2.6 HW5

2.6.1 problem description

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2.6.2 problem 1

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).$$(m_1+m_2)y^{\prime \prime }+y^{\prime }\mu +ky=f\left(t\right)$$

Since $y(t)=A\sin \left(\omega t\right)$ hence $$y(t)=\mathrm{Re}\left(\frac{A}{i}{e}^{i\omega t}\right)$$ Let $$f(t)=\mathrm{Re}\left(\frac{\hat{F}}{i}{e}^{i\left(\omega t\right)}\right)$$ Where $\hat{F}$ is the complex amplitude of the force. Now we substitute all these in the differential equation above.$$\begin{array}{rl}{y}^{\prime }& =\mathrm{Re}\left(\omega A{e}^{i\omega t}\right)\\ y^{\prime \prime }& =\mathrm{Re}\left(i{\omega }^{2}A{e}^{i\omega t}\right)\end{array}$$

$$\begin{array}{rl}(m_1+m_2)y^{\prime \prime }+y^{\prime }\mu +ky& =\mathrm{Re}\left(\frac{\hat{F}}{i}{e}^{i\left(\omega t\right)}\right)\\ \mathrm{Re}\left(i{\omega }^{2}A{e}^{i\omega t}\right)(m_1+m_2)+\mathrm{Re}\left(\omega A{e}^{i\omega t}\right)\mu +k\mathrm{Re}\left(\frac{A}{i}{e}^{i\omega t}\right)& =\mathrm{Re}\left(\frac{\hat{F}}{i}{e}^{i\left(\omega t\right)}\right)\\ \mathrm{Re}\left[(i{\omega }^2(m_1+m_2)+\omega \mu +\frac{1}{i}k)A{e}^{i\omega t}\right]& =\mathrm{Re}\left(\frac{\hat{F}}{i}{e}^{i\left(\omega t\right)}\right)\\ (i{\omega }^2(m_1+m_2)+\omega \mu +\frac{1}{i}k)A& =\frac{\hat{F}}{i}\end{array}$$

Hence $$\hat{F}=(-{\omega }^2(m_1+m_2)+i\omega \mu +k)A$$

$k=3.2\times {10}^3$ Nm$,\mu =40$ Ns/m,$A=0.02$ meter. When $\omega =75$rad/sec the above becomes$$\begin{array}{rl}\hat{F}& =(-{75}^2\left(1.5\right)+i75\times 40+3.2\times {10}^3)0.02\\ & =-104.75+60.0i\end{array}$$

Hence $\mathrm{Re}\left(\hat{F}\right)=-104.75$N and the phase is ${\tan }^{-1}(-\frac{60.0}{104.75})=2.62$ rad/sec.

When $\omega =85$$$\begin{array}{rl}\hat{F}& =1.5(-{85}^2+i85\frac{40}{1.5}+2133.3)0.02\\ & =-152.75+68.0i\end{array}$$

$\mathrm{Re}\left(\hat{F}\right)=-152.75$ N and the phase is ${\tan }^{-1}(-\frac{68}{152.75})=2.722$ rad/sec.

2.6.3 problem 2

2.6.3.1 Part(a)

Force transmitted to floor is given by$$F_{tr}=c{z}^{\prime }+kz$$

Let $f(t)=F\cos \left(\omega t\right)=\mathrm{Re}\left(F{e}^{i\omega t}\right)=\mathrm{Re}\left(F{e}^{i\omega t}\right)$ 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 $z_{ss}=\mathrm{Re}\left(\frac{F}{k}\left|D\right|e^{i(\omega t-\varphi )}\right)$ where $\varphi ={\tan }^{-1}\left(\frac{2\zeta r}{1-r^2}\right)$ and $\left|D\right|=\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2\zeta r\right)}^2}}$. and $r=\frac{\omega }{{\omega }_n}$ Hence $z^{\prime }=\mathrm{Re}\left(i\omega \frac{F}{k}\left|D\right|e^{i(\omega t-\varphi )}\right)=\mathrm{Re}\left(\omega \frac{F}{k}\left|D\right|e^{i(\omega t-\varphi +\frac{\pi }{2})}\right)$. Therefore$$F_{tr}=c\mathrm{Re}\left(\omega \frac{F}{k}\left|D\right|e^{i(\omega t-\varphi +\frac{\pi }{2})}\right)+k\mathrm{Re}\left(\frac{F}{k}\left|D\right|e^{i(\omega t-\varphi )}\right)$$

