2.7 HW6

2.7.1 problem description

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2.7.2 problem 1

3.41 in text: A periodic disturbance consists of a sequence of exponentially pulse repeated at intervals $T$, such that $Q(t)=F{e}^{\frac{-\lambda t}{T}}$ for $0<t<T$, and $Q(t\pm T)=Q\left(t\right)$. The parameter $\lambda$ is nondimensional. Determine the complex Fourier series representing the force. Evaluate the first $5$ coefficients when $\lambda =0.1,1,10$. What does this reveal regarding the influence of $\lambda$ on the frequency spectrum?

Let $\tilde{Q}(t)$ be the Fourier series approximation to $Q(t)$ given by$$\tilde{Q}(t)=\frac{1}{2}{\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}{F}_n{e}^{in\frac{2\pi }{T}t}$$

Where$$\begin{array}{rl}{F}_n& =\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}}Q(t)e^{-in\frac{2\pi }{T}t}dt\\ & =\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}}F{e}^{\frac{-\lambda t}{T}}{e}^{-in\frac{2\pi }{T}t}dt=\frac{2F}{T}{\displaystyle \underset{0}{\overset{T}{\int }}}{e}^{-t(in\frac{2\pi }{T}-\frac{\lambda }{T})}dt=\frac{2F}{T}{\left(\frac{e^{-t(in\frac{2\pi }{T}-\frac{\lambda }{T})}}{in\frac{2\pi }{T}-\frac{\lambda }{T}}\right)}_0^T\\ & =\frac{2F}{in2\pi -\lambda }(e^{-T(in\frac{2\pi }{T}-\frac{\lambda }{T})}-1)\\ & =\frac{2F}{in2\pi -\lambda }(e^{-in2\pi }{e}^{-\lambda }-1)\end{array}$$

But $e^{-in2\pi }=1$, hence $$F_n=\frac{2F}{in2\pi -\lambda }(e^{-\lambda }-1)$$

Hence Eq ?? becomes$$\begin{array}{rl}\tilde{Q}(t)& =\frac{1}{2}{\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}\frac{2F}{in2\pi -\lambda }(e^{-\lambda }-1)e^{in\frac{2\pi }{T}t}\\ & =F{\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}\frac{(e^{-\lambda }-1)}{in2\pi -\lambda }{e}^{in\frac{2\pi }{T}t}\\ & =F{\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}\frac{1-e^{-\lambda }}{\lambda +in2\pi }{e}^{in\frac{2\pi }{T}t}\end{array}$$

For $n=-2,-1,0,1,2$ we obtain$$\begin{array}{rl}\tilde{Q}(t)& =F{\displaystyle \sum _{n=-2}^2}\frac{1-e^{-\lambda }}{\lambda +in2\pi }{e}^{in\frac{2\pi }{T}t}\\ & =F(\frac{1-e^{-\lambda }}{\lambda -i4\pi }{e}^{-i\frac{4\pi }{T}t}+\frac{1-e^{-\lambda }}{\lambda -i2\pi }{e}^{-i\frac{2\pi }{T}t}+\frac{1-e^{-\lambda }}{\lambda }+\frac{1-e^{-\lambda }}{\lambda +i2\pi }{e}^{i\frac{2\pi }{T}t}+\frac{1-e^{-\lambda }}{\lambda +i4\pi }{e}^{i\frac{4\pi }{T}t})\end{array}$$

For $\lambda =0.1$$$\begin{array}{rl}\tilde{Q}(t)& =F(\frac{1-e^{-0.1}}{0.1-i4\pi }{e}^{-i\frac{4\pi }{T}t}+\frac{1-e^{-0.1}}{0.1-i2\pi }{e}^{-i\frac{2\pi }{T}t}+\frac{1-e^{-0.1}}{0.1}+\frac{1-e^{-0.1}}{0.1+i2\pi }{e}^{i\frac{2\pi }{T}t}+\frac{1-e^{-0.1}}{0.1+i4\pi }{e}^{i\frac{4\pi }{T}t})\\ & =F\{(6.026\times {10}^{-5}+7.572\times {10}^{-3}i)e^{-i\frac{4\pi }{T}t}\\ & +(2.41\times {10}^{-4}+1.514\times {10}^{-2}i)e^{-i\frac{2\pi }{T}t}\\ & +0.952\\ & +(2.4099\times {10}^{-4}-1.5142\times {10}^{-2}i)e^{i\frac{2\pi }{T}t}\\ & +(6.026\times {10}^{-5}-7.572\times {10}^{-3}i)e^{i\frac{4\pi }{T}t}\}\end{array}$$

For $\lambda =1$$$\begin{array}{rl}\tilde{Q}(t)& =F(\frac{1-e^{-1}}{1-i4\pi }{e}^{-i\frac{4\pi }{T}t}+\frac{1-e^{-1}}{1-i2\pi }{e}^{-i\frac{2\pi }{T}t}+\frac{1-e^{-1}}{1}+\frac{1-e^{-1}}{1+i2\pi }{e}^{i\frac{2\pi }{T}t}+\frac{1-e^{-1}}{1+i4\pi }{e}^{i\frac{4\pi }{T}t})\\ & =F\{(0.00398+0.05i)e^{-i\frac{4\pi }{T}t}+(0.016+0.098i)e^{-i\frac{2\pi }{T}t}+0.632+(0.016+0.098i)e^{i\frac{2\pi }{T}t}+(0.00398+0.05i)e^{i\frac{4\pi }{T}t}\}\end{array}$$

For $\lambda =10$$$\begin{array}{rl}\tilde{Q}(t)& =F(\frac{1-e^{-10}}{10-i4\pi }{e}^{-i\frac{4\pi }{T}t}+\frac{1-e^{-10}}{10-i2\pi }{e}^{-i\frac{2\pi }{T}t}+\frac{1-e^{-10}}{10}+\frac{1-e^{-10}}{10+i2\pi }{e}^{i\frac{2\pi }{T}t}+\frac{1-e^{-10}}{10+i4\pi }{e}^{i\frac{4\pi }{T}t})\\ & =F\{(3.877\times {10}^{-2}+4.872\times {10}^{-2}i)e^{-i\frac{4\pi }{T}t}\\ & +(7.169\times {10}^{-2}+4.505\times {10}^{-2}i)e^{-i\frac{2\pi }{T}t}\\ & +0.1\\ & +(7.169\times {10}^{-2}-4.505\times {10}^{-2}i)e^{i\frac{2\pi }{T}t}\\ & +(3.877\times {10}^{-2}-4.872\times {10}^{-2}i)e^{i\frac{4\pi }{T}t}\}\end{array}$$

We notice that as $\lambda$ became larger, the DC term became smaller. Since the $DC$ term represents average value of the whole signal, then we can say that as $\lambda$ gets larger, then the average becomes smaller. This means the energy of the signal becomes smaller as $\lambda$ becomes larger.

