User:EGM4313.f14.Team1.Linehan/Report 2

=Report 2=

Problem Statement
Find the general solution of the following equation,


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$$ 9 y'' - 30 y' + 25 y = 0 $$
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find the characteristic equation and derive in detail that the first homogeneous solution is,


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$$ y_{h,1}(x) = e^{5x/3} $$
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Then follow in the method of variation of parameters to derive the 2nd homogeneous solution that is linearly independent from the 1st homogeneous solution, using the following trial solution for the 2nd homogeneous solution:


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$$ y_{h,2}(x) = u(x) y_{h,1}(x) = u(x) e^{5x/3} $$
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Solution

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$$9y''-30y'+25y=0$$
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The solution to a linear, second order ODE (one in the form ay"+by'+cy=0) is in the form y(x)=y1(x)+y2(x),

where y1(x) is in the form C1erx, and. In other words, y1(x) and y2(x) are linearly independent to each other.

y(x)=C1er1x+C2er2x,

where r1 and r2 are the solutions to the characteristic equation, which has the form ar2+br+c=0.


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$$y(x)=y_1(x)+y_2(x)$$
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$$y_1(x)=C_1 e^{r x}$$
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$$y_2(x)=C_2 e^{r x}$$
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Because a=9, b=-30, and c=25, the characteristic equation is


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$$9r^2-30r+25=0$$
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Solve the characteristic equation using the quadratic formula.


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$$r=\frac{30 \pm \sqrt{30^2-4(9)(725)}}{2(9)}$$
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The discriminant of the characteristic equation is 0, so this equation only has one root: 5/3.

In order to find the other solution, y2(x), we must do reduction of order.


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$$r=\frac{30 \pm \cancelto{0}\sqrt{900-900}}{18}$$
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$$r=\frac{5}{3}$$
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$$y_1(x)=C_1 e^{\frac{5x}{3}}$$
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Because y1(x) and y2(x) must be linearly independent, y2(x) is in the form u(x)y1(x),

where u(x) is any non-constant function of x. If u(x) was a constant,

then y1(x) and y2(x) would be multiples of each other and, therefore, not linearly independent.


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$$y_2(x)=u(x)y_1(x)=C_1 u(x) e^{\frac{5x}{3}}$$
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Take the first and second derivatives of y2(x) to be able to solve for u(x).


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$$y_2'(x)=C_1 u'(x) e^{\frac{5x}{3}} + \frac{5}{3} C_1 u(x) e^{\frac{5x}{3}}$$
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$$y_2(x)=C_1 u(x) e^{\frac{5x}{3}} + \frac{10}{3} C_1 u'(x) e^{\frac{5x}{3}} + \frac{25}{9} C_1 u(x) e^{\frac{5x}{3}}$$
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y1(x) and y2(x) also have the property of both being solutions to the differential equation.

Restate the differential equation with y2(x), y2'(x), and y2"(x) substituted in for y(x), y'(x), and y"(x).


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$$9y_2''(x)-30y_2'(x)+25y_2(x)=0$$
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Substitute the values for y2(x), y2'(x), and y2"(x) into the differential equation.


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$$9C_1 u''(x) e^{\frac{5x}{3}} + 30C_1 u'(x) e^{\frac{5x}{3}} + 25C_1 u(x) e^{\frac{5x}{3}} - 30C_1 u'(x) e^{\frac{5x}{3}} - 50C_1 u(x) e^{\frac{5x}{3}} + 25u(x)y_1(x)=C_1 u(x) e^{\frac{5x}{3}}=0$$
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Combine like terms the terms with u'(x) and u(x) cancel, leaving only the first term, with u"(x).


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$$9C_1 u''(x) e^{\frac{5x}{3}}=0$$
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Divide both sides of the equation by $$9C_1 e^{\frac{5x}{3}}$$ to isolate u"(x).


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$$u''(x)=0$$
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Integrate both sides of the equation to get u'(x) equaling a new constant, denoted as C2.


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$$\int \! u''(x)\,dx = \int \! 0\,dx$$
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$$u'(x)=C_2$$
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Integrate both sides of the equation again to get u(x) equaling C2 plus a third constant, denoted as C3. We have now reached u(x).


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$$\int \! u'(x)\,dx = \int \! C_2\,dx$$
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$$u(x)=C_2 x + C_3$$
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Substitute u(x) into the definition of y(x). Distribute the elements of u(x) to get three terms, two exponential with different arbitrary constant,

and one exponential multiplied by a third arbitrary constant and x.


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$$y(x)=C_1 e^{\frac{5x}{3}} + C_2 x e^{\frac{5x}{3}} + C_3 e^{\frac{5x}{3}}$$
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Because C1 and C3 are arbitrary in the solution, they can be combined into a new constant, denoted as C4.

This is the solution of the given differential equation.


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$$y(x)=C_4 e^{\frac{5x}{3}} + C_2 x e^{\frac{5x}{3}}, C_4=C_1+C_3$$
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Problem Statement
Problem 5: What are the frequencies of vibration of a body of mass $$ m = 5kg $$ (i) on a spring of modulus $$ k_1 = 20 \frac{nt}{m} $$, (ii) on a spring of modulus $$ k_2 = 45 \frac{nt}{m} $$ (iii) on the two springs in parallel?

Problem 7: Find frequency of oscillation of a pendulum of length L, neglecting air resistance and the weight of the rod, and assuming $$ \theta $$ to be small that $$ sin\theta = \theta $$.

Solution

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$$ \omega_o = \sqrt{\frac{k}{m}} $$
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$$ f = \frac{\omega_o}{2\pi} $$ Solve for $$ \omega_o $$.
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$$ \omega_o = 2\pi f $$ Substitute $$ \omega_o $$ into first equation.
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$$ 2\pi f = \sqrt{\frac{k}{m}} $$
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$$ f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} $$     (1)
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Problem 5(i)

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$$k_1 = 20 \frac{nt}{m}$$
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From eq. 1


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$$ f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} $$
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Plug in values given in the initial problem. (Repeat for steps (i)-(iii)).
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$$ f = \frac{1}{2\pi}\sqrt{\frac{20 \frac{nt}{m}}{5 kg}} $$
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$$ f = \frac{1}{2\pi}\sqrt{4} $$
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$$ f = 0.3183 Hz $$
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Problem 5(ii)
$$k_2 = 45\frac{nt}{m}$$

from eq. 1


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$$ f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} $$
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$$ f = \frac{1}{2\pi}\sqrt{\frac{45 \frac{nt}{m}}{5 kg}} $$
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$$ f = \frac{1}{2\pi}\sqrt{9} $$
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$$ f = 0.47746 Hz $$
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Problem 5(iii)
Because the springs are in parallel the spring constants are added together. Therefore rewrite eq. 1:


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$$ f = \frac{1}{2\pi}\sqrt{\frac{k_1 + k_2}{m}} $$
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$$ f = \frac{1}{2\pi}\sqrt{\frac{20 \frac{nt}{m} + 45 \frac{nt}{m}}{5 kg}} $$
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$$ f = \frac{1}{2\pi}\sqrt{13} $$
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$$ f = 0.57384 Hz $$
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Problem 7



