2.2 HW1

2.2.1 problem description

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2.2.2 Problem 1 (1.1 book)

$k_3$ and $k_2$ are in parallel, hence the effective stiffness is $$k_{23}=k_2+k_3$$ $k_{23}$ and $k_1$ are now in series, hence the effective stiffness is  $$\frac{1}{k_{123}}=\frac{1}{k_1}+\frac{1}{k_{23}}=\frac{k_{23}+k_1}{k_1{k}_{23}}=\frac{k_2+k_3+k_1}{k_1(k_2+k_3)}=\frac{k_2+k_3+k_1}{k_1{k}_2+k_1{k}_3}$$ Therefore $$k_{123}=\frac{k_1{k}_2+k_1{k}_3}{k_2+k_3+k_1}$$ $k_{123}$ and $k_4$ are now in parallel, hence the effective stiffness is $$\begin{array}{rl}{k}_{1234}& =k_4+k_{123}\\ & =k_4+\frac{k_1{k}_2+k_1{k}_3}{k_2+k_3+k_1}\end{array}$$

Hence the final effective stiffness is$$k_{eq}=\frac{k_4(k_2+k_3+k_1)+k_1{k}_2+k_1{k}_3}{k_2+k_3+k_1}$$

2.2.3 Problem 2

We start by drawing a free body diagram and taking displacement of mass from the static equilibrium position. Let the displacement of the mass be $x$ and positive pointing upwards.

Let ${\mathrm{△}}_1$ be the downward deflection at right end of the bottom beam. Let ${\mathrm{△}}_2$ be the downward deflection at right end of top beam. The free body diagram is

Applying equilibrium of vertical forces $\sum F_v=0$ for mass $m$ and noting that inertial forces opposes motion, results in the equation of motion $$m{x}^{\prime \prime }+k(x-{\mathrm{△}}_1)=F$$ To find an expression for ${\mathrm{△}}_1$ in terms of $x,$ we apply equilibrium of vertical forces at the right end of the lower beam¹ $$k(x-{\mathrm{△}}_1)=k_b{\mathrm{△}}_1+k({\mathrm{△}}_1-{\mathrm{△}}_2)$$ Similarly, applying equilibrium of vertical forces at the right end of the top beam $$k({\mathrm{△}}_1-{\mathrm{△}}_2)=k_b{\mathrm{△}}_2$$ Solving for ${\mathrm{△}}_1,{\mathrm{△}}_2$ from Eqs ??,?? (2 equations, 2 unknowns) gives $${\mathrm{△}}_1=\frac{k(k+k_b)}{k^2+3k{k}_b+k_b^2}x$$ Substituting the above value into Eq ?? results in the equation of motion$$m{x}^{\prime \prime }+kx(1-\frac{k(k+k_b)}{k^2+3k{k}_b+k_b^2})=F$$

2.2.4 Problem 3

Assuming periodic motion, the period is $T=6$ ms, or $6\times {10}^{-3}$ sec. Hence $\omega =\frac{\pi }{3}$ rad/ms Representing this as a cosine signal with phase gives$$x(t)=A\cos (\omega t+\theta )$$ Then$$\begin{array}{rl}x(t)& =\mathrm{Re}[A+\cos (\omega t+\theta )]\\ & =\mathrm{Re}[A{e}^{i\theta }{e}^{i\omega t}]\\ & =\mathrm{Re}[\overline{A}{e}^{i\omega t}]\end{array}$$

Where now $\hat{A}=A{e}^{i\theta }$.Using phasor diagram

Hence from the diagram we see that for $x(t_0)$ to be zero when $t_0=1$ ms, we need to have $$\omega t_0+\theta =\frac{\pi }{2}$$ But $\omega =\frac{\pi }{3}$ rad/ms, hence $$\theta =\frac{\pi }{2}-\frac{\pi }{3}=\frac{\pi }{6}$$ To find $A$ we see that the maximum absolute value of $x(t)$ is 20 mm hence $A=20$ mm or $20\times {10}^{-3}$ meter. The equation of $x(t)$ when substituting all numerical values becomes

$$x(t)=20\cos (\frac{\pi }{3}t+\frac{\pi }{6})$$ Where units used are radians, milliseconds and mm. This is a plot of the above function

parms = f -> 1/(6 10^-3);
Plot[0.02 Cos[2 Pi f t + (Pi/6)] /. parms, {t,0,0.005},
 AxesLabel -> {t,x[t]}, ImageSize -> 300]

2.2.4.1 part(a)