2.7.2.1 Verification using Matlab ffteasy.m

From above, we found for $\lambda =1$$$\begin{array}{rl}{F}_n& =\frac{2F}{in2\pi -\lambda }(e^{-\lambda }-1)\\ & =\frac{2F}{in2\pi -1}(e^{-1}-1)\end{array}$$

and the first $5$ found to be

$n$	$F_n$
$-2$	$0.00398+0.05i$
$-1$	$0.016+0.098i$
$0$	$0.632$
$1$	$0.016-0.098i$
$2$	$0.00398-0.05i$

To verify the result with ffteasy.m using $\lambda =1$, Using $F=1$, and using $T=1$. This below shows the result for $F_0,F_1,F_2$ and we see that the DC term $F_0$ agrees, and that complex component of $F_1,F_2$ also agrees. The real parts are little larger than what I obtained using the above. This might be a scaling issue, and I was not able to determine the reason for it at this time.

EDU>> T=1; del=0.01; t=0:del:T; lambda=1; xt=exp(-lambda*t/T);
EDU>> (1/length(t))*fft_easy(xt,t)

ans =

   0.6326 + 0.0000i
   0.0190 - 0.0986i
   0.0072 - 0.0502i

2.7.3 problem 2

We are given that $m=1200$ kg, $f=5$ Hz, $\zeta =0.4$ and$$z(x)=\{\begin{array}{lll}x-5x^2& & 0<x<0.2\\ 0& & 0.2<x<4\end{array}$$

A plot of $z(x)$ for first $20$ meters is

z[x_] := Piecewise[{{x - 5 x^2, 0 <= x < 0.2}, {0, 0.2 <= x <= 4}}]
z[x_] /; x > 4 := z[Mod[x, 4]];
Table[{x, z[x]}, {x, 0, 21, .1}];
ListLinePlot[%, PlotRange -> {All, {0, .07}}, Frame -> True,
 FrameLabel -> {{"z(x) hight or road (mm)", None}, {"meter",
    "bumps on road"}}]

We need to be able to express $z(t)$ as $\mathrm{Re}\left\{Z{e}^{i\frac{2\pi }{T}t}\right\}$ where $T$ is the period of the function $z(t)$. Hence we need to represent $z(x)$ as Fourier series approximation then replace $x=vt$ and use the result.

The period $T=4$ meter. Let $\tilde{z}(x)$ be the Fourier series approximation to $z(x)$, hence$$\tilde{z}(x)=\frac{1}{2}{F}_0+\mathrm{Re}\left({\displaystyle \sum _{n=1}^N}{F}_n{e}^{in\frac{2\pi }{T}x}\right)$$

Where$$F_n=\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}}z(x)e^{-in\frac{2\pi }{T}x}dx=\frac{1}{2}{\displaystyle \underset{0}{\overset{2/10}{\int }}}(x-5x^2)e^{-in\frac{\pi }{2}x}dx=\frac{1}{2}{\displaystyle \underset{0}{\overset{2/10}{\int }}}x{e}^{-in\frac{\pi }{2}x}dx-\frac{5}{2}{\displaystyle \underset{0}{\overset{2/10}{\int }}}{x}^2{e}^{-in\frac{\pi }{2}x}dx$$

Using integration by parts $\int udv=uv-\int vdu$, letting $u=x$ and $dv=e^{-in\frac{\pi }{2}x}$ then $v=\int e^{-in\frac{\pi }{2}x}dx=\frac{i{e}^{-in\frac{\pi }{2}x}}{n\frac{\pi }{2}}$hence$$\begin{array}{rl}{\displaystyle \underset{0}{\overset{2/10}{\int }}}x{e}^{-in\frac{\pi }{2}x}dx& ={x\frac{i{e}^{-in\frac{\pi }{2}x}}{n\frac{\pi }{2}}|}_0^{\frac{2}{10}}-{\displaystyle \underset{0}{\overset{2/10}{\int }}}\frac{i{e}^{-in\frac{\pi }{2}x}}{n\frac{\pi }{2}}dx\\ & =\frac{2}{10}\frac{i{e}^{-in\frac{\pi }{2}\frac{2}{10}}}{n\frac{\pi }{2}}-\frac{2}{n\pi }{\displaystyle \underset{0}{\overset{2/10}{\int }}}i{e}^{-in\frac{\pi }{2}x}dx\\ & =\frac{4}{10}\frac{i{e}^{-in\frac{\pi }{10}}}{n\pi }-\frac{i2}{n\pi }{\left(\frac{e^{-in\frac{\pi }{2}x}}{-in\frac{\pi }{2}}\right)}_0^{\frac{2}{10}}\\ & =\frac{4}{10}\frac{i{e}^{-in\frac{\pi }{10}}}{n\pi }+\frac{4}{n^2{\pi }^2}{\left(e^{-in\frac{\pi }{2}x}\right)}_0^{\frac{2}{10}}\\ & =\frac{4}{10}\frac{i{e}^{-in\frac{\pi }{10}}}{n\pi }+\frac{4}{n^2{\pi }^2}(e^{-in\frac{\pi }{2}\frac{2}{10}}-1)\\ & =\frac{4i}{10n\pi }{e}^{-in\frac{\pi }{10}}+\frac{4}{n^2{\pi }^2}{e}^{-in\frac{\pi }{10}}-\frac{4}{n^2{\pi }^2}\\ & =e^{-in\frac{\pi }{10}}(\frac{4}{n^2{\pi }^2}+\frac{2i}{5n\pi })-\frac{4}{n^2{\pi }^2}\end{array}$$

Now we do the second integral ${\displaystyle \underset{0}{\overset{2/10}{\int }}}{x}^2{e}^{-in\frac{\pi }{2}x}dx$.