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$$ mL\theta^' = -mg\sin \theta $$ As stated in the problem $$ \sin \theta = \theta $$, so the $$ \sin $$ is discarded.
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$$ mL\theta^' = -mg\theta $$ Tangential Component of $$ W = mg $$
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$$ \theta^' + \omega_o^2 \theta = 0 $$
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$$ \frac{\omega_o}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{g}{L}} $$
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$$ f = \frac{\omega_o}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{g}{L}} $$
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$$ f = \frac{1}{2\pi}\sqrt{\frac{g}{L}} $$
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Problem Statement

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$$ \overline{(z_1+z_2)} = \overline z_1 + \overline z_2 $$
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$$ \overline{(z_1-z_2)} = \overline z_1 - \overline z_2 $$
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$$ \overline{(z_1z_2)} = \overline z_1\overline z_2 $$
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$$ \overline{\frac{z_1}{z_2}} = \frac{\overline z_1}{\overline z_2} $$     (1)
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Problem 4: Verify (1) for $$ z_1 = -11 + 10i, z_2 = -1 + 4i $$

Let $$ z_1 = -2 + 11i, z_2 = 2 - i $$ showing details of your work, find, in the form $$ x+iy:$$

Problem 8: $$ z_1z_2, \overline{(z_1z_2)} $$

Problem 4

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$$z_1=-11+10i, z_2=-1+4i$$
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For complex equations, the overbar means the complex conjugate, which switches the + sign to a -, and vice versa.


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$$\overline{a+bi}=a-bi$$
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Proof 1

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$$\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}$$ (2)
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Substitute z1 and z2 into (2).


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$$\overline{(-11+10i)+(-1+4i)} = \overline{-11+10i} + \overline{-1+4i}$$
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Combine like terms on the left side of the equation. Apply the overbars to the complex numbers on the right side of the equation.


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$$\overline{-12+14i} = (-11-10i) + (-1-4i)$$
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Combine like terms on the right side of the equation. Apply the overbar to the complex number on the left side of the equation. The two sides of the equation are equal, proving (2).


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$$-12-14i = -12-14i \checkmark$$
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Proof 2

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$$\overline{z_1-z_2}=\overline{z_1}-\overline{z_2}$$ (3)
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Substitute z1 and z2 into (3).


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$$\overline{(-11+10i)-(-1+4i)} = \overline{-11+10i} - \overline{-1+4i}$$
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Combine like terms on the left side of the equation. Apply the overbars to the complex numbers on the right side of the equation.


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$$\overline{-10+6i} = (-11-10i) - (-1-4i)$$
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Combine like terms on the right side of the equation. Apply the overbar to the complex number on the left side of the equation. The two sides of the equation are equal, proving (3).


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$$-10-6i = -10-6i \checkmark$$
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Proof 3

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$$\overline{z_1 z_2}=(\overline{z_1}) (\overline{z_2})$$ (4)
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Substitute z1 and z2 into (4).


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$$\overline{(-11+10i)(-1+4i)}=(\overline{-11+10i}) (\overline{-1+4i})$$
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Use the FOIL method on the left side of the equation. Apply the overbars to the complex numbers on the right side of the equation.


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$$\overline{11-44i-10i+40i^2}=(-11-10i)(-1-4i)$$
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Combine like terms on the left side of the equation. Knowing that $$i^2=-1$$, replace the $$i^2$$ on the left side of the equation with -1. Use the FOIL method on the right side of the equation.


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$$\overline{11-54i-40}=11+44i+10i+40i^2$$
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Combine like terms on the left and right sides of the equation.Knowing that $$i^2=-1$$, replace the $$i^2$$ on the right side of the equation with -1.


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$$\overline{-29-54i}=11+54i-40$$
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Combine like terms on the right side of the equation. Apply the overbar to the left side of the equation. The two sides of the equation are equal, proving (4).


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$$-29+54i=-29+54i \checkmark$$
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Proof 4

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$$\overline{(\frac{z_1}{z_2})} = \frac{\overline {z_1}}{\overline {z_2}}$$ (5)
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Substitute z1 and z2 into (5).


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$$\overline{(\frac{-11+10i}{-1+4i})} = \frac{\overline{-11+10i}}{\overline {-1+4i}}$$
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To divide complex numbers, multiply the fraction on the left side of the equation by the conjugate of its denominator.

Apply the overbars to the numerator and denominator of the fraction on the right side of the equation.


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$$\overline{(\frac{-11+10i}{-1+4i}\frac{-1-4i}{-1-4i})} = \frac{-11-10i}{-1-4i}$$
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As before, multiply the fraction on the right side of the equation by the conjugate of its denominator.

Use the FOIL method on the numerators and denominators on the left side of the equation to get just one fraction on that side.


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$$\overline{(\frac{11+44i-10i-40i^2}{1+4i-4i-16i^2})} = \frac{-11-10i}{-1-4i} \frac{-1+4i}{-1+4i}$$
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Combine like terms on the left side of the equation. Knowing that $$i^2=-1$$, replace the $$i^2$$ in both complex numbers on the

left side of the equation with -1. Use the FOIL method on the numerators and denominators on the right side of the equation to get just one fraction on that side.


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$$\overline{(\frac{11+34i+40}{1+16})} = \frac{11-44i+10i-40i^2}{1-4i+4i-16i^2}$$
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Combine like terms on either side of the equation. Knowing that $$i^2=-1$$, replace the $$i^2$$ in both complex numbers on the right side of the equation with -1.


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$$\overline{(\frac{51+34i}{17})} = \frac{11-34i+40}{1+16}$$
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Combine like terms on the right side of the equation. Simplify the fraction on the left side of the equation.


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$$\overline{3+2i} = \frac{54-34i}{17}$$
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Simplify the fraction on the right side of the equation. Apply the overbar to the complex number on the left side of the equation. The two sides of the equation are equal, proving (5).


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$$3-2i = 3-2i \checkmark $$
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Problem 8

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$$z_1=-2+11i, z_2=2-i$$
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$$z_1 z_2=?$$ (6)
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$$\overline{z_1 z_2}=?$$ (7)
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Substitute z1 and z2 into (6).


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$$z_1 z_2 = (-2+11i)(2-i)$$
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Use the FOIL method on the right side of the equation.


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$$z_1 z_2 = -4+2i+22i-11i^2$$
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Combine like terms on the right side of the equation. Knowing that $$i^2=-1$$, replace the $$i^2$$ on the right side of the equation with -1.


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$$z_1 z_2 = -4+24i+11$$
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Combine like terms on the right side of the equation. The right side cannot be simplified any further, so the answer to (6) has been reached.


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$$z_1 z_2 = 7+24i$$
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Substitute the answer gotten for (6), $$7+24i$$, into (7).


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$$\overline{z_1 z_2}=\overline{7+24i}$$
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Apply the overbar to the right side of the equation. The right side cannot be further simplified, so the answer to (7) has been reached.


