T+U\right ) =0 \] We found \(T,U\) in part (a), therefore the above becomes\begin{align*} \frac{d}{dt}\left ( \frac{1}{2}m\left ( r_{1}\dot{\theta }\right ) ^{2}+\frac{1}{2}J_{o}\dot{\theta }^{2}+\frac{1}{2}k\left ( r_{2}\theta \right ) ^{2}\right ) & =0\\ mr_{1}^{2}\dot{\theta }\ddot{\theta }+J_{o}\dot{\theta }\ddot{\theta }+kr_{2}^{2}\theta \dot{\theta } & =0 \end{align*}
For non trivial motion \(\dot{\theta }\neq 0\) for all time, hence we can divide throughout by \(\dot{\theta }\) and obtain\begin{align*} mr_{1}^{2}\ddot{\theta }+J_{o}\ddot{\theta }+kr_{2}^{2}\theta & =0\\ \ddot{\theta }\left ( mr_{1}^{2}+J_{o}\right ) +kr_{2}^{2}\theta & =0\\ \ddot{\theta }+\frac{kr_{2}^{2}}{mr_{1}^{2}+J_{o}}\theta & =0 \end{align*}
The above is the equation of motion.
First part
The first step is to determine damping ratio \(\zeta \). This is done using logarithmic decrement.
Since \(X_{1.5}=\frac{1}{4}X_{1}\) and \(X_{2}=\frac{1}{4}X_{1.5}\) then \begin{align*} X_{2} & =\frac{1}{4}\left ( \frac{1}{4}X_{1}\right ) \\ & =\frac{1}{16}X_{1} \end{align*}
Using\[ \frac{X_{1}}{X_{2}}=\frac{e^{-\zeta \omega _{n}t_{1}}}{e^{-\zeta \omega _{n}\left ( t_{1}+t_{2}\right ) }}\] Where \(t_{2}=t_{1}+\tau _{d}\) and \(\tau _{d}\) is damped period. Therefore the above becomes\begin{align*} \frac{X_{1}}{\frac{1}{16}X_{1}} & =\frac{e^{-\zeta \omega _{n}t_{1}}}{e^{-\zeta \omega _{n}\left ( t_{1}+\tau _{d}\right ) }}=\frac{e^{-\zeta \omega _{n}t_{1}}}{e^{-\zeta \omega _{n}t_{1}}e^{-\zeta \omega _{n}\tau _{d}}}=e^{\zeta \omega _{n}\tau _{d}}\\ \ln \left ( 16\right ) & =\zeta \omega _{n}\tau _{d} \end{align*}
Taking log of both sides gives
\begin{equation} \ln \left ( 16\right ) =\zeta \omega _{n}\tau _{d}\tag{1} \end{equation}
But \begin{align*} \tau _{d} & =\frac{2\pi }{\omega _{d}}\\ & =\frac{2\pi }{\omega _{n}\sqrt{1-\zeta ^{2}}} \end{align*}
And (1) simplifies to\begin{align*} \ln \left ( 16\right ) & =\zeta \omega _{n}\frac{2\pi }{\omega _{n}\sqrt{1-\zeta ^{2}}}\\ 2.7726 & =\frac{2\pi \zeta }{\sqrt{1-\zeta ^{2}}} \end{align*}
Squaring both sides and solving for \(\zeta \) gives \begin{align*} \left ( 2.7726\right ) ^{2}\left ( 1-\zeta ^{2}\right ) & =4\pi ^{2}\zeta ^{2}\\ \zeta ^{2}\left ( 4\pi ^{2}+7.6873\right ) & =7.6873\\ \zeta ^{2} & =\frac{7.6873}{4\pi ^{2}+7.6873} \end{align*}
Taking the positive root results in\begin{align*} \zeta & =\sqrt{\frac{7.6873}{4\pi ^{2}+7.6873}}\\ & =0.40371 \end{align*}