Integration by parts, $\int udv=uv-\int vdu$, letting $u=x^2$ and $dv=e^{-in\frac{\pi }{2}x}$ then $v=\frac{i{e}^{-in\frac{\pi }{2}x}}{n\frac{\pi }{2}}$ hence$$\begin{array}{rl}{\displaystyle \underset{0}{\overset{2/10}{\int }}}{x}^2{e}^{-in\frac{\pi }{2}x}dx& ={\left[x^2\frac{i{e}^{-in\frac{\pi }{2}x}}{n\frac{\pi }{2}}\right]}_0^{\frac{2}{10}}-{\displaystyle \underset{0}{\overset{2/10}{\int }}}2x\frac{i{e}^{-in\frac{\pi }{2}x}}{n\frac{\pi }{2}}dx\\ & =\frac{8}{100}\frac{i{e}^{-in\frac{\pi }{10}}}{n\pi }-\frac{4i}{n\pi }{\displaystyle \underset{0}{\overset{2/10}{\int }}}x{e}^{-in\frac{\pi }{2}x}dx\end{array}$$

But ${\displaystyle \underset{0}{\overset{2/10}{\int }}}x{e}^{-in\frac{\pi }{2}x}dx$ was solved before and its results is Eq 2.1, hence$$\begin{array}{rl}{\displaystyle \underset{0}{\overset{2/10}{\int }}}{x}^2{e}^{-in\frac{\pi }{2}x}dx& =\frac{4}{100}\frac{i{e}^{-in\frac{\pi }{10}}}{n\frac{\pi }{2}}-\frac{4i}{n\pi }(e^{-in\frac{\pi }{10}}(\frac{4}{n^2{\pi }^2}+\frac{2i}{5n\pi })-\frac{4}{n^2{\pi }^2})\\ & =\frac{8i}{100n\pi }{e}^{-in\frac{\pi }{10}}-e^{-in\frac{\pi }{10}}(\frac{16i}{n^3{\pi }^3}-\frac{8}{5n^2{\pi }^2})+\frac{16i}{n^3{\pi }^3}\\ & =e^{-in\frac{\pi }{10}}(\frac{8i}{100n\pi }-\frac{16i}{n^3{\pi }^3}+\frac{8}{5n^2{\pi }^2})+\frac{16i}{n^3{\pi }^3}\end{array}$$

Putting all the above together, we obtain $F_n$ as$$\begin{array}{rl}{F}_n& =\frac{1}{2}{\displaystyle \underset{0}{\overset{2/10}{\int }}}x{e}^{-in\frac{\pi }{2}x}dx-\frac{5}{2}{\displaystyle \underset{0}{\overset{2/10}{\int }}}{x}^2{e}^{-in\frac{\pi }{2}x}dx\\ & =\frac{1}{2}[e^{-in\frac{\pi }{10}}(\frac{4}{n^2{\pi }^2}+\frac{2i}{5n\pi })-\frac{4}{n^2{\pi }^2}]-\frac{5}{2}[e^{-in\frac{\pi }{10}}(\frac{8i}{100n\pi }-\frac{16i}{n^3{\pi }^3}+\frac{8}{5n^2{\pi }^2})+\frac{16i}{n^3{\pi }^3}]\\ & =e^{-in\frac{\pi }{10}}(\frac{2}{n^2{\pi }^2}+\frac{i}{5n\pi })-\frac{2}{n^2{\pi }^2}-e^{-in\frac{\pi }{10}}(\frac{20i}{100n\pi }-\frac{40i}{n^3{\pi }^3}+\frac{20}{5n^2{\pi }^2})-\frac{40i}{n^3{\pi }^3}\\ & =e^{-in\frac{\pi }{10}}[\frac{2}{n^2{\pi }^2}+\frac{i}{5n\pi }-\frac{20i}{100n\pi }+\frac{40i}{n^3{\pi }^3}-\frac{4}{n^2{\pi }^2}]-\frac{2}{n^2{\pi }^2}-\frac{40i}{n^3{\pi }^3}\\ & =e^{-in\frac{\pi }{10}}(\frac{40i}{n^3{\pi }^3}-\frac{2}{n^2{\pi }^2})-\frac{2}{n^2{\pi }^2}-\frac{40i}{n^3{\pi }^3}\end{array}$$

Now $$F_0=\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}}z(x)dx=\frac{1}{2}{\displaystyle \underset{0}{\overset{2/10}{\int }}}(x-5x^2)dx=\frac{1}{300}$$ Hence$$\begin{array}{rl}\tilde{z}(x)& =\frac{1}{2}{F}_0+\mathrm{Re}\left({\displaystyle \sum _{n=1}^N}{F}_n{e}^{in\frac{2\pi }{T}x}\right)\\ & =\frac{1}{600}+\mathrm{Re}\left({\displaystyle \sum _{n=1}^N}(e^{-in\frac{\pi }{10}}(\frac{40i}{n^3{\pi }^3}-\frac{2}{n^2{\pi }^2})-\frac{2}{n^2{\pi }^2}-\frac{40i}{n^3{\pi }^3})e^{in\frac{\pi }{2}x}\right)\\ & =\frac{1}{600}+\mathrm{Re}({\displaystyle \sum _{n=1}^N}{e}^{i(\frac{n\pi }{2}x-\frac{n\pi }{10})}(\frac{40i}{n^3{\pi }^3}-\frac{2}{n^2{\pi }^2})-e^{in\frac{\pi }{2}x}(\frac{2}{n^2{\pi }^2}+\frac{40i}{n^3{\pi }^3}))\\ & =\frac{1}{600}+\mathrm{Re}({\displaystyle \sum _{n=1}^N}\frac{-40}{n^3{\pi }^3}\frac{1}{i}{e}^{i(\frac{n\pi }{2}x-\frac{n\pi }{10})}-\frac{2}{n^2{\pi }^2}{e}^{i(\frac{n\pi }{2}x-\frac{n\pi }{10})}-\frac{2}{n^2{\pi }^2}{e}^{in\frac{\pi }{2}x}+\frac{40}{n^3{\pi }^3}\frac{1}{i}{e}^{in\frac{\pi }{2}x})\end{array}$$