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$$\overline{z_1 z_2}=7-24i$$
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Problem Statement

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$$ \mathbf A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \ \mathbf B = \begin{bmatrix} 5 & 7 \\ 6 & 8 \end{bmatrix} , \ \mathbf C = \begin{bmatrix} 1 & 3 \\ 5 & 6 \end{bmatrix} $$
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verify numerically the following identities:


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$$ ( \mathbf A \mathbf B )^T = \mathbf B^T \mathbf A^T $$     (1)
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$$ ( \mathbf A \mathbf B \mathbf C )^T = \mathbf C^T \mathbf B^T \mathbf A^T $$     (2)
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Solution
Prove


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$$ ( \mathbf A \mathbf B )^T = \mathbf B^T \mathbf A^T $$ First Solve for
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$$ ( \mathbf A \mathbf B )^T
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$$
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Multiply Matrix A and Matrix B
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$$ \mathbf A \mathbf B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 7 \\ 6 & 8 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf A \mathbf B = \begin{bmatrix} (1*5)+ (2*6) & (1*7) + (2*8) \\ (3*5) + (4*6) & (3*7) + (4*8) \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf A \mathbf B = \begin{bmatrix} 17 & 23 \\ 39 & 53 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Transpose Product of Matrix A and B multiplication


 * {| style="width:100%" border="0"

$$ ( \mathbf A \mathbf B )^T = \begin{bmatrix} 17 & 39 \\ 23 & 53 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$ Second solve for
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf B^T \mathbf A^T
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Transpose Matrix of Matrix A


 * {| style="width:100%" border="0"

$$ \mathbf A^T = \begin{bmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Create transpose matrix with values from Matrix A


 * {| style="width:100%" border="0"

$$ \mathbf A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Transpose Matrix of Matrix B


 * {| style="width:100%" border="0"

$$ \mathbf B^T = \begin{bmatrix} b_{11} & b_{21} \\ b_{12} & b_{22} \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$ Create transpose matrix with values from Matrix B
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf B^T = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Multiply Transpose Matrix B and Transpose Matrix A


 * {| style="width:100%" border="0"

$$ \mathbf B^T \mathbf A^T = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf B^T \mathbf A^T = \begin{bmatrix} (5*1) + (6*2) & (5*3) + (6*4) \\ (7*1) + (8*2) & (7*3) + (8*4) \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf B^T \mathbf A^T = \begin{bmatrix} 17 & 39 \\ 23 & 53 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Set solutions equal to each other

 * {| style="width:100%" border="0"

$$ ( \mathbf A \mathbf B )^T = \begin{bmatrix} 17 & 39 \\ 23 & 53 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf B^T \mathbf A^T = \begin{bmatrix} 17 & 39 \\ 23 & 53 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 17 & 39 \\ 23 & 53 \end{bmatrix} = \begin{bmatrix} 17 & 39 \\ 23 & 53 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Thus,


 * {| style="width:100%" border="0"

$$ ( \mathbf A \mathbf B )^T = \mathbf B^T \mathbf A^T \checkmark
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Proof 2
Prove


 * {| style="width:100%" border="0"

$$ ( \mathbf A \mathbf B \mathbf C )^T = \mathbf C^T \mathbf B^T \mathbf A^T
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

First solve for


 * {| style="width:100%" border="0"

$$ ( \mathbf A \mathbf B \mathbf C )^T
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Using Properties of Matrix Multiplication


 * {| style="width:100%" border="0"

$$ ( \mathbf A \mathbf B \mathbf C ) = ( \mathbf A \mathbf B)(\mathbf C)
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Perform Matrix Multiplication


 * {| style="width:100%" border="0"

$$ \mathbf A \mathbf B = \mathbf M_1' = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 7 \\ 6 & 8 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf M_1' = \begin{bmatrix} (1*5) + (2*6) & (1*7) + (2*8) \\ (3*5) + (4*6) & (3*7) + (4*8) \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf M_1' = \begin{bmatrix} 17 & 23 \\ 39 & 53 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Multiply Product Matrix M' and Matrix C


 * {| style="width:100%" border="0"

$$ \mathbf M_1'\mathbf C = \begin{bmatrix} 17 & 23 \\ 39 & 53 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 5 & 6 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf M_1'\mathbf C = \mathbf M_1'' = \begin{bmatrix} (17*1) + (23*5)  & (17*3) + (23*6) \\ (39*1) + (53*5) & (39*3) + (53*6) \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf M_1'' = \begin{bmatrix} (17*1) + (23*5) & (17*3) + (23*6) \\ (39*1) + (53*5) & (39*3) + (53*6) \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf M_1'' = \begin{bmatrix} 132 & 189 \\ 304 & 435 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Create Transpose Matrix of M


 * {| style="width:100%" border="0"

$$ \mathbf M_1''^ = \mathbf M_1'\mathbf C = ( \mathbf A \mathbf B \mathbf C ) = \begin{bmatrix} 132  & 189 \\ 304 & 435 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ ( \mathbf A \mathbf B \mathbf C )^T = \begin{bmatrix} 132  & 304 \\ 189 & 435 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Second solve for


 * {| style="width:100%" border="0"

$$ \mathbf C^T \mathbf B^T \mathbf A^T
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf C^T = \begin{bmatrix} 1 & 5 \\ 3 & 6 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf B^T = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Perform Matrix Multiplication with CT and BT


 * {| style="width:100%" border="0"

$$ \mathbf C^T \mathbf B^T = \mathbf M_2' = \begin{bmatrix} 1 & 5 \\ 3 & 6 \end{bmatrix} \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf M_2' = \begin{bmatrix} (1*5) + (5*7) & (1*6) + (5*8) \\ (3*5) + (6*7) & (3*6) + (6*8) \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf M_2' = \begin{bmatrix} 40 & 46 \\ 57 & 66 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Multiply M'2 and Matrix AT
 * {| style="width:100%" border="0"

$$ \mathbf M_2' \mathbf A^T = \mathbf M_2'' = \begin{bmatrix} 40 & 46 \\ 57 & 66 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf M_2'' = \begin{bmatrix} (40*1) + (46*2) & (40*4) + (46*4) \\ (57*1) + (66*2) & (57*3) + (66*4) \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf M_2'' = \begin{bmatrix} 132 & 304 \\ 189 & 435 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

And recall that
 * {| style="width:100%" border="0"

$$ \mathbf M_1'' = \begin{bmatrix} 132 & 304 \\ 189 & 435 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Since
 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 132 & 304 \\ 189 & 435 \end{bmatrix} = \begin{bmatrix} 132  & 304 \\ 189 & 435 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Then
 * {| style="width:100%" border="0"

$$ \mathbf M_1 = \mathbf M_2
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

And so


 * {| style="width:100%" border="0"

$$ ( \mathbf A \mathbf B \mathbf C )^T = \mathbf C^T \mathbf B^T \mathbf A^T \checkmark $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Problem Statement
Consider the stiffness matrix of the 2-DOF system:



Spring Constants
 * {| style="width:100%" border="0"

$$ k_1 = k_2 = k_3 = 5 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Mass Coefficients
 * {| style="width:100%" border="0"

$$ m_1 = m_2 = 1.3 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Dampening Coefficients
 * {| style="width:100%" border="0"

$$ c_1 = c_2 = c_3 = 2 $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Solve the standard eigenvalue problem of the stiffness matrix to find :

(1) the eigenvalues of this stiffness matrix;

(2) the eigenvectors of this stiffness matrix;

(3) verify the orthogonality of the eigenvectors.