At $t=0$, from ?? $x(0)=\mathrm{Re}[\hat{A}]=A\cos (\theta )=20\cos (\frac{\pi }{6})$ hence    $$x(0)=17.321\text{ mm}$$ From ?? $x^{\prime }(t)=\mathrm{Re}[\omega \hat{A}{e}^{i\omega t}]$ hence $x^{\prime }(0)=\mathrm{Re}[\omega \hat{A}]=\omega A\cos (\theta )=20\frac{\pi }{3}\cos (\frac{\pi }{6})$ giving $$x^{\prime }(0)=18.138\text{ m/sec}$$

2.2.4.2 part(b)

This can be solved using calculus² $$\begin{array}{rl}{x}^{\prime }(t)& =-2\pi fA\sin (2\pi ft+\theta )\\ 0& =-2\pi fA\sin (2\pi ft+\frac{\pi }{6})\\ & =-\left(\frac{2\pi }{6\times {10}^{-3}}\right)(20\times {10}^{-3})\sin (\frac{2\pi }{6\times {10}^{-3}}t+\frac{\pi }{6})\\ 0& =\sin (\frac{2\pi }{6\times {10}^{-3}}t+\frac{\pi }{6})\end{array}$$

We solve for $t$ and find $t$=2.5 ms. But this can be solved more easily by looking at the phasor diagram

The minimum $x(t)$ (in negative sense and not in absolute value sense) occurs when $\omega t_{min}+\theta =\pi$, hence $t_{min}=\frac{\pi -\theta }{\omega }$, therefore $$t_{min}=2.5$$

2.2.4.3 part(c)

This  is solved in a similar way by treating the speed as the rotating vector in complex plan. Since $x^{\prime }(t)=\mathrm{Re}\left(A\omega e^{i(\omega t+\theta +\frac{\pi }{2})}\right)$ then in complex plan as follows

The difference is that the velocity vector has phase of $\theta +\frac{\pi }{2}$ instead of $\theta$ as was the case with the position vector, and the amplitude is $A\omega$ instead of $A$. Hence the first time the speed vector will have the maximum value is when $$\theta +\frac{\pi }{2}+\omega t=2\pi$$ Hence $$\begin{array}{rl}t& =\frac{2\pi -\frac{\pi }{2}-\theta }{\omega }\\ & =\frac{2\pi -\frac{\pi }{2}-\frac{\pi }{6}}{\frac{\pi }{3}}\end{array}$$

Hence $t=4$ ms and the amplitude is given by $A\omega =20\frac{\pi }{3}$ hence $A\omega =20.944$ meter/sec

2.2.4.4 part(d)

Now treating the acceleration as the rotating vector in complex plan$$\begin{array}{rl}x(t)& =\mathrm{Re}\left(A{e}^{i(\theta +\omega t)}\right)\\ x^{\prime }(t)& =\mathrm{Re}\left(iA\omega e^{i(\theta +\omega t)}\right)\\ x^{\prime \prime }(t)& =\mathrm{Re}(-A{\omega }^2{e}^{i(\theta +\omega t)})\end{array}$$

But $-1=e^{i\pi }$ This adds a $\pi$ to the phase resulting in$$x^{\prime \prime }(t)=\mathrm{Re}\left(A{\omega }^2{e}^{i(\theta +\omega t+\pi )}\right)$$ Representing $x^{\prime \prime }(t)$ in complex plan gives

The first time the $x^{″}(t)$ vector will have the maximum value is when $$\theta +\pi +\omega t=2\pi$$ Hence $$\begin{array}{rl}t& =\frac{2\pi -\pi -\theta }{\omega }\\ & =\frac{\pi -\frac{\pi }{6}}{\frac{\pi }{6}}\end{array}$$

Hence $t=2.5\text{ms}$ and the amplitude is $$\begin{array}{rl}A{\omega }^2& =20\text{mm}{\left(\frac{\pi }{3}\text{rad/msec}\right)}^2\\ & =21.933\times {10}^3{\text{meter/sec}}^2\end{array}$$

2.2.5 Problem 4 (2.5 book)

2.2.5.1 part(a)

The function of the signal is converted to complex exponential. A $\sin$ or $\cos$ can be used to represent the signal as long as we are consistent. Assuming the signal is $x(t)=A\cos (\omega t+\theta )$, plotting the general representation of the position vector in complex plan gives