But $x=vt$, hence$$\tilde{z}(t)=\frac{1}{600}+\mathrm{Re}({\displaystyle \sum _{n=1}^N}\frac{-40}{n^3{\pi }^3}\frac{1}{i}{e}^{i(\frac{n\pi v}{2}t-\frac{n\pi }{10})}-\frac{2}{n^2{\pi }^2}{e}^{i(\frac{n\pi v}{2}t-\frac{n\pi }{10})}-\frac{2}{n^2{\pi }^2}{e}^{in\frac{\pi v}{2}t}+\frac{40}{n^3{\pi }^3}\frac{1}{i}{e}^{in\frac{\pi v}{2}t})$$

Therefore the forcing frequency is $n{\varpi }_1=n\frac{\pi v}{2}$ or from $2\pi f_1=\frac{\pi v}{2}$, hence $f_1=\frac{v}{4}$Hz$.$The above can be written as$$\begin{array}{rl}\tilde{z}(t)& =\frac{1}{600}+{\displaystyle \sum _{n=1}^N}\mathrm{Re}\left(\frac{-40}{n^3{\pi }^3}\frac{1}{i}{e}^{i(\frac{n\pi v}{2}t-\frac{n\pi }{10})}\right)-{\displaystyle \sum _{n=1}^N}\mathrm{Re}\left(\frac{2}{n^2{\pi }^2}{e}^{i(\frac{n\pi v}{2}t-\frac{n\pi }{10})}\right)\\ & -{\displaystyle \sum _{n=1}^N}\mathrm{Re}\left(\frac{2}{n^2{\pi }^2}{e}^{in\frac{\pi v}{2}t}\right)+{\displaystyle \sum _{n=1}^N}\mathrm{Re}\left(\frac{40}{n^3{\pi }^3}\frac{1}{i}{e}^{in\frac{\pi v}{2}t}\right)\\ & \\ & =\frac{1}{600}+{\displaystyle \sum _{n=1}^N}\frac{-40}{n^3{\pi }^3}\sin (n{\varpi }_{1}t-\frac{n\pi }{10})-{\displaystyle \sum _{n=1}^N}\frac{2}{n^2{\pi }^2}\cos (n{\varpi }_{1}t-\frac{n\pi }{10})\\ & -{\displaystyle \sum _{n=1}^N}\frac{2}{n^2{\pi }^2}\cos \left(n{\varpi }_{1}t\right)+{\displaystyle \sum _{n=1}^N}\frac{40}{n^3{\pi }^3}\sin \left(n{\varpi }_{1}t\right)\\ & \\ & =\frac{1}{600}-\frac{40}{{\pi }^3}{\displaystyle \sum _{n=1}^N}\frac{1}{n^3}\sin (n{\varpi }_{1}t-\frac{n\pi }{10})-\frac{2}{{\pi }^2}{\displaystyle \sum _{n=1}^N}\frac{1}{n^2}\cos (n{\varpi }_{1}t-\frac{n\pi }{10})\\ & -\frac{2}{{\pi }^2}{\displaystyle \sum _{n=1}^N}\frac{1}{n^2}\cos \left(n{\varpi }_{1}t\right)+\frac{40}{{\pi }^3}{\displaystyle \sum _{n=1}^N}\frac{1}{n^3}\sin \left(n{\varpi }_{1}t\right)\end{array}$$

Where ${\varpi }_1=\frac{\pi v}{2}$

To verify the above, here is a plot for different number of fourier series terms showing that approximation improves as $N$ increases. This was done for $v=5m/s$ and for $5$ seconds.

2.7.3.1 Part(a)

The equation of motion is$$\begin{array}{rl}m{y}^{\prime \prime }+c(y^{\prime }-z^{\prime })+k(y-z)& =0\\ \text{(2.1)}& m{y}^{\prime \prime }+c{y}^{\prime }+ky& =c{z}^{\prime }+kz\end{array}$$

From earlier, we found that fourier series approximation to $z\left(t\right)$ is$$\begin{array}{rl}z(t)& =\frac{1}{600}+\mathrm{Re}({\displaystyle \sum _{n=1}^{\mathrm{\infty }}}\frac{-40}{n^3{\pi }^3}\frac{1}{i}{e}^{i(n\varpi t-\frac{n\pi }{10})}-\frac{2}{n^2{\pi }^2}{e}^{i(n\varpi t-\frac{n\pi }{10})}-\frac{2}{n^2{\pi }^2}{e}^{in\varpi t}+\frac{40}{n^3{\pi }^3}\frac{1}{i}{e}^{in\varpi t})\\ & =\frac{1}{600}+\mathrm{Re}({\displaystyle \sum _{n=1}^{\mathrm{\infty }}}\frac{-40}{n^3{\pi }^3}{e}^{-i\frac{n\pi }{10}}\frac{1}{i}{e}^{in\varpi t}-\frac{2}{n^2{\pi }^2}{e}^{-i\frac{n\pi }{10}}{e}^{in\varpi t}-\frac{2}{n^2{\pi }^2}{e}^{in\varpi t}+\frac{40}{n^3{\pi }^3}\frac{1}{i}{e}^{in\varpi t})\\ & =\frac{1}{600}+\mathrm{Re}\left({\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{e}^{in\varpi t}[\frac{-40}{n^3{\pi }^3}{e}^{-i(\frac{n\pi }{10}+\frac{\pi }{2})}-\frac{2}{n^2{\pi }^2}{e}^{-i\frac{n\pi }{10}}-\frac{2}{n^2{\pi }^2}+\frac{40}{n^3{\pi }^3}{e}^{-i\frac{\pi }{2}}]\right)\end{array}$$