Solution
Element 1




 * {| Italic textstyle="width:100%" border="0"

$$ f_A^{(1)} = k_1 ( -d_B + d_A ) + c_1 ( -d'_B + d'_A ) = 5 ( -d_B + d_A ) + 2 ( -d'_B + d'_A ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_B^{(1)} = k_1 ( d_B - d_A ) + c_1 ( d'_B - d'_A ) = 5 ( d_B - d_A ) + 2 ( d'_B - d'_A ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Rearrange Terms in Equations to Prepare to Change to Matrix Form


 * {| style="width:100%" border="0"

$$ f_A^{(1)} = k_1 ( d_A - d_B ) + c_1 ( d'_A - d'_B ) = 5 ( d_A - d_B ) + 2 ( d'_A - d'_B ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_B^{(1)} = k_1 ( -d_A + d_B ) + c_1 ( -d'_A + d'_B ) = 5 ( -d_A + d_B ) + 2 ( -d'_A + d'_B ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Equations in Matrix Form


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} f_A^{(1)} \\ f_B^{(1)} \end{Bmatrix} = \begin{bmatrix} k_1 & - k_1 \\ -k_1 &  k_1 \end{bmatrix} \begin{Bmatrix} d_A \\ d_B \end{Bmatrix} + \begin{bmatrix} c_1 & - c_1 \\ -c_1 & c_1 \end{bmatrix} \begin{Bmatrix} d'_A \\ d'_B \end{Bmatrix} = \begin{bmatrix} 5 & - 5 \\ -5 &  5 \end{bmatrix} \begin{Bmatrix} d_A \\ d_B \end{Bmatrix} + \begin{bmatrix} 2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_A \\ d'_B \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |


 * }
 * }

Element 2 Start with FBD:



Force matrix follows same topology as in element 1.


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} f_B^{(2)} \\ f_C^{(2)} \end{Bmatrix} = \begin{bmatrix}   k_2 & - k_2 \\ -k_2 &   k_2 \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} + \begin{bmatrix}  c_2 & - c_2 \\ -c_2 & c_2 \end{bmatrix} \begin{Bmatrix} d'_B \\ d'_C \end{Bmatrix} = \begin{bmatrix}   5 & - 5 \\ -5 &   5 \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} + \begin{bmatrix}  2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_B \\ d'_C \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Element 3

Start with FBD:



Force matrix follows same topology as in elements 1 and 2.


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} f_C^{(3)} \\ f_D^{(3)} \end{Bmatrix} = \begin{bmatrix}   k_3 & - k_3 \\ -k_3 &   k_3 \end{bmatrix} \begin{Bmatrix} d_C \\ d_D \end{Bmatrix} + \begin{bmatrix}  c_3 & - c_3 \\ -c_3 & c_3 \end{bmatrix} \begin{Bmatrix} d'_C \\ d'_D \end{Bmatrix} = \begin{bmatrix}   5 & - 5 \\ -5 &   5 \end{bmatrix} \begin{Bmatrix} d_C \\ d_D \end{Bmatrix} + \begin{bmatrix}  2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_C \\ d'_D \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Mass B



$$ \sum_{} f = F_B(t)- m_Bd''-f_B^{(1)}-f_B^{(2)} = 0 $$


 * {| style="width:100%" border="0"

$$ f_B^{(1)} = \begin{bmatrix} -k_1d_A + k_1d_B \end{bmatrix} + \begin{bmatrix} -c_1d'_A + c_1d'_B \end{bmatrix} = \begin{bmatrix} -5d_A + 5d_B \end{bmatrix} + \begin{bmatrix} -2d'_A + 2d'_B \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_B^{(2)} = \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} = \begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Sub in For Forces


 * {| style="width:100%" border="0"

$$ F_B(t) = m_Bd''_B + (\begin{bmatrix} -k_1d_A + k_1d_B \end{bmatrix} + \begin{bmatrix} -c_1d'_A + c_1d'_B \end{bmatrix}) + (\begin{bmatrix} k_2d_B - k_2d_C \end{bmatrix} + \begin{bmatrix} c_2d'_B - c_2d'_C \end{bmatrix}) $$  (1)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$ = 1.3d''_B + (\begin{bmatrix} -5d_A + 5d_B \end{bmatrix} + \begin{bmatrix} -2d'_A + 2d'_B \end{bmatrix}) + (\begin{bmatrix} 5d_B - 5d_C \end{bmatrix} + \begin{bmatrix} 2d'_B - 2d'_C \end{bmatrix}) $$  (1)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Mass C



$$ \sum_{} f = 0 = F_C(t)- m_Cd''_C- f_C^{(2)} - f_C^{(3)} $$


 * {| style="width:100%" border="0"

$$ f_C^{(2)} = \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} = \begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_C^{(3)} = \begin{bmatrix} k_3d_C - k_3d_D \end{bmatrix} + \begin{bmatrix} c_3d'_C - c_3d'_D \end{bmatrix} = \begin{bmatrix} 5d_C - 5d_D \end{bmatrix} + \begin{bmatrix} 2d'_C - 2d'_D \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Sub in For Forces


 * {| style="width:100%" border="0"

$$ F_C(t) = m_Cd''_C + (\begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix}) + (\begin{bmatrix} k_3d_C - k_3d_D \end{bmatrix} + \begin{bmatrix} c_3d'_C - c_3d'_D \end{bmatrix}) $$ (2)
 * style="width:95%" |
 * style="width:95%" |
 * 


 * }
 * {| style="width:100%" border="0"

$$ = 1.3d''_C + (\begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix}) + (\begin{bmatrix} 5d_C - 5d_D \end{bmatrix} + \begin{bmatrix} 2d'_C - 2d'_D \end{bmatrix}) $$ (2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">


 * }

Boundary Conditions for Both Equations of Motion


 * {| style="width:100%" border="0"

$$ d_A(t) = d'_A(t) = 0 $$
 * style="width:95%" |
 * style="width:95%" |


 * }
 * }


 * {| style="width:100%" border="0"

$$ d_D(t) = d'_D(t) = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Equations of Motion After Substituting in Boundary Conditions


 * {| style="width:100%" border="0"

$$ F_B(t) = m_Bd_B + k_1d_B + c_1d'_b + \begin{bmatrix} k_2d_B - k_2d_C \end{bmatrix} + \begin{bmatrix} c_2d'_b - c_2d'_C \end{bmatrix} = 1.3d_B + 5d_B + 2d'_b + \begin{bmatrix} 5d_B - 5d_C \end{bmatrix} + \begin{bmatrix} 2d'_b - 2d'_C \end{bmatrix} $$   (3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$ F_C(t) = m_Cd_C + \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} + k_3d_C + c_3d'_C = 1.3d_C + \begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} + 5d_C + 2d'_C $$   (4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Into Matrix Form


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} F_B(t) \\ F_C(t) \end{Bmatrix} = \begin{bmatrix}   m_B & 0 \\ 0 &  m_C \end{bmatrix} \begin{bmatrix} d_B \\ d_C \end{bmatrix} + \begin{bmatrix}  c_1 + c_2 & - c_2 \\ -c_2 & c_2 + c_3 \end{bmatrix} \begin{bmatrix} d'_B \\ d'_C \end{bmatrix} + \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 + k_3 \end{bmatrix} \begin{bmatrix} d_b \\ d_c \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Sub In Values Below into Equation of Motion
 * {| style="width:100%" border="0"

$$ k_1 = k_2 = k_3 = 5 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ m_1 = m_2 = 1.3 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ c_1 = c_2 = c_3 = 2 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ m_B = m_1 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ m_C = m_2 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Final Equation of Motion with All Values Substituted In