The complex representation of the position vector is $$x(t)=\mathrm{Re}\left[A{e}^{i(\omega t+\theta )}\right]$$ We are given that $\omega =\frac{2\pi }{T}=\frac{2\pi }{16}$, and since $x(t_0)$ has first zero at $t_0=5.5$ ms this means from looking at the above diagram that $$\theta +\omega t_0=\frac{\pi }{2}$$ Hence $\theta =\frac{\pi }{2}-(\omega t_0)=\frac{\pi }{2}-(\frac{\pi }{8}\frac{55}{10})$ which gives $$\theta =\frac{-3\pi }{16}\text{radians}$$ Hence the signal is $$\begin{array}{rl}x(t)& =\mathrm{Re}\left[A{e}^{i(\omega t+\theta )}\right]\\ & =\mathrm{Re}\left[1.2e^{i(\frac{\pi }{8}t-\frac{3\pi }{16})}\right]\\ & =\mathrm{Re}\left[1.2e^{-i\frac{3\pi }{16}}{e}^{i\frac{\pi }{8}t}\right]\\ & =\mathrm{Re}\left[\hat{A}{e}^{i\frac{\pi }{8}t}\right]\end{array}$$

Where $\hat{A}=1.2e^{-i\frac{3\pi }{16}}$ is the complex amplitude in polar coordinates. In rectangular coordinates it becomes $$\begin{array}{rl}\hat{A}& =1.2e^{-i\frac{3\pi }{16}}\\ & =1.2(\cos \left(\frac{3\pi }{16}\right)-i\sin \left(\frac{3\pi }{16}\right))\\ & =1.2(0.831-i0.5556)\\ & =\boxed{0.9977-i0.6667}\end{array}$$

Hence $$\begin{array}{rl}x(t)& =\mathrm{Re}\left[(0.998-i0.668)(\cos \frac{\pi }{8}t+i\sin \frac{\pi }{8}t)\right]\\ & =\mathrm{Re}[(0.998\cos \frac{\pi }{8}t+0.668\sin \frac{\pi }{8}t)+\\ & i(0.998\sin \frac{\pi }{8}t-0.668\cos \frac{\pi }{8}t)]\end{array}$$

Here is a plot of the signal for 20 ms

w = Pi/8;
f = 1.2 Cos[w t - 3 Pi/16];
Plot[f, {t, 0, 20}, AxesLabel -> {t, x[t]},
                    ImageSize -> 300,
                    GridLines -> Automatic,
     GridLinesStyle->{{Dashed,Gray},{Dashed,Gray}},
     PlotStyle -> Red]

2.2.5.2 part(b)

From above it was found that$$x(t)=\mathrm{Re}\left[A{e}^{i(\omega t+\theta )}\right]$$ Hence $$\begin{array}{rl}{x}^{\prime }(t)& =\mathrm{Re}\left[i\omega A{e}^{i(\omega t+\theta )}\right]\\ & =\mathrm{Re}\left[e^{i\frac{\pi }{2}}\omega A{e}^{i\theta }{e}^{i\omega t}\right]\\ & =\mathrm{Re}\left[\omega A{e}^{i(\frac{\pi }{2}+\theta )}{e}^{i\omega t}\right]\\ & =\mathrm{Re}\left[\hat{A}{e}^{i\omega t}\right]\end{array}$$

Where $\hat{A}=\omega A{e}^{i(\frac{\pi }{2}+\theta )}$ Replacing numerical values gives $\hat{A}=\frac{\pi }{8}\left(1.2\right)e^{i(\frac{\pi }{2}-\frac{3}{16}\pi )}=0.471e^{i0.983}$ and $$\begin{array}{rl}{x}^{\prime }(t)& =\mathrm{Re}\left[0.471e^{i0.983}{e}^{i\omega t}\right]\\ & =\mathrm{Re}\left[0.471e^{i0.983}{e}^{i\frac{\pi }{8}t}\right]\\ & =\mathrm{Re}\left[0.471e^{i0.983}{e}^{i0.3923t}\right]\end{array}$$

In rectangular coordinates, the above becomes

$$\begin{array}{rcl}{x}^{\prime }(t)& =& \mathrm{Re}[0.471(\cos 0.983+i\sin 0.983)\\ & & (\cos 0.3923t+i\sin 0.3923t)]\\ & =& \mathrm{Re}\left[(0.261+0.392i)(\cos 0.392t+i\sin 0.392t)\right]\\ & =& \mathrm{Re}[(0.261\cos 0.392t-0.392\sin 0.392t)\\ & & +i(0.261\sin 0.392t+0.392\cos 0.392t)]\end{array}$$

2.2.5.3 part(c)

To find the maximum rate of the signal$$x^{\prime }(t)=\mathrm{Re}\left[\hat{A}{e}^{i\omega t}\right]$$

Then the maximum $x^{\prime }(t)$ is $\left|\hat{A}\right|$ which is$$\begin{array}{rl}\left|\hat{A}\right|& =|0.261+0.392i|\\ & =\sqrt{{0.261}^2+{0.392}^2}\\ & =0.471\end{array}$$

Hence maximum $x^{\prime }(t)$ is $0.471$ v/ms or $471$ volt/sec.