Let $$Z_n=\frac{-40}{n^3{\pi }^3}{e}^{-i(\frac{n\pi }{10}+\frac{\pi }{2})}-\frac{2}{n^2{\pi }^2}{e}^{-i\frac{n\pi }{10}}-\frac{2}{n^2{\pi }^2}+\frac{40}{n^3{\pi }^3}{e}^{-i\frac{\pi }{2}}$$ Then above can be simplified to$$z(t)=\frac{1}{600}+\mathrm{Re}\left({\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{e}^{in\varpi t}{Z}_n\right)$$

Where $\varpi =\frac{\pi v}{2}$, hence$$z^{\prime }(t)=\mathrm{Re}\left({\displaystyle \sum _{n=1}^{\mathrm{\infty }}}in\varpi e^{in\varpi t}{Z}_n\right)$$

Hence, let $$y_{ss}(t)=\mathrm{Re}{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{Y}_n{e}^{in\varpi t}$$

Hence Eq 2.1 becomes$$\begin{array}{rl}{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}-m{n}^2{\varpi }^2{Y}_n{e}^{in\varpi t}+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}icn\varpi Y_n{e}^{in\varpi t}+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}k{Y}_n{e}^{in\varpi t}& ={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}icn\varpi e^{in\varpi t}{Z}_n+\frac{k}{600}+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}k{e}^{in\varpi t}{Z}_n\\ {\displaystyle \sum _{n=1}^{\mathrm{\infty }}}(-m{n}^2{\varpi }^2+icn\varpi +k)Y_n{e}^{in\varpi t}& ={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}(icn\varpi +k)Z_n{e}^{in\varpi t}+\frac{k}{600}\\ {\displaystyle \sum _{n=1}^{\mathrm{\infty }}}(-m{n}^2{\varpi }^2+icn\varpi +k)Y_n{e}^{in\varpi t}& =\frac{k}{600}+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}(icn\varpi +k)Z_n{e}^{in\varpi t}\end{array}$$

Hence$$Y_n=\frac{(icn\varpi +k)}{-m{\left(n\varpi \right)}^2+icn\varpi +k}{Z}_n$$

Let $$\begin{array}{rl}D(r_n,\zeta )& =\frac{icn\varpi +k}{-m{\left(n\varpi \right)}^2+icn\varpi +k}\\ & =\frac{i2\zeta m{\omega }_{nat}n\varpi +{\omega }_{nat}^{2}m}{-m{\left(n\varpi \right)}^2+i2\zeta m{\omega }_{nat}n\varpi +{\omega }_{nat}^{2}m}\\ & =\frac{i2\zeta n\frac{\varpi }{{\omega }_{nat}}+1}{-{\left(n\frac{\varpi }{{\omega }_{nat}}\right)}^2+i2\zeta n\frac{\varpi }{{\omega }_{nat}}+1}\\ & =\frac{1+i2\zeta r_n}{(1-r_n^2)+i2\zeta r_n}\end{array}$$

Where in the above $r_n=\frac{n\varpi }{{\omega }_{nat}}$ where $\varpi$ is $\frac{2\pi }{T}$ which means it is the fundamental frequency of the forcing function and ${\omega }_{nat}$ is the natural frequency.

Then Eq ?? becomes$$Y_n=D(r_n,\zeta )Z_n$$

And the steady state solution $y_{ss}(t)$ becomes$$y_{ss}(t)=\frac{k}{600}+\mathrm{Re}\left({\displaystyle \sum _{n=1}^{\mathrm{\infty }}}D(r_n,\zeta )Z_n{e}^{in\varpi t}\right)$$

Now we can answer the question. When $c=0$ then $D(r_n,\zeta )$ reduces to $\frac{k}{-m{\left(n\varpi \right)}^2+k}=\frac{1}{1-{\left(n\frac{\varpi }{{\omega }_{nat}}\right)}^2}=\frac{1}{1-r_n^2}$, hence$$y_{ss}(t)=\frac{k}{600}+\mathrm{Re}\left({\displaystyle \sum _{n=1}^{\mathrm{\infty }}}\frac{1}{1-r_n^2}{Z}_n{e}^{in\varpi t}\right)$$

So the displacement $y_{ss}(t)$ will be resonant when $r_n=1$ or $\frac{n\pi v}{2{\omega }_{nat}}=1$ or $v=\frac{2{\omega }_{nat}}{n\pi }$

Hence$$v=\frac{2\left(2\pi 5\right)}{n\pi }=\frac{20}{n}$$

Hence $v=20,10,5,2.5,1.25,\cdots$ meter/sec will each cause resonance. To verify, here is a plot of $y_{ss}(t)$ with no damper for speed near resonance $v=19.99$ and comparing this for speeds away from resonance speed. This plot shows that when speed $v$ is close to any of the above speeds, then the displacement $y_{ss}(t)$ becomes very large. Once the speed is away from those values, then $y_{ss}(t)$ quickly comes down to steady state $F/k$ value.

2.7.4 problem 3

The function is periodic with period $T=2\tau$$$f(t)=\begin{array}{lll}\frac{P}{\tau }t& & 0<t<\tau \\ 0& & \tau <t<2\tau \end{array}$$

and $f(t\pm T)=f(t)$. Let $\tilde{f}\left(t\right)$ be the Fourier series approximation to $f(t)$, hence$$\tilde{f}(t)=\frac{1}{2}{F}_0+\mathrm{Re}\left({\displaystyle \sum _{n=1}^N}{F}_n{e}^{in\frac{2\pi }{T}x}\right)$$

Where$$\begin{array}{rl}{F}_n& =\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}}f(t)e^{-in\frac{2\pi }{T}t}dt\\ & =\frac{2}{2\tau }{\displaystyle \underset{0}{\overset{\tau }{\int }}}\frac{P}{\tau }t{e}^{-in\frac{\pi }{\tau }t}dt\\ & =\frac{P}{{\tau }^2}{\displaystyle \underset{0}{\overset{\tau }{\int }}}t{e}^{-in\frac{\pi }{\tau }t}dt\end{array}$$