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} F_B(t) \\ F_C(t) \end{Bmatrix} = \begin{bmatrix}   1.3 & 0 \\ 0 &  1.3 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} + \begin{bmatrix}  4 & -2 \\ -2 & 4 \end{bmatrix} \begin{bmatrix} d'_1 \\ d'_2 \end{bmatrix} + \begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Must now start with undamped free vibration equation:


 * {| style="width:100%" border="0"

$$ \mathbf M \mathbf d'' + \mathbf K \mathbf d = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Standard eigenvalue problem dictates that the mass matrix is equal to the identity matrix, thus the undamped free vibration equation becomes,


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{Bmatrix} d_1(t) \\ d_2(t) \end{Bmatrix} +
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} \begin{Bmatrix} d_1(t) \\ d_2(t) \end{Bmatrix} =

\begin{Bmatrix} 0 \\ 0 \end{Bmatrix} $$
 * }
 * }

Now we must assume a trial solution taking the form,


 * {| style="width:100%" border="0"

$$ \mathbf d (t) = e^{\alpha t} \boldsymbol \phi $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

taking the second derivative of the trial solution recovers,


 * {| style="width:100%" border="0"

$$ \mathbf d'' (t) = \alpha^2 e^{\alpha t} \boldsymbol \phi $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Substituting this back into the undamped free vibration equation yields,


 * {| style="width:100%" border="0"

$$ \mathbf M ( \alpha^2 e^{\alpha t} \boldsymbol \phi ) + \mathbf K (e^{\alpha t} \boldsymbol \phi) = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

combining like terms,


 * {| style="width:100%" border="0"

$$ [ \alpha^2 \mathbf M + \mathbf K] e^{\alpha t} \boldsymbol \phi = \mathbf 0 $$ Noting that $$ e^{\alpha t} $$ is never zero thus permits
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ [ \mathbf M \alpha^2 + \mathbf K ] \boldsymbol \phi = \mathbf 0 \Rightarrow \mathbf K \boldsymbol \phi = - \alpha^2 \mathbf M \boldsymbol \phi $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

where we then restrict $$ \alpha $$ to be imaginary and let,


 * {| style="width:100%" border="0"

$$ \beta := -\alpha^2 > 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

the standard eigenvalue problem is now written as:


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} \begin{Bmatrix} \phi_1 \\ \phi_2 \end{Bmatrix} =
 * style="width:95%" |
 * style="width:95%" |

\beta \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{Bmatrix} \phi_1 \\ \phi_2 \end{Bmatrix} $$
 * }
 * }

which can be rewritten in the form


 * {| style="width:100%" border="0"

$$ [ \mathbf K - \beta \mathbf M ] \boldsymbol \phi = \mathbf 0 $$ We then know to obtain non-zero solutions we must set the determinant of the bracketed portion to zero.
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \det \begin{bmatrix} 10 - \beta & -5 \\ -5 & 10 - \beta \end{bmatrix} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

thus yields the following quadratic


 * {| style="width:100%" border="0"

$$ \beta^2 -20\beta + 75 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

exposing two roots or eigenvalues as


 * {| style="width:100%" border="0"

$$ \beta_1 = 5 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \beta_2 = 15 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

with the eigenvalues known we can now solve the standard eigenvalue equation


 * {| style="width:100%" border="0"

$$ \mathbf K \boldsymbol \phi = \beta \mathbf M \boldsymbol \phi $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

with the values inputted with indices i= 1,2 indicating the two eigen pairs,


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} \begin{Bmatrix} \phi_{i1} \\ \phi_{i2} \end{Bmatrix} =
 * style="width:95%" |
 * style="width:95%" |

\beta_i \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{Bmatrix} \phi_{i1} \\ \phi_{i2} \end{Bmatrix}, \ i=1,2 $$
 * }
 * }

starting with the 1st pair we know that $$ \beta = 5, \phi_1 = \begin{Bmatrix} \phi_{11} \\ \phi_{12} \end{Bmatrix} $$ thus,


 * {| style="width:100%" border="0"

$$ ( \mathbf K - \beta_1 \mathbf M ) \boldsymbol \phi_1 = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 10 - 5 & -5 \\ -5 & 10 - 5\end{bmatrix} \begin{Bmatrix} \phi_{11} \\
 * style="width:95%" |
 * style="width:95%" |

\phi_{12} \end{Bmatrix} = \begin{bmatrix} 5 & -5 \\ -5 & 5 \end{bmatrix} \begin{Bmatrix} \phi_{11} \\

\phi_{12} \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} $$
 * }
 * }

We set first component of first eigenvector $$ \phi_{11} = 1 $$ then we find $$  \phi_{12} $$ from other equation and thus $$  \phi_{12}=-1 $$ and the first eigenvector is


 * {| style="width:100%" border="0"

$$ \boldsymbol \phi_1 = \begin{Bmatrix} 1 \\ -1 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

repeating the same process for the 2nd eigen pair where $$ \beta_2 = 15, \ \boldsymbol \phi_2 = \begin{Bmatrix} \phi_{21} \\ \phi_{22} \end{Bmatrix} $$ results in the second eigenvector.


 * {| style="width:100%" border="0"

$$ \boldsymbol \phi_2 = \begin{Bmatrix} 1 \\ 1 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

To verify orthogonality we can write


 * {| style="width:100%" border="0"

$$ \boldsymbol \phi_1^T \mathbf M \boldsymbol \phi_2 = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

thus


 * {| style="width:100%" border="0"

$$ \lfloor 1 -1 \rfloor \begin{bmatrix} 1 & 0 \\ 0 & -1
 * style="width:95%" |
 * style="width:95%" |

\end{bmatrix} \begin{Bmatrix} 1 \\ 1 \end{Bmatrix} = \lfloor 1 -1 \rfloor \begin{Bmatrix} 1 \\

1 \end{Bmatrix} = 0 $$
 * }
 * }

Problem Statement
Solve the general eigenvalue problem of the stiffness matrix for the same system as Pb 2.5 to find :

(1) the eigenvalues of this stiffness matrix;

(2) the eigenvectors of this stiffness matrix;

(3) verify the orthogonality of the eigenvectors with respect to the mass matrix;

(4) verify whether the damping matrix is diagonalizable;

(4) find the normalized eigenvectors of the model;

(5) create an animation of normalized eigenvectors in desmos compare to the animation of the normalized eigenvectors in example on Pb 53-36

Solution
Eigenvalues

Referring back to the work from R1.6, we found that


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |


 * {| Italic textstyle="width:100%" border="0"

$$ \mathbf M \mathbf d'' + \mathbf C \mathbf d' + \mathbf K \mathbf d = F(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| Italic textstyle="width:100%" border="0"

$$ \mathbf M = \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| Italic textstyle="width:100%" border="0"

$$ \mathbf C = \begin{bmatrix} 4 & -2 \\ -2 & 4 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| Italic textstyle="width:100%" border="0"

$$ \mathbf K = \begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| Italic textstyle="width:100%" border="0"

$$ \mathbf F = \begin{Bmatrix} f_1(t) \\ f_2(t) \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| Italic textstyle="width:100%" border="0"

$$ \mathbf d = \begin{Bmatrix} d_1 \\ d_2 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * }
 * }

The combined equation


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \end{Bmatrix} + \begin{bmatrix} 4 & -2 \\ -2 & 4 \end{bmatrix} \begin{Bmatrix} d'_1 \\ d'_2 \end{Bmatrix} + \begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \end{Bmatrix} = \begin{Bmatrix} f_1(t) \\ f_2(t) \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Generalized Eigenvalue Problem (GEP)


 * {| style="width:100%" border="0"

$$ \mathbf K \phi = \beta \mathbf M \phi $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ [ \mathbf K - \beta \mathbf M ] \phi = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

To have zero solutions for phi would be trivial. Therefore, we must have:


 * {| style="width:100%" border="0"

$$ \det [ \mathbf K - \beta \mathbf M ] = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Evaluate what we are finding the determinant of.