Now that \(\zeta \) is know, \(\omega _{n}\) can be found, since we are told that \(\tau _{d}=2\) seconds. Using \[ \tau _{d}=\frac{2\pi }{\omega _{n}\sqrt{1-\zeta ^{2}}}\] Then solving for\(\ \omega _{n}\) from the above gives\begin{align*} 2 & =\frac{2\pi }{\omega _{n}\sqrt{1-0.40371^{2}}}\\ \omega _{n} & =\frac{\pi }{\sqrt{1-0.40371^{2}}}\\ & =3.4339\text{ rad/sec} \end{align*}
Now we are ready to find the stiffness coefficient \(k\) and damping coefficient \(c\). Using\[ \zeta =\frac{c}{2\omega _{n}m}\] Then\begin{align*} c & =2\zeta \omega _{n}m\\ & =2\left ( 0.40371\right ) \left ( 3.4339\right ) \left ( 200\right ) \\ & =554.52\text{ N-s/m} \end{align*}
But since \[ \omega _{n}^{2}=\frac{k}{m}\] Then \(k\) is now found\begin{align*} k & =\omega _{n}^{2}m\\ & =\left ( 3.4339\right ) ^{2}\left ( 200\right ) \\ & =2358.3\text{ N/m} \end{align*}
Second part
Maximum displacement occurs at time \(t_{1}\) as given by (from textbook)\[ \sin \omega _{d}t_{1}=\sqrt{1-\zeta ^{2}}\] Hence\begin{align*} \omega _{d}t_{1} & =\arcsin \left ( \sqrt{1-\zeta ^{2}}\right ) \\ t_{1} & =\frac{1}{\omega _{n}\sqrt{1-\zeta ^{2}}}\arcsin \left ( \sqrt{1-\zeta ^{2}}\right ) \\ & =\frac{1}{3.4339\sqrt{1-0.40371^{2}}}\arcsin \left ( \sqrt{1-0.40371^{2}}\right ) \\ & =0.36772\text{ sec} \end{align*}
Since\begin{equation} x\left ( t\right ) =Xe^{-\zeta \omega _{n}t}\sin \left ( \omega _{d}t\right ) \tag{2} \end{equation}
Then at maximum displacement, where \(x=0.25\) m, the above becomes\begin{align*} x_{\max }\left ( t_{1}\right ) & =Xe^{-\zeta \omega _{n}t_{1}}\sin \left ( \omega _{d}t_{1}\right ) \\ \frac{x_{\max }e^{\zeta \omega _{n}t_{1}}}{\sin \left ( \omega _{d}t_{1}\right ) } & =X \end{align*}
Plug-in numerical values to solve for maximum displacement \(X\) gives\begin{align*} X & =\frac{0.25\exp \left ( 0.40371\times 3.4339\times 0.36772\right ) }{\sin \left ( \left ( 3.4339\sqrt{1-0.40371^{2}}\right ) \left ( 0.36772\right ) \right ) }\\ & =0.45495\text{ m} \end{align*}
From (2), the velocity is found\begin{align*} \dot{x}\left ( t\right ) & =-\zeta \omega _{n}Xe^{-\zeta \omega _{n}t}\sin \left ( \omega _{d}t\right ) +Xe^{-\zeta \omega _{n}t}\omega _{d}\cos \left ( \omega _{d}t\right ) \\ & =Xe^{-\zeta \omega _{n}t}\left ( \omega _{d}\cos \left ( \omega _{d}t\right ) -\zeta \omega _{n}\sin \left ( \omega _{d}t\right ) \right ) \end{align*}
At \(t=0\) the above gives\begin{align*} \dot{x}\left ( 0\right ) & =X\omega _{d}\\ & =X\left ( \omega _{n}\sqrt{1-\zeta ^{2}}\right ) \end{align*}
Plug-in in numerical values\begin{align*} \dot{x}\left ( 0\right ) & =0.45495\left ( 3.4339\sqrt{1-0.40371^{2}}\right ) \\ & =1.4293\text{ m/s} \end{align*}