Using integration by parts $\int udv=uv-\int vdu$, letting $u=t$ and $dv=e^{-in\frac{\pi }{\tau }t}$ then $v=\int e^{-in\frac{\pi }{\tau }t}dt=\frac{i{e}^{-in\frac{\pi }{\tau }t}}{n\frac{\pi }{\tau }}$hence$$\begin{array}{rl}{F}_n& =\frac{P}{{\tau }^2}[{\left(t\frac{i{e}^{-in\frac{\pi }{\tau }t}}{n\frac{\pi }{\tau }}\right)}_0^{\tau }-\frac{i}{n\frac{\pi }{\tau }}{\displaystyle \underset{0}{\overset{\tau }{\int }}}{e}^{-in\frac{\pi }{\tau }t}dt]\\ & =\frac{P}{{\tau }^2}[\left(\tau \frac{i{e}^{-in\frac{\pi }{\tau }\tau }}{n\frac{\pi }{\tau }}\right)-\frac{i}{n\frac{\pi }{\tau }}{\left(\frac{e^{-in\frac{\pi }{\tau }t}}{-in\frac{\pi }{\tau }}\right)}_0^{\tau }]\\ & =\frac{P}{{\tau }^2}[\left({\tau }^2\frac{i{e}^{-in\pi }}{n\pi }\right)+\frac{{\tau }^2}{n^2{\pi }^2}{\left(e^{-in\frac{\pi }{\tau }t}\right)}_0^{\tau }]\\ & =\frac{P}{{\tau }^2}[\left({\tau }^2\frac{i{e}^{-in\pi }}{n\pi }\right)+\frac{{\tau }^2}{n^2{\pi }^2}(e^{-in\pi }-1)]\end{array}$$

$e^{-in\pi }=\cos \left(n\pi \right)={(-1)}^n$, hence$$F_n=\frac{P}{{\tau }^2}[\left({\tau }^2\frac{i{(-1)}^n}{n\pi }\right)+\frac{{\tau }^2}{n^2{\pi }^2}({(-1)}^n-1)]$$

Hence for even $n$$$\begin{array}{rl}{F}_n& =\frac{P}{{\tau }^2}\left[\left({\tau }^2\frac{i}{n\pi }\right)\right]\\ & =P\frac{i}{n\pi }\end{array}$$

and for odd $n$$$\begin{array}{rl}{F}_n& =\frac{P}{{\tau }^2}[(-{\tau }^2\frac{i}{n\pi })-2\frac{{\tau }^2}{n^2{\pi }^2}]\\ & =-\frac{P}{n\pi }(\frac{2}{n\pi }+i)\end{array}$$

$$\begin{array}{rl}{F}_0& =\frac{P}{{\tau }^2}{\displaystyle \underset{0}{\overset{\tau }{\int }}}tdt\\ & =\frac{P}{{\tau }^2}{\left(\frac{t^2}{2}\right)}_0^{\tau }=\frac{P}{{\tau }^2}\left(\frac{{\tau }^2}{2}\right)\\ & =\frac{P}{2}\end{array}$$

Now Eq ?? becomes$$\begin{array}{rl}\tilde{f}(t)& =\frac{1}{2}{F}_0+\mathrm{Re}\left({\displaystyle \sum _{n=1}^N}{F}_n{e}^{in\frac{2\pi }{T}x}\right)\\ & =\frac{1}{2}{F}_0+\mathrm{Re}({\displaystyle \sum _{evenn}}{F}_n{e}^{in\frac{2\pi }{T}t}+{\displaystyle \sum _{oddn}}{F}_n{e}^{in\frac{2\pi }{T}t})\\ & =\frac{p}{4}+\mathrm{Re}({\displaystyle \sum _{evenn}}P\frac{i}{n\pi }{e}^{in\frac{2\pi }{T}t}+{\displaystyle \sum _{oddn}}-\frac{P}{n\pi }(\frac{2}{n\pi }+i)e^{in\frac{2\pi }{T}t})\\ & =\frac{p}{4}+\mathrm{Re}(\frac{P}{\pi }{\displaystyle \sum _{evenn}}\frac{i}{n}{e}^{in\frac{2\pi }{T}t}-\frac{P}{\pi }{\displaystyle \sum _{oddn}}\frac{1}{n}(\frac{2}{n\pi }+i)e^{in\frac{2\pi }{T}t})\\ & =\frac{P}{4}+\mathrm{Re}(\frac{P}{\pi }{\displaystyle \sum _{evenn}}\frac{i}{n}{e}^{in\frac{2\pi }{T}t}-\frac{P}{\pi }{\displaystyle \sum _{oddn}}(\frac{2}{n^2\pi }+\frac{i}{n})e^{in\frac{2\pi }{T}t})\end{array}$$

To verify, here is a plot of the above, using $P=1$ and $\tau =0.5$ sec for $t=0\cdots 2$ seconds. This shows as more terms are added, the approximation becomes very close to the function. At $N=40$ the approximation appears very good.

Now we need to write $f(t)$ as sum of exponential to answer the question. $$\tilde{f}(t)=\frac{1}{2}{F}_0+\mathrm{Re}\left({\displaystyle \sum _{n=1}^N}{F}_n{e}^{in\frac{2\pi }{T}x}\right)$$

where $\varpi$ is the fundamental frequency of the force given by $\frac{2\pi }{T}=\frac{2\pi }{2\tau }=\frac{\pi }{\tau }$

Hence, let $y_{ss}={\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}{Y}_n{e}^{in\varpi t}$, then$$\begin{array}{rl}\mathrm{Re}(m{\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}-{\left(n\varpi \right)}^2{Y}_n{e}^{in\varpi t}+c{\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}in\varpi Y_n{e}^{in\varpi t}+k{\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}{Y}_n{e}^{in\varpi t})& =\frac{1}{2}{F}_0+\mathrm{Re}\left({\displaystyle \sum _{n=1}^N}{F}_n{e}^{in\frac{2\pi }{T}x}\right)\\ {\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}(-m{\left(n\varpi \right)}^2+icn\varpi +k)Y_n{e}^{in\varpi t}& =\frac{1}{2}{F}_0+\mathrm{Re}\left({\displaystyle \sum _{n=1}^N}{F}_n{e}^{in\frac{2\pi }{T}x}\right)\end{array}$$