 * {| style="width:100%" border="0"

$$ \mathbf K - \beta \mathbf M = \begin{bmatrix} K_{11} & K_{12} \\ K_21 & K_22 \end{bmatrix} - \beta \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix} = \begin{bmatrix} K_11 - m_1 \beta & K_12 \\ K_21 & K_22 - m_2 \beta \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plugging in the given values gives the following equation.


 * {| style="width:100%" border="0"

$$ \mathbf K - \beta \mathbf M = \begin{bmatrix} 10 - 1.3 \beta & -5 \\ -5 & 10 - 1.3 \beta \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The determinant of this can then be calculated.


 * {| style="width:100%" border="0"

$$ \det \begin{bmatrix} 10 - \beta 1.3 & -5 \\ -5 & 10 - \beta 1.3 \end{bmatrix} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The following quadratic is produced.


 * {| style="width:100%" border="0"

$$ 1.69 \beta^2 -26\beta + 75 = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Two roots or eigenvalues are found from the quadratic.


 * {| style="width:100%" border="0"

$$ \beta_1 = \frac{50}{13} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \beta_2 = \frac{150}{13} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Eigenvectors

From the GEP and the two eigenvaules we just found,


 * {| style="width:100%" border="0"

$$ \mathbf K \phi = \beta \mathbf M \phi $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} \begin{Bmatrix} \phi_{i1} \\ \phi_{i2} \end{Bmatrix} = \beta_i \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} \phi_{i1} \\ \phi_{i2} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

where i = 1, 2

When i = 1 and when setting the first coefficient in each eigenvector equal to one as advised in the problem


 * {| style="width:100%" border="0"

$$ ( \mathbf K - \beta \mathbf M ) \phi_1 = \begin{bmatrix} 10 - 1.3 \beta_1 & -5 \\ -5 & 10 - 1.3 \beta_1 \end{bmatrix} \begin{Bmatrix} 1 \\ \phi_{12} \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now $$ \phi_{12} $$ can be solved for to get


 * {| style="width:100%" border="0"

$$ \phi_{12} = -1 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Therefore, the first eigenvector is


 * {| style="width:100%" border="0"

$$ \mathbf \phi_1 = \begin{Bmatrix} 1 \\ -1 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The second eigenvector is found the same way as the first, except $$ \beta_2 $$ is used. This yields


 * {| style="width:100%" border="0"

$$ \mathbf \phi_2 = \begin{Bmatrix} 1 \\ 1 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Normalized Eigenvectors

Modal mass of mode 1:


 * {| style="width:100%" border="0"

$$ \mathbf \phi_1^{T} \mathbf M \mathbf \phi_1 = \begin{Bmatrix} 1 & -1 \end{Bmatrix} \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} 1 \\ -1 \end{Bmatrix} = 2.6 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

1st normalized eigenvector:


 * {| style="width:100%" border="0"

$$ \bar{\mathbf \phi_1} = \frac{\mathbf \phi_1}{\sqrt{\bar{m_1}}} = \frac{1}{\sqrt{2.6}} \begin{Bmatrix} 1 \\ -1 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Doing the same process to find the 2nd normalized eigenvector yields,


 * {| style="width:100%" border="0"

$$ \bar{\mathbf \phi_2} = \frac{1}{\sqrt{2.6}} \begin{Bmatrix} 1 \\ 1 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Orthogonality

Use the dot product to find if the eigenvectors are orthogonal to eachother.


 * {| style="width:100%" border="0"

$$ \mathbf \phi_1 \cdot \mathbf \phi_2 = \begin{Bmatrix} 1 \\ -1 \end{Bmatrix} \cdot \begin{Bmatrix} 1 \\ 1 \end{Bmatrix} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Check if the eigenvectors are orthogonal to the mass matrix.


 * {| style="width:100%" border="0"

$$ \mathbf \phi_1^{T} \mathbf M \mathbf \phi_2 = \begin{Bmatrix} 1 & -1 \end{Bmatrix} \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} 1 \\ 1 \end{Bmatrix} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Therefore, the eigenvectors are orthogonal to the mass matrix.

Verify whether the Damping Matrix is Diagonalizable

The damping matrix, as found in R1.6, is


 * {| style="width:100%" border="0"

\mathbf C = \begin{bmatrix} 4 & -2 \\ -2 & 4 \end{bmatrix} $$
 * }
 * }

The transpose of this matrix is as follows.


 * {| style="width:100%" border="0"

\mathbf C^T = \begin{bmatrix} 4 & -2 \\ -2 & 4 \end{bmatrix} $$
 * }
 * }

It can be seen that the damping matrix and it's transpose are equivalent.

Now it must be checked that $$ \mathbf C \mathbf C^T = \mathbf C^T \mathbf C $$.

Since $$ \mathbf C = \mathbf C^T $$ the multiplication of these two matrices in either order will yield the same result.


 * {| style="width:100%" border="0"

\begin{bmatrix} 4 & -2 \\ -2 & 4 \end{bmatrix} \begin{bmatrix} 4 & -2 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 12 & -16 \\ -16 & 12 \end{bmatrix} $$
 * }
 * }

Therefore, the damping matrix is diagonalizable.

Animations

video Mode 1 link: https://www.desmos.com/calculator/5r8hvjup0h

video Mode 2 link: https://www.desmos.com/calculator/0osgcwje7g

Problem Statement
Derive the matrix equation of motion for a 2-DOF system with additional elements. similar to Pb 2.6, with the same numerical values of the parameters, but for the following 2-DOF system:



in this system, there is an additional elastic bar in parallel with element 2 (or BC), which consists of a spring and a damper in parallel.

the numerical values of the additional parameters E,A,L (Young’s modulus, area of cross-section, length) are:


 * {| style="width:100%" border="0"

$$ E = 2, \ A = \frac{1}{2} , \ L = 5 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Element 1
Begin with a free body diagram for element 1.