Where $c=2\zeta {\omega }_{n}m$ and When $F_{tr}=2\times {10}^3$N . We now solve for $k$ from$$2\times {10}^3\ge 2\zeta {\omega }_{n}m\mathrm{Re}\left(\omega \frac{F}{k}\left|D\right|e^{i(\omega t-\varphi +\frac{\pi }{2})}\right)+k\mathrm{Re}\left(\frac{F}{k}\left|D\right|e^{i(\omega t-\varphi )}\right)$$

Taking the maximum case for RHS where exponential are unity magnitude, hence$$\begin{array}{rl}2\times {10}^3& =2\zeta {\omega }_{n}m\omega \frac{F}{k}\left|D\right|+F\left|D\right|\\ & =(2\zeta {\omega }_{n}m\omega \left(\frac{F}{k}\right)+F)\left|D\right|\\ & =\frac{F(1+2\zeta {\omega }_n\frac{m}{k}\omega )}{\sqrt{{(1-r^2)}^2+{\left(2\zeta r\right)}^2}}\end{array}$$

Where $r=\frac{\omega }{{\omega }_n}=\frac{\omega }{\sqrt{\frac{k}{m}}}$. Hence the above becomes$$2\times {10}^3=\frac{F(1+2\zeta {\omega }_n\frac{m}{k}\omega )}{\sqrt{{(1-\frac{{\omega }^2}{\frac{k}{m}})}^2+{\left(2\zeta \frac{\omega }{\sqrt{\frac{k}{m}}}\right)}^2}}$$

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$$2\times {10}^3=\frac{20\times {10}^3(1+2\left(0.03\right)\sqrt{\frac{k}{450}}\frac{450}{k}\left(60\pi \right))}{\sqrt{{(1-\frac{450{\left(60\pi \right)}^2}{k})}^2+{\left(2\left(0.03\right)\frac{60\pi }{\sqrt{\frac{k}{450}}}\right)}^2}}$$

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.

2.6.3.2 part(b)

The total displacement is given by$$\begin{array}{rl}z(t)& =z_{transient}(t)+z_{ss}(t)\\ & =e^{-\zeta {\omega }_{n}t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)+\mathrm{Re}\left(\frac{F}{k}\left|D\right|e^{i(\omega t-\varphi )}\right)\end{array}$$

Where $$z_{transient}(t)=e^{-\zeta {\omega }_{n}t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)$$

Assuming at $t=0$ the system is relaxed hence $z(0)=0$ and $z^{\prime }(0)=0$ we can determine $A,B$ from Eq ??$.$

At $t=0,$

$$\begin{array}{rl}z(0)& =0\\ & =A+\mathrm{Re}\left(\frac{F}{k}\left|D\right|e^{-i\varphi }\right)\end{array}$$

Hence $$A=-\mathrm{Re}\left(\frac{F}{k}\left|D\right|e^{-i\varphi }\right)$$

and $$\begin{array}{rl}{z}^{\prime }(t)& =-\zeta {\omega }_n{e}^{-\zeta {\omega }_{n}t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)+e^{-\zeta {\omega }_{n}t}(-{\omega }_{d}A\sin {\omega }_{d}t+{\omega }_{d}B\cos {\omega }_{d}t)\\ & +\mathrm{Re}\left(\omega \frac{F}{k}\left|D\right|e^{i(-\varphi +\frac{\pi }{2})}\right)\end{array}$$

Hence at $t=0$$$\begin{array}{rl}{z}^{\prime }(0)& =0\\ & =-\zeta {\omega }_{n}A+{\omega }_{d}B+\mathrm{Re}\left(\omega \frac{F}{k}\left|D\right|e^{i(-\varphi +\frac{\pi }{2})}\right)\end{array}$$

Hence$$\begin{array}{rl}B& =\frac{\zeta {\omega }_n}{{\omega }_d}A-\frac{1}{{\omega }_d}\mathrm{Re}\left(\omega \frac{F}{k}\left|D\right|e^{i(-\varphi +\frac{\pi }{2})}\right)\\ & =-\frac{\zeta {\omega }_n}{{\omega }_d}\mathrm{Re}\left(\frac{F}{k}\left|D\right|e^{-i\varphi }\right)-\frac{1}{{\omega }_d}\mathrm{Re}\left(\omega \frac{F}{k}\left|D\right|e^{i(-\varphi +\frac{\pi }{2})}\right)\end{array}$$