Hence$$\begin{array}{rl}{Y}_n& =\frac{F_n}{k}\frac{1}{(1-{\left(n\frac{\varpi }{{\omega }_{nat}}\right)}^2)+i2\zeta n\frac{\varpi }{{\omega }_{nat}}}\\ & =\frac{F_n}{k}\frac{1}{(1-{\left(nr\right)}^2)+i2\zeta nr}\end{array}$$

Hence$$y_{ss}=\frac{1}{2}{F}_0+\mathrm{Re}\left({\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{Y}_n{e}^{in\varpi t}\right)$$

Finding $Y_n$ for $\tau =\frac{\pi }{3{\omega }_{nat}}$

where $r=\frac{\varpi }{{\omega }_{nat}}$. When $\zeta =0.04$ and $\tau =\frac{\pi }{3{\omega }_{nat}}$, hence now $r=\frac{2\pi }{\left(2\tau \right){\omega }_{nat}}=\frac{2\pi }{\left(2\frac{\pi }{3{\omega }_{nat}}\right){\omega }_{nat}}=3$, therefore$$\begin{array}{rl}{Y}_n& =\frac{F_n}{k}\frac{1}{(1-{\left(3n\right)}^2)+i6\left(0.04\right)n}\\ & =\frac{F_n}{k}\frac{1}{(1-9n^2)+i0.24n}\end{array}$$

The largest $Y_n$ will occur when the denominator of the above is smallest. Plotting the modulus of the denominator $\sqrt{{(1-9n^2)}^2+{\left(0.24n\right)}^2}$ for different $n$ values shows that $n=1$ is the values which makes it minimum.

This happens since for any $n>1$ the denominator will become larger due to $n^2$ and hence $Y_n$ will become smaller. So $n=1$ will be used.

For $n=1$, we obtain$$Y_1=\frac{F_1}{k}\frac{1}{(1-9)+i6\left(0.04\right)}$$

But $F_1=-\frac{P}{\pi }(\frac{2}{\pi }+i)$, hence$$\begin{array}{rl}{Y}_1& =\frac{-\frac{P}{\pi }(\frac{2}{\pi }+i)}{k}\frac{1}{(1-9)+i6\left(0.04\right)}=\frac{-P}{\pi k}\frac{(\frac{2}{\pi }+i)}{-8+i0.24}\\ & =\frac{P}{\pi k}\frac{\frac{2}{\pi }+i}{8-i0.24}=\frac{P}{\pi k}\frac{(\frac{2}{\pi }+i)(8+i0.24)}{(8-i0.24)(8+i0.24)}\\ & =\frac{P}{\pi k}(0.075759+0.12727i)\end{array}$$

Therefore$$Y_1=\frac{P}{k}(0.024115+0.0405i)$$

Here is a list of $Y_n$ for $n=1\cdots 10$ with the phase and magnitude of each (this was done for $\frac{p}{k}=1$)

From the above we see that most of the energy in the response will be contained in $Y_1$ and adding more terms will not have large effect on the response shape. This is confirmed by the plot that follows.

Plot for the steady state

Since $$y_{ss}=\frac{1}{2}{F}_0+\mathrm{Re}\left({\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{Y}_n{e}^{in\varpi t}\right)$$

Where now $r=\frac{\varpi }{{\omega }_{nat}}$. When $\zeta =0.04$ and $\tau =\frac{\pi }{3{\omega }_{nat}}$, hence now $r=\frac{2\pi }{\left(2\tau \right){\omega }_{nat}}=\frac{2\pi }{\left(2\frac{\pi }{3{\omega }_{nat}}\right){\omega }_{nat}}$ therefore $r=3$$$\begin{array}{rl}{y}_{ss}& =\frac{p}{4}+\mathrm{Re}({\displaystyle \sum _{n=1,3,5\cdots }^{\mathrm{\infty }}}{Y}_n{e}^{in\varpi t}+{\displaystyle \sum _{n=2,4,6\cdots }^{\mathrm{\infty }}}{Y}_n{e}^{in\varpi t})\\ & =\frac{p}{4}+\mathrm{Re}({\displaystyle \sum _{n=1,3,5\cdots }^{\mathrm{\infty }}}\frac{F_{n_{odd}}}{k}\frac{1}{(1-{\left(nr\right)}^2)+i2\zeta nr}{e}^{in\varpi t}+{\displaystyle \sum _{n=2,4,6\cdots }^{\mathrm{\infty }}}\frac{F_{n_{even}}}{k}\frac{1}{(1-{\left(nr\right)}^2)+i2\zeta nr}{e}^{in\varpi t})\\ & =\frac{p}{4}+\mathrm{Re}({\displaystyle \sum _{n=1,3,5\cdots }^{\mathrm{\infty }}}\frac{-\frac{P}{n\pi }(\frac{2}{n\pi }+i)}{k}\frac{1}{(1-{\left(nr\right)}^2)+i2\zeta nr}{e}^{in\varpi t}+{\displaystyle \sum _{n=2,4,6\cdots }^{\mathrm{\infty }}}\frac{P\frac{i}{n\pi }}{k}\frac{1}{(1-{\left(nr\right)}^2)+i2\zeta nr}{e}^{in\varpi t})\\ & =\frac{p}{4}+\frac{p}{k}\mathrm{Re}({\displaystyle \sum _{n=1,3,5\cdots }^{\mathrm{\infty }}}-\frac{\frac{1}{n\pi }(\frac{2}{n\pi }+i)}{(1-{\left(nr\right)}^2)+i2\zeta nr}{e}^{in\varpi t}+{\displaystyle \sum _{n=2,4,6\cdots }^{\mathrm{\infty }}}\frac{\frac{i}{n\pi }}{(1-{\left(nr\right)}^2)+i2\zeta nr}{e}^{in\varpi t})\end{array}$$