Using simple statics, write the equations of equilibrium and solve for $$f_A^{(1)} $$and $$f_B^{(1)}$$ (a subscript $$ _A$$ denotes the node and a superscript in parenthesis $$^{(1)}$$ denotes the element)


 * {| Italic textstyle="width:100%" border="0"

$$ f_A^{(1)} = k_1 ( -d_B + d_A ) + c_1 ( -d'_B + d'_A ) = 5 ( -d_B + d_A ) + 2 ( -d'_B + d'_A ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_B^{(1)} = k_1 ( d_B - d_A ) + c_1 ( d'_B - d'_A ) = 5 ( d_B - d_A ) + 2 ( d'_B - d'_A ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Rearrange the terms in the equation in preparation to write it in matrix form:


 * {| style="width:100%" border="0"

$$ f_A^{(1)} = k_1 ( d_A - d_B ) + c_1 ( d'_A - d'_B ) = 5 ( d_A - d_B ) + 2 ( d'_A - d'_B ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_B^{(1)} = k_1 ( -d_A + d_B ) + c_1 ( -d'_A + d'_B ) = 5 ( -d_A + d_B ) + 2 ( -d'_A + d'_B ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Here is the force matrix for element 1:


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} f_A^{(1)} \\ f_B^{(1)} \end{Bmatrix} = \begin{bmatrix} k_1 & - k_1 \\ -k_1 &  k_1 \end{bmatrix} \begin{Bmatrix} d_A \\ d_B \end{Bmatrix} + \begin{bmatrix} c_1 & - c_1 \\ -c_1 & c_1 \end{bmatrix} \begin{Bmatrix} d'_A \\ d'_B \end{Bmatrix} = \begin{bmatrix} 5 & - 5 \\ -5 &  5 \end{bmatrix} \begin{Bmatrix} d_A \\ d_B \end{Bmatrix} + \begin{bmatrix} 2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_A \\ d'_B \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |


 * }
 * }

Element 2
Begin with a Free Body Diagram for Element 2:



Using simple statics, write the equations of equilibrium and solve for $$f_B^{(2)} $$and $$f_C^{(2)}$$

(a subscript $$ _A$$ denotes the node and a superscript in parenthesis $$^{(1)}$$ denotes the element).

This is very similar to the analysis done for Element 1, except that there is one more component to account for the elastic rod in parallel.


 * {| style="width:100%" border="0"

$$ f_B^{(2)} = k_2 ( d_B - d_C ) + c_2 ( d'_B - d'_C ) + \frac{EA}{L}( d_B - d_C ) = 5 ( d_B - d_C ) + 2 ( d'_B - d'_C ) + \frac{1}{5}( d_B - d_C ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_C^{(2)} = k_2 ( -d_B + d_C ) + c_2 ( -d'_B + d'_C )+ \frac{EA}{L}( -d_B + d_C ) = 5 ( -d_B + d_C ) + 2 ( -d'_B + d'_C ) + \frac{1}{5}( -d_B + d_C ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Writing in matrix form:


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} f_B^{(2)} \\ f_C^{(2)} \end{Bmatrix} = \begin{bmatrix}   k_2 & - k_2 \\ -k_2 &   k_2 \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} + \begin{bmatrix}  c_2 & - c_2 \\ -c_2 & c_2 \end{bmatrix} \begin{Bmatrix} d'_B \\ d'_C \end{Bmatrix} + \begin{bmatrix}   \frac{EA}{L} & -\frac{EA}{L} \\ -\frac{EA}{L} &  \frac{EA}{L} \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} = \begin{bmatrix}  5 & - 5 \\ -5 &   5 \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} + \begin{bmatrix}  2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_B \\ d'_C \end{Bmatrix} + \begin{bmatrix}  \frac{1}{5} & -\frac{1}{5} \\ -\frac{1}{5} &  \frac{1}{5} \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Element 3
Begin with a Free Body Diagram for Element 3:



Follow the same exact steps from analysis of Element 1. The resulting matrix is almost exactly the same.


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} f_C^{(3)} \\ f_D^{(3)} \end{Bmatrix} = \begin{bmatrix}   k_3 & - k_3 \\ -k_3 &   k_3 \end{bmatrix} \begin{Bmatrix} d_C \\ d_D \end{Bmatrix} + \begin{bmatrix}  c_3 & - c_3 \\ -c_3 & c_3 \end{bmatrix} \begin{Bmatrix} d'_C \\ d'_D \end{Bmatrix} = \begin{bmatrix}   5 & - 5 \\ -5 &   5 \end{bmatrix} \begin{Bmatrix} d_C \\ d_D \end{Bmatrix} + \begin{bmatrix}  2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_C \\ d'_D \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Mass B
Begin with a Free Body Diagram for Mass B:



Determine the forces acting on mass B:

$$ \sum_{} f = F_B(t)- m_Bd''-f_B^{(1)}-f_B^{(2)} = 0 $$


 * {| style="width:100%" border="0"

$$ f_B^{(1)} = \begin{bmatrix} -k_1d_A + k_1d_B \end{bmatrix} + \begin{bmatrix} -c_1d'_A + c_1d'_B \end{bmatrix} = \begin{bmatrix} -5d_A + 5d_B \end{bmatrix} + \begin{bmatrix} -2d'_A + 2d'_B \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_B^{(2)} = \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} + \begin{bmatrix} -\frac{EA}{L}d_B + \frac{EA}{L}d_C \end{bmatrix} = \begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} + \begin{bmatrix} -0.2d_B + 0.2d_C \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Sum the forces acting on mass B and plug in the constants into the equation of equilibrium:


 * {| style="width:100%" border="0"

$$ F_B(t) = m_Bd''_B + (\begin{bmatrix} -k_1d_A + k_1d_B \end{bmatrix} + \begin{bmatrix} -c_1d'_A + c_1d'_B \end{bmatrix}) + (\begin{bmatrix} k_2d_B - k_2d_C \end{bmatrix} + \begin{bmatrix} c_2d'_B - c_2d'_C \end{bmatrix} + \begin{bmatrix} \frac{EA}{L}d_B - \frac{EA}{L}d_C \end{bmatrix}) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ = 1.3d''_B + (\begin{bmatrix} -5d_A + 5d_B \end{bmatrix} + \begin{bmatrix} -2d'_A + 2d'_B \end{bmatrix}) + (\begin{bmatrix} 5d_B - 5d_C \end{bmatrix} + \begin{bmatrix} 2d'_B - 2d'_C \end{bmatrix} + \begin{bmatrix} 0.2d_B - 0.2d_C \end{bmatrix}) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Mass C
Begin with a Free Body Diagram for Mass C:



Determine the forces acting on mass C:

$$ \sum_{} f = 0 = F_C(t)- m_Cd''_C- f_C^{(2)} - f_C^{(3)} $$


 * {| style="width:100%" border="0"

$$ f_C^{(2)} = \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} + \begin{bmatrix} -\frac{EA}{L}d_B + \frac{EA}{L}d_C \end{bmatrix}= \begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} + \begin{bmatrix} -\frac{1}{5}d_B + \frac{1}{5}d_C \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_C^{(3)} = \begin{bmatrix} k_3d_C - k_3d_D \end{bmatrix} + \begin{bmatrix} c_3d'_C - c_3d'_D \end{bmatrix} = \begin{bmatrix} 5d_C - 5d_D \end{bmatrix} + \begin{bmatrix} 2d'_C - 2d'_D \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Sum the forces acting on mass C and plug in the constants into the equation of equilibrium:


 * {| style="width:100%" border="0"

$$ F_C(t) = m_Cd''_C + (\begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} + \begin{bmatrix} -\frac{EA}{L}d_B + \frac{EA}{L}d_C \end{bmatrix}) + (\begin{bmatrix} k_3d_C - k_3d_D \end{bmatrix} + \begin{bmatrix} c_3d'_C - c_3d'_D \end{bmatrix}) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">