Therefore the displacement is$$\begin{array}{rl}z(t)& =e^{-\zeta {\omega }_{n}t}[-\mathrm{Re}\left(\frac{F}{k}\left|D\right|e^{-i\varphi }\right)\cos {\omega }_{d}t+\{-\frac{\zeta {\omega }_n}{{\omega }_d}\mathrm{Re}\left(\frac{F}{k}\left|D\right|e^{-i\varphi }\right)-\frac{1}{{\omega }_d}\mathrm{Re}\left(\omega \frac{F}{k}\left|D\right|e^{i(-\varphi +\frac{\pi }{2})}\right)\}\sin {\omega }_{d}t]\\ & +\mathrm{Re}\left(\frac{F}{k}\left|D\right|e^{i(\omega t-\frac{\pi }{2}-\varphi )}\right)\end{array}$$

Hence expressed in $\sin$ and $\cos$$$\begin{array}{rl}z(t)& =-(\frac{F}{k}\left|D\right|\cos \varphi )e^{-\zeta {\omega }_{n}t}\cos {\omega }_{d}t+e^{-\zeta {\omega }_{n}t}\{-\frac{\zeta {\omega }_n}{{\omega }_d}(\frac{F}{k}\left|D\right|\cos \varphi )-\frac{1}{{\omega }_d}(\omega \frac{F}{k}\left|D\right|\sin \varphi )\}\sin {\omega }_{d}t\\ & +\frac{F}{k}\left|D\right|\cos (\omega t-\varphi )\\ & =-(\frac{F}{k}\left|D\right|\cos \varphi )e^{-\zeta {\omega }_{n}t}\cos {\omega }_{d}t+e^{-\zeta {\omega }_{n}t}\frac{F\left|D\right|}{{\omega }_{d}k}(-\zeta {\omega }_n\cos \varphi -\omega \sin \varphi )\sin {\omega }_{d}t+\frac{F}{k}\left|D\right|\cos (\omega t-\varphi )\\ & =\frac{F}{k}\left|D\right|e^{-\zeta {\omega }_{n}t}[-\cos \varphi \cos {\omega }_{d}t+\frac{1}{{\omega }_d}(-\zeta {\omega }_n\cos \varphi -\omega \sin \varphi )\sin {\omega }_{d}t]+\frac{F}{k}\left|D\right|\cos (\omega t-\varphi )\end{array}$$

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$$\begin{array}{rl}z(t)& =\frac{F}{k}\left|D\right|e^{-\zeta {\omega }_{n}t}[-\cos \varphi \cos {\omega }_{n}t+\frac{1}{{\omega }_n}(-\zeta {\omega }_n\cos \varphi -\omega \sin \varphi )\sin {\omega }_{n}t]+\frac{F}{k}\left|D\right|\cos (\omega t-\varphi )\\ & =\frac{F}{k}\left|D\right|e^{-\zeta {\omega }_{n}t}[-\cos \varphi \cos {\omega }_{n}t-(\zeta \cos \varphi +\frac{\omega }{{\omega }_n}\sin \varphi )\sin {\omega }_{n}t]+\frac{F}{k}\left|D\right|\cos (\omega t-\varphi )\end{array}$$

This is the amplitude. In the above $\left|D\right|=\frac{1}{\sqrt{{(1-{\left(\frac{\omega }{{\omega }_n}\right)}^2)}^2+{\left(2\zeta \left(\frac{\omega }{{\omega }_n}\right)\right)}^2}}$, and $\varphi ={\tan }^{-1}\frac{2\zeta \frac{\omega }{{\omega }_n}}{1-{\left(\frac{\omega }{{\omega }_n}\right)}^2}.$ 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 $z(t)$ 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

2.6.3.3 Part(c)

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.