Now let $r=3$, $\zeta =0.04$. Normalizing the equation for $\varpi =1$ which implies $\tau =\pi$ and $k=1$ and $p=1$, then the above becomes$$y_{ss}=\frac{1}{4}+\mathrm{Re}({\displaystyle \sum _{n=1,3,5\cdots }^{\mathrm{\infty }}}-\frac{\frac{1}{n\pi }(\frac{2}{n\pi }+i)}{(1-{\left(3n\right)}^2)+i2\left(0.04\right)3n}{e}^{int}+{\displaystyle \sum _{n=2,4,6\cdots }^{\mathrm{\infty }}}\frac{\frac{i}{n\pi }}{(1-{\left(3n\right)}^2)+i2\left(0.04\right)3n}{e}^{int})$$

Here is a plot of the above for $t=0\cdots 20$ seconds for different values of $n$

We see from the above plot, that $y_{ss}(t)$ does not change too much as more terms are added, since when $r=3$, then $Y_n$ for $n=1$ contains most of the energy, hence adding more terms did not have an effect.

Repeating the calculations for $\tau =\frac{3\pi }{{\omega }_{nat}}$

$r=\frac{\varpi }{{\omega }_{nat}}$. When $\zeta =0.04$ and $\tau =\frac{3\pi }{{\omega }_{nat}}$, hence now $r=\frac{2\pi }{\left(2\tau \right){\omega }_{nat}}=\frac{2\pi }{\left(2\frac{3\pi }{{\omega }_{nat}}\right){\omega }_{nat}}=\frac{1}{3}$, therefore$$\begin{array}{rl}{Y}_n& =\frac{F_n}{k}\frac{1}{(1-{\left(nr\right)}^2)+i2\zeta nr}\\ & =\frac{F_n}{k}\frac{1}{(1-{\left(\frac{1}{3}n\right)}^2)+i\frac{2}{3}\left(0.04\right)n}\\ & =\frac{F_n}{k}\frac{1}{(1-\frac{n^2}{9})+i0.0267n}\end{array}$$

The largest $Y_n$ will occur when the denominator of the above is smallest. Similar to above, we can either find $n$ which minimizes the denominator (by taking derivative and setting it to zero and solve for $n$) or we can make a plot and see how the function behaves. Making a plot shows this

From the above we see that the smallest value of the denominator happens when $n=3$.

so using $n=3$ we find$$\begin{array}{rl}{Y}_3& =\frac{F_3}{k}\frac{1}{(1-{\left(3r\right)}^2)+i2\zeta 3r}\\ & =\frac{F_3}{k}\frac{1}{(1-{\left(3\frac{1}{3}\right)}^2)+i2\left(0.04\right)3\frac{1}{3}}\\ & =\frac{F_3}{k}\frac{1}{i0.08}\end{array}$$

But $F_n=-\frac{P}{n\pi }(\frac{2}{n\pi }+i)$, hence $$F_3=-\frac{P}{3\pi }(\frac{2}{3\pi }+i)$$

Therefore$$Y_3=\frac{-\frac{P}{3\pi }(\frac{2}{3\pi }+i)}{k}\frac{1}{i0.08}$$

Hence$$Y_3=\frac{p}{k}(-1.3263+0.28145i)$$

Here is a list of $Y_n$ for $n=1\cdots 10$ with the phase and magnitude of each (this was done for $\frac{p}{k}=1$)

We see from the above that $\left|Y_3\right|$ is the largest harmonic.

Plot for the steady state

Since $$y_{ss}=\frac{1}{2}{F}_0+\mathrm{Re}\left({\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{Y}_n{e}^{in\varpi t}\right)$$

Where now $r=\frac{\varpi }{{\omega }_{nat}}$. When $\zeta =0.04$ and $\tau =\frac{3\pi }{{\omega }_{nat}}$, hence now $r=\frac{2\pi }{\left(2\tau \right){\omega }_{nat}}=\frac{2\pi }{\left(2\frac{3\pi }{{\omega }_{nat}}\right){\omega }_{nat}}=\frac{1}{3}$,  therefore from above$$y_{ss}=\frac{p}{4}+\frac{p}{k}\mathrm{Re}({\displaystyle \sum _{n=1,3,5\cdots }^{\mathrm{\infty }}}-\frac{1}{n\pi }(\frac{2}{n\pi }+i)\frac{1}{(1-{\left(nr\right)}^2)+i2\zeta nr}{e}^{in\varpi t}+{\displaystyle \sum _{n=2,4,6\cdots }^{\mathrm{\infty }}}\frac{i}{n\pi }\frac{1}{(1-{\left(nr\right)}^2)+i2\zeta nr}{e}^{in\varpi t})$$

Now let $r=\frac{1}{3}$, $\zeta =0.04$, and assuming $\tau =0.5$ then $\varpi =\frac{2\pi }{2\tau }=\frac{\pi }{0.5},$ and assuming $k=1$, then the above becomes$$\begin{array}{rl}{y}_{ss}& =\frac{1}{4}+\frac{1}{k}\mathrm{Re}({\displaystyle \sum _{n=1,3,5\cdots }^{\mathrm{\infty }}}-\frac{1}{n\pi }(\frac{2}{n\pi }+i)\frac{1}{(1-{\left(n\frac{1}{3}\right)}^2)+i2\left(0.04\right)\frac{1}{3}n}{e}^{in\frac{\pi }{0.5}t})\\ & +\frac{1}{k}\mathrm{Re}\left({\displaystyle \sum _{n=2,4,6\cdots }^{\mathrm{\infty }}}\frac{i}{n\pi }\frac{1}{(1-{\left(n\frac{1}{3}\right)}^2)+i2\left(0.04\right)\frac{1}{3}n}{e}^{in\frac{\pi }{0.5}t}\right)\end{array}$$

Here is a plot of the above for $t=0\cdots 20$ seconds for different values of $n$

We see now that after $n=3$ that the response did not change much by adding more terms, this is because more of the energy are contained in the first 3 harmonics with $Y_n$ being the the largest.

2.7.5 Key solution for HW 6

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