Maximum velocity in simple harmonic motion occurs when $x(t)=0$. This occurs at $t=5.5$ ms and at $8$ ms henceforth. Hence maximum speed occurs at $$t=5.5+n(8)$$ for $n=0,1,2,\cdots$ this results in$$t=5.5,13,5,21.5,\cdots \text{ms}$$ Here is a plot of $x^{\prime }(t)$ in units of volt/ms

f = 0.261 Cos[0.392 t] - 0.392 Sin[0.392 t];
Plot[f, {t, 0, 30},
 AxesLabel -> {Row[{t, "(ms)"}], x'[t]},
 ImageSize -> 300, GridLines -> Automatic,
 GridLinesStyle -> {{Dashed, Gray}, {Dashed, Gray}},
 PlotStyle -> Red]

2.2.6 Problem 5 (2.8 book)

2.2.6.1 part(a)

This is a plot of the signal

f = 0.01 Sin[50 t] - 0.02 Cos[50 t - 0.3 Pi];
Plot[f, {t, 0, 0.2},
        AxesLabel -> {Row[{t , " (sec)"}], x[t]},
        ImageSize -> 300,
        GridLines -> Automatic,
        GridLinesStyle->{{Dashed,Gray},{Dashed,Gray}},
        PlotStyle -> Red]

$$\begin{array}{rl}q& =0.01\sin \left(50t\right)-0.02\cos (50t-0.3\pi )\\ & =\mathrm{Re}[\frac{0.01}{i}{e}^{i50t}-0.02e^{i(50t-0.3\pi )}]\\ & =\mathrm{Re}[0.01e^{-i\frac{\pi }{2}}{e}^{i50t}-0.02e^{i50t}{e}^{-i0.3\pi }]\\ & =\mathrm{Re}\left[(0.01e^{-i\frac{\pi }{2}}-0.02e^{-i0.3\pi })e^{i50t}\right]\\ & =\mathrm{Re}\left[\hat{A}{e}^{i50t}\right]\end{array}$$

Hence the complex amplitude is $$\hat{A}=0.01e^{-i\frac{\pi }{2}}-0.02e^{-i0.3\pi }$$

2.2.6.2 part(b)

From above, we see that $$\omega =50\text{ rad/sec}$$ Hence $f=\frac{50}{2\pi }$Hz, or the period $T=\frac{2\pi }{50}=0.126\sec$, therefore the time period separating the zeros is $\frac{0.126}{2}=0.063\sec$ or $63$ ms

2.2.6.3 part(c)

The complex phase $\hat{A}$ can be found by adding the vector $0.01e^{-i\frac{\pi }{2}}$ and $-0.02e^{-i\frac{3\pi }{10}}$ by completing the parallelogram as shown in this diagram. $\hat{A}=-0.02\cos 0.7\pi +i(-0.01+0.02\sin 0.7\pi )$, hence the angle $\alpha$ that $\hat{A}$ makes with the horizontal is$$\begin{array}{rl}{\tan }^{-1}\left(\frac{-0.01+0.02\sin 0.7\pi }{-0.02\cos 0.7\pi }\right)& =\arctan \left(0.526\right)\\ & =0.484\text{radian}\\ & =27.73\text{degree}\end{array}$$

and the amplitude is $$\sqrt{{(-0.01+0.02\sin 0.7\pi )}^2+{(0.02\cos 0.7\pi )}^2}=0.0133\text{V}$$ To find the earliest time $q$ will be zero, we need to find the time the complex position vector will take to rotate and reach the imaginary axis.

Hence we need to solve $$\begin{array}{rl}\pi -\alpha +\omega t_0& =\frac{3}{2}\pi \\ t_0& =\frac{\frac{3}{2}\pi -\pi +0.48402}{50}\\ & =0.0411\text{ s}\end{array}$$

Therefore$$t=41.1\text{ ms}$$

2.2.6.4 part(d)

The largest value of $q$ is the absolute value of its complex amplitude. We found this above as $$\left|\hat{A}\right|=0.0133\text{Volt}$$ To find when this occur first time, the time the position vector will align with the real axis in the positive direction is found. Hence solving for $t_0$ from $$\begin{array}{rl}\pi -\alpha +\omega t_0& =2\pi \\ t_0& =\frac{2\pi -\pi +0.484}{50}\end{array}$$

Gives $t=72.5\text{ms}$. Another way would be to take derivative of $qt)$ and set that to zero and solve for first $t$ which satisfy the equation.