2.6.4 Problem 3

Let $\epsilon =50mm=0.05m$ be the distance of the unbalance mass $m$. Let $M=80kg$ be the mass of the motor. The equation of motion is given by$$\begin{array}{rl}(M+m)y^{\prime \prime }+c{y}^{\prime }+ky& =m\epsilon {\mathrm{\Omega }}^2\sin \left(\mathrm{\Omega }t\right)\\ y^{\prime \prime }+2\zeta {\omega }_n{y}^{\prime }+{\omega }_n^{2}y& =\frac{m}{m+M}\epsilon {\mathrm{\Omega }}^2\mathrm{Re}\left(\frac{1}{i}{e}^{i\mathrm{\Omega }t}\right)\end{array}$$

Where ${\omega }_n=\sqrt{\frac{k}{m+M}}$ and $\zeta =\frac{c}{2(M+m){\omega }_n}$. Let $y=\mathrm{Re}\left(\frac{Y}{i}{e}^{i\mathrm{\Omega }t}\right)$. This leads to$$Y=\frac{m}{m+M}\frac{\epsilon {\mathrm{\Omega }}^2}{{\omega }_n^2-{\mathrm{\Omega }}^2+2i\zeta {\omega }_n\mathrm{\Omega }}$$

Since static deflection is $40mm$, then $$\begin{array}{rl}\frac{(M+m)g}{k}& =0.04\\ k& =\frac{(M+m)g}{0.04}\end{array}$$

But ${\omega }_n^2=\frac{k}{m+M}=\frac{(M+m)g}{0.04(M+m)}=\frac{g}{0.04}$, hence ${\omega }_n=\sqrt{\frac{9.81}{0.04}}=15.66$ rad/sec or $2.492$ Hz.

2.6.4.1 part(a)

Since at steady state the displacement is $10$ mm, then $\mathrm{\Omega }=2\pi \frac{145}{60}=15.184$ $$or $2.4167$ Hz hence$$\begin{array}{rl}y& =\mathrm{Re}\left(\frac{Y}{i}{e}^{i\mathrm{\Omega }t}\right)=\mathrm{Re}\left(\frac{\epsilon m}{m+M}\frac{r^2}{(1-r^2+2i\zeta r)}{e}^{i(\mathrm{\Omega }t-\frac{\pi }{2})}\right)\\ & =\frac{\epsilon m{r}^2}{m+M}\frac{1}{\sqrt{({(1-r^2)}^2+{\left(2\zeta r\right)}^2)}}\mathrm{Re}\left(e^{-i\varphi }{e}^{i(\mathrm{\Omega }t-\frac{\pi }{2})}\right)\end{array}$$

Where $\varphi ={\tan }^{-1}\left(\frac{2\zeta r}{1-r^2}\right)$. $r=\frac{\mathrm{\Omega }}{{\omega }_n}=\frac{15.184}{15.66}=$$0.9696$ hence the above becomes, at steady state$$\begin{array}{rl}0.01& =\frac{\left(0.05\right)m}{m+80}\frac{{0.9696}^2}{\sqrt{{(1-{0.9696}^2)}^2+{\left(2\zeta 0.9696\right)}^2}}\mathrm{Re}\left(e^{i(15.184t-\frac{\pi }{2}-\varphi )}\right)\\ & =\frac{\left(0.05\right)m}{m+80}\frac{{0.9696}^2}{\sqrt{{(1-{0.9696}^2)}^2+{\left(2\zeta 0.9696\right)}^2}}\sin (15.184t-\varphi )\end{array}$$

We are now told that at $\mathrm{\Omega }=15.184$ and when $\mathrm{\Omega }t={75}^0$ then the displacement is zero, hence$$0=\frac{\left(0.05\right)m}{m+80}\frac{{0.9696}^2}{\sqrt{{(1-{0.9696}^2)}^2+{\left(2\zeta 0.9696\right)}^2}}\sin ({75}^0-\varphi )$$

or$$\begin{array}{rl}\sin ({75}^0-\varphi )& =0\\ {75}^0-\varphi & =0\\ \varphi & ={75}^0\end{array}$$

Since $\varphi ={\tan }^{-1}\left(\frac{2\zeta r}{1-r^2}\right)$ then $$75\left(\frac{\pi }{180}\right)={\tan }^{-1}\left(\frac{2\zeta 0.9696}{1-{0.9696}^2}\right)$$

Hence$$\begin{array}{rl}{\tan }^{-1}\left(\frac{2\zeta 0.9696}{1-{0.9696}^2}\right)& =1.3090\\ \frac{2\zeta 0.9696}{1-{0.9696}^2}& =\tan \left(1.3090\right)\\ \frac{2\zeta 0.9696}{1-{0.9696}^2}& =3.7321\end{array}$$

Hence $\zeta =0.11523$

2.6.4.2 Part(b)

From Eq ??$$0.01=\frac{\left(0.05\right)m}{m+80}\frac{{0.9696}^2}{\sqrt{{(1-{0.9696}^2)}^2+{\left(2\zeta 0.9696\right)}^2}}\sin (15.184t-\varphi )$$

The maximum amplitude is when$$0.01=\frac{\left(0.05\right)m}{m+80}\frac{{0.9696}^2}{\sqrt{{(1-{0.9696}^2)}^2+{\left(2\zeta 0.9696\right)}^2}}$$

But $\zeta =0.11523$, hence we now solve for $m$$$0.01=\frac{\left(0.05\right)m}{m+80}\frac{{0.9696}^2}{\sqrt{{(1-{0.9696}^2)}^2+{\left(2\left(0.11523\right)0.9696\right)}^2}}$$ Hence $$\boxed{m=4.1kg}$$

Hence $\epsilon m=\left(0.05\right)\left(4.1\right)=$$0.20$ kg meter

2.6.4.3 Part(c)

since

$$y=\frac{\epsilon m{r}^2}{m+M}\frac{1}{\sqrt{({(1-r^2)}^2+{\left(2\zeta r\right)}^2)}}\mathrm{Re}\left(e^{i(\mathrm{\Omega }t-\frac{\pi }{2}-\varphi )}\right)$$

As $\mathrm{\Omega }$ becomes much larger than ${\omega }_n$ then ${(1-r^2)}^2\to r^4$ . Now dividing numerator and denominator by $r^2$ gives$$\begin{array}{rl}y& =\frac{\epsilon m}{m+M}\frac{1}{\sqrt{\frac{(r^4+{\left(2\zeta r\right)}^2)}{r^4}}}\sin (\mathrm{\Omega }t-\varphi )\\ & =\frac{\epsilon m}{m+M}\frac{1}{\sqrt{(1+\frac{4{\zeta }^2}{r^2})}}\sin (\mathrm{\Omega }t-\varphi )\end{array}$$

as $r$ becomes large then $\frac{4{\zeta }^2}{r^2}\to 0$ hence$$y\simeq \frac{\epsilon m}{m+M}\sin (\mathrm{\Omega }t-\varphi )$$

The smallest possible amplitude is$$\begin{array}{rl}\left|y\right|& =\frac{0.20}{4.1+80}\\ & =2.3781\times {10}^{-3}\text{ meter}\end{array}$$

or$$\left|y\right|=2.38\text{ mm}$$

2.6.5 problem 4

2.6.5.1 Part(a)

(note: total mass of system includes the small unbalanced masses) Since static deflection is $8.5mm$, then $$\begin{array}{rl}\frac{Mg}{k}& =0.0085\\ k& =\frac{Mg}{0.0085}\end{array}$$

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

2.6.5.2 Part(b)

The equation of motion is (angle $\mathrm{\Omega }$ is now measured from horizontal, anti-clock wise positive)$$M{y}^{\prime \prime }+c{y}^{\prime }=2m\epsilon {\mathrm{\Omega }}^2\sin \left(\mathrm{\Omega }t\right)=\mathrm{Re}\left(\frac{1}{i}2m\epsilon {\mathrm{\Omega }}^2{e}^{i\left(\mathrm{\Omega }t\right)}\right)$$

Let $y(t)=\mathrm{Re}\left(\frac{1}{i}Y{e}^{i\mathrm{\Omega }t}\right)$ hence $y^{\prime }(t)=\mathrm{Re}\left(Y\mathrm{\Omega }{e}^{i\mathrm{\Omega }t}\right)$,$y^{\prime \prime }(t)=\mathrm{Re}\left(iY{\mathrm{\Omega }}^2{e}^{i\mathrm{\Omega }t}\right)$, hence the above becomes$$\begin{array}{rl}\mathrm{Re}\left(iY{\mathrm{\Omega }}^2{e}^{i\mathrm{\Omega }t}\right)+\frac{c}{M}\mathrm{Re}\left(Y\mathrm{\Omega }{e}^{i\mathrm{\Omega }t}\right)& =\mathrm{Re}\left(\frac{1}{i}\frac{2m\epsilon {\mathrm{\Omega }}^2}{M}{e}^{i\mathrm{\Omega }t}\right)\\ \mathrm{Re}\left((i{\mathrm{\Omega }}^2+\frac{c\mathrm{\Omega }}{M})Y{e}^{i\mathrm{\Omega }t}\right)& =\mathrm{Re}\left(\frac{1}{i}\frac{2m\epsilon {\mathrm{\Omega }}^2}{M}{e}^{i\mathrm{\Omega }t}\right)\\ [i{\mathrm{\Omega }}^2+\frac{c\mathrm{\Omega }}{M}]Y& =\frac{1}{i}\frac{2m\epsilon {\mathrm{\Omega }}^2}{M}\\ Y& =\frac{1}{i}\frac{\frac{2m\epsilon {\mathrm{\Omega }}^2}{M}}{i{\mathrm{\Omega }}^2+\frac{c\mathrm{\Omega }}{M}}\\ & =\frac{2m\epsilon {\mathrm{\Omega }}^2}{ic\mathrm{\Omega }-M{\mathrm{\Omega }}^2}\end{array}$$