2.2.7 Problem 6 (2.10 book)

$$\begin{array}{rl}{x}_1& =8\sin (10t-\frac{5}{6}\pi )\\ x_2& =12\cos (10t+\varphi )\end{array}$$

Let $\omega =10$, hence$$\begin{array}{rl}x& =x_1+x_2\\ & =\mathrm{Re}\left[\frac{8}{i}{e}^{i(\omega t-\frac{5}{6}\pi )}\right]+\mathrm{Re}\left[12e^{i(\omega t+\varphi )}\right]\\ & =\mathrm{Re}[\frac{8}{i}{e}^{i(\omega t-\frac{5}{6}\pi )}+12e^{i(\omega t+\varphi )}]\\ & =\mathrm{Re}[8e^{-i\frac{\pi }{2}}{e}^{i(\omega t-\frac{5}{6}\pi )}+12e^{i(\omega t+\varphi )}]\\ & =\mathrm{Re}[8e^{-i\frac{\pi }{2}}{e}^{i\omega t}{e}^{-i\frac{5}{6}\pi }+12e^{i\omega t}{e}^{i\varphi }]\\ & =\mathrm{Re}\left[(8e^{-i\left(\frac{4}{3}\pi \right)}+12e^{i\varphi })e^{i\omega t}\right]\\ & =\mathrm{Re}\left[\hat{A}{e}^{i\omega t}\right]\end{array}$$

Where $$\begin{array}{rl}\hat{A}& =8e^{-i\left(\frac{4}{3}\pi \right)}+12e^{i\varphi }\\ & =(-4+6.928i)+12(\cos \varphi +i\sin \varphi )\\ & =(-4+12\cos \varphi )+i(6.928+\sin \varphi )\end{array}$$

Hence Eq ?? becomes$$x=\mathrm{Re}\left[\{(-4+12\cos \varphi )+i(6.928+\sin \varphi )\}{e}^{i\omega t}\right]$$ To convert to $\sin$ we multiply and divide by $i$ hence$$\begin{array}{rl}x& =\mathrm{Re}\left[\{(-4+12\cos \varphi )+i(6.928+\sin \varphi )\}i\frac{e^{i\omega t}}{i}\right]\\ & =\mathrm{Re}\left[\{-(6.928+\sin \varphi )+i(-4+12\cos \varphi )\}\frac{e^{i\omega t}}{i}\right]\end{array}$$

The complex number $-(6.928+\sin \varphi )+i(-4+12\cos \varphi )$ can be written in polar form as $k{e}^{i\beta }$ where $K=\sqrt{{(6.928+\sin \varphi )}^2+{(-4+12\cos \varphi )}^2}$ and $\beta ={\tan }^{-1}\left(\frac{-4+12\cos \varphi }{-(6.928+\sin \varphi )}\right)$, hence Eq ?? becomes$$\begin{array}{rl}x& =\mathrm{Re}\left[k{e}^{i\beta }\frac{e^{i\omega t}}{i}\right]\\ & =\mathrm{Re}\left[k\frac{e^{i(\omega t+\beta )}}{i}\right]\\ & =k\sin (\omega t+\beta )\end{array}$$

or in full form $$\begin{array}{c}x=\sqrt{{(6.928+\sin \varphi )}^2+{(-4+12\cos \varphi )}^2}\hfill \\ \hfill \sin (\omega t+{\tan }^{-1}\left(\frac{-4+12\cos \varphi }{-(6.928+\sin \varphi )}\right))\end{array}$$ For pure sine function we need $\frac{-4+12\cos \varphi }{-(6.928+\sin \varphi )}=0$ or $12\cos \varphi =4$ or $\cos \varphi =\frac{1}{3}$, hence$$\begin{array}{rl}\varphi & =1.23096\text{ radian}\\ & ={70.529}^{\circ }\end{array}$$

The amplitude can also be found from the complex amplitude above when $\varphi =1.23096$ as follows$$\begin{array}{rl}|8e^{-i\left(\frac{4}{3}\pi \right)}+12e^{i1.23096}|& =|-6.592\times {10}^{-6}+18.242i|\\ & =\sqrt{{(-6.592\times {10}^{-6})}^2+{\left(18.242\right)}^2}\\ & =18.242\end{array}$$

2.2.8 Key solution for HW 1

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¹$k_b$ is beam stiffness against vertical displacement at the end and is given as $k_b=\frac{3EI}{L^3}$

²Taking derivative of $x(t)$ and setting the result to zero and solving for $t$