Hence$$\begin{array}{rl}{y}_{ss}(t)& =\mathrm{Re}\left(\frac{1}{i}Y{e}^{i\left(\mathrm{\Omega }t\right)}\right)\\ & =\mathrm{Re}\left(\frac{1}{i}\frac{2m\epsilon {\mathrm{\Omega }}^2}{ic\mathrm{\Omega }-M{\mathrm{\Omega }}^2}{e}^{i\left(\mathrm{\Omega }t\right)}\right)\end{array}$$

Now we are told when $\mathrm{\Omega }t=\frac{\pi }{2}$ (upright position) then $y=0$(since it passes static equilibrium). At this moment $\mathrm{\Omega }=2\pi \frac{900}{60}=$$94.248$ rad/sec$,$ At this moment the centripetal forces equal the damping force downwards (since the mass was moving upwards). Hence $$\boxed{m\epsilon {\mathrm{\Omega }}^2=c{y}^{\prime }(t)}$$

But from above we found that $$\begin{array}{rl}{y}^{\prime }(t)& =\mathrm{Re}\left(\frac{2m\epsilon {\mathrm{\Omega }}^2}{ic\mathrm{\Omega }-M{\mathrm{\Omega }}^2}\mathrm{\Omega }{e}^{i\mathrm{\Omega }t}\right)\\ & =\mathrm{Re}\left(\frac{\left(0.5\right){94.248}^2}{ic\left(94.248\right)-200{\left(94.248\right)}^2}94.248e^{i\frac{\pi }{2}}\right)\\ & =\mathrm{Re}\left(\frac{8.3718\times {10}^5}{94.248ic-1.7765\times {10}^6}{e}^{i\frac{\pi }{2}}\right)\end{array}$$

Hence$$\begin{array}{rl}m\epsilon {\mathrm{\Omega }}^2& =c|y^{\prime }(t)|\\ \left(0.5\right){94.248}^2& =c\frac{8.3718\times {10}^5}{\sqrt{{\left(94.248c\right)}^2+{(1.7765\times {10}^6)}^2}}\\ 4441.3& =8.3718\times {10}^5\frac{c}{\sqrt{8882.7c^2+3.1560\times {10}^{12}}}\end{array}$$

Solving numerically for $c$ gives$$c=1.0882\times {10}^4\text{ N second per meter}$$

2.6.5.3 Part(c)

When $\mathrm{\Omega }=\left(2\pi \frac{1000}{60}\right)=104.72$ rad/sec or $16.667$ Hz. From$$\begin{array}{rl}y& =\mathrm{Re}\left(\frac{1}{i}\frac{2m\epsilon {\mathrm{\Omega }}^2}{ic\mathrm{\Omega }-M{\mathrm{\Omega }}^2}{e}^{i\left(\mathrm{\Omega }t\right)}\right)\\ & =\mathrm{Re}\left(\frac{2\left(0.5\right){\left(104.72\right)}^2}{i(1.0882\times {10}^4)104.72-200{\left(104.72\right)}^2}{e}^{i(104.72t-\frac{\pi }{2})}\right)\\ & =\mathrm{Re}\left(\frac{10966.}{i1.1396\times {10}^6-2.1933\times {10}^6}{e}^{i(104.72t-\frac{\pi }{2})}\right)\\ \left|y\right|& =4.4\text{ mm}\end{array}$$

2.6.6 Key solution for HW 5

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