 * }
 * {| style="width:100%" border="0"

$$ = 1.3d''_C + (\begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} + \begin{bmatrix} -\frac{1}{5}d_B + \frac{1}{5}d_C \end{bmatrix}) + (\begin{bmatrix} 5d_C - 5d_D \end{bmatrix} + \begin{bmatrix} 2d'_C - 2d'_D \end{bmatrix}) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">


 * }

Apply the boundary conditions to both equations of motion (this will eliminate some terms):


 * {| style="width:100%" border="0"

$$ d_A(t) = d'_A(t) = 0 $$
 * style="width:95%" |
 * style="width:95%" |


 * }
 * }


 * {| style="width:100%" border="0"

$$ d_D(t) = d'_D(t) = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Equations of motion after substituting in boundary conditions:


 * {| style="width:100%" border="0"

$$ F_B(t) = m_Bd_B + k_1d_B + c_1d'_b + \begin{bmatrix} k_2d_B - k_2d_C \end{bmatrix} + \begin{bmatrix} c_2d'_B - c_2d'_C \end{bmatrix} + \begin{bmatrix} \frac{EA}{L}d_B - \frac{EA}{L}d_C \end{bmatrix} = 1.3d_B + 5d_B + 2d'_b + \begin{bmatrix} 5d_B - 5d_C \end{bmatrix} + \begin{bmatrix} 2d'_B - 2d'_C \end{bmatrix} + \begin{bmatrix} \frac{1}{5}d_B - \frac{1}{5}d_C \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$ F_C(t) = m_Cd_C + \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} + k_3d_C + c_3d'_C + \begin{bmatrix} -\frac{EA}{L}d_B + \frac{EA}{L}d_C \end{bmatrix}= 1.3d_C + \begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} + 5d_C + 2d'_C + \begin{bmatrix} -\frac{1}{5}d_B + \frac{1}{5}d_C \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now, assembling these final two equations into matrix form:


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} F_B(t) \\ F_C(t) \end{Bmatrix} = \begin{bmatrix}   m_B & 0 \\ 0 &  m_C \end{bmatrix} \begin{bmatrix} d_B \\ d_C \end{bmatrix} + \begin{bmatrix}  c_1 + c_2 & - c_2 \\ -c_2 & c_2 + c_3 \end{bmatrix} \begin{bmatrix} d'_B \\ d'_C \end{bmatrix} + \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 + k_3 \end{bmatrix} \begin{bmatrix} d_B \\ d_C \end{bmatrix} + \begin{bmatrix} \frac{EA}{L} & -\frac{EA}{L} \\ -\frac{EA}{L} & \frac{EA}{L} \end{bmatrix} \begin{bmatrix} d_B \\ d_C \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Substitute the values in the matrix:


 * {| style="width:100%" border="0"

$$ k_1 = k_2 = k_3 = 5 $$
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 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ m_1 = m_2 = 1.3 $$
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 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ c_1 = c_2 = c_3 = 2 $$
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 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ m_B = m_1 $$
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 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ m_C = m_2 $$
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 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ E = 2 $$
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 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ A = 0.5 $$
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 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ L = 5 $$ Final equation of motion with all values substituted in:
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} F_B(t) \\ F_C(t) \end{Bmatrix} = \begin{bmatrix}   1.3 & 0 \\ 0 &  1.3 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} + \begin{bmatrix}  4 & -2 \\ -2 & 4 \end{bmatrix} \begin{bmatrix} d'_1 \\ d'_2 \end{bmatrix} + \begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} + \begin{bmatrix} 0.2 & -0.2 \\ -0.2 & 0.2 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Problem Statement
Derive the matrix equation of motion for a 2-DOF system with additional elements. similar to Pb 2.6, with the same numerical values of the parameters, but for the following 2-DOF system:



in this system, two elastic bars are put in series first, then this pair of elastic bars in series is then put in parallel with element 3 (or CD), which consists of a spring and a damper in parallel.

the numerical values of the additional parameters E,A,L (Young’s modulus, area of cross-section, length) are:


 * {| style="width:100%" border="0"

$$ E_1 = 2, \ A_1 = 1 / 2 , \ L_1 = 5 $$
 * style="width:95%" |
 * style="width:95%" |
 * }


 * {| style="width:100%" border="0"

$$ E_2 = 2, \ A_2 = 1 / 2 , \ L_2 = 4 $$
 * style="width:95%" |
 * style="width:95%" |


 * }

Element 1
Begin with a drawing of the free body diagram for element 1:



Using simple statics, write the equations of equilibrium and solve for $$f_A^{(1)} $$and $$f_B^{(1)}$$ (a subscript $$ _A$$ denotes the node and a superscript in parenthesis $$^{(1)}$$ denotes the element)


 * {| Italic textstyle="width:100%" border="0"

$$ f_A^{(1)} = k_1 ( -d_B + d_A ) + c_1 ( -d'_B + d'_A ) = 5 ( -d_B + d_A ) + 2 ( -d'_B + d'_A ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_B^{(1)} = k_1 ( d_B - d_A ) + c_1 ( d'_B - d'_A ) = 5 ( d_B - d_A ) + 2 ( d'_B - d'_A ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Rearrange the terms in the equations to prepare to write it in matrix form:


 * {| style="width:100%" border="0"

$$ f_A^{(1)} = k_1 ( d_A - d_B ) + c_1 ( d'_A - d'_B ) = 5 ( d_A - d_B ) + 2 ( d'_A - d'_B ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_B^{(1)} = k_1 ( -d_A + d_B ) + c_1 ( -d'_A + d'_B ) = 5 ( -d_A + d_B ) + 2 ( -d'_A + d'_B ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The equations in matrix form are:


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} f_A^{(1)} \\ f_B^{(1)} \end{Bmatrix} = \begin{bmatrix} k_1 & - k_1 \\ -k_1 &  k_1 \end{bmatrix} \begin{Bmatrix} d_A \\ d_B \end{Bmatrix} + \begin{bmatrix} c_1 & - c_1 \\ -c_1 & c_1 \end{bmatrix} \begin{Bmatrix} d'_A \\ d'_B \end{Bmatrix} = \begin{bmatrix} 5 & - 5 \\ -5 &  5 \end{bmatrix} \begin{Bmatrix} d_A \\ d_B \end{Bmatrix} + \begin{bmatrix} 2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_A \\ d'_B \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |


 * }
 * }

Element 2
Begin with a free body diagram for element 2:



Summing the forces and following the same steps as explained for element 1, the matrix for element 2 is:


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} f_B^{(2)} \\ f_C^{(2)} \end{Bmatrix} = \begin{bmatrix}   k_2 & - k_2 \\ -k_2 &   k_2 \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} + \begin{bmatrix}  c_2 & - c_2 \\ -c_2 & c_2 \end{bmatrix} \begin{Bmatrix} d'_B \\ d'_C \end{Bmatrix} = \begin{bmatrix}   5 & - 5 \\ -5 &   5 \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} + \begin{bmatrix}  2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_B \\ d'_C \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Element 3
Begin with a free body diagram for element 3:



Taking a step back, let's analyze the system to see how we can simplify it.

The equation for an elastic rod is:


 * {| Italic textstyle="width:100%" border="0"

$$ F= \frac{EA}{L}*x $$
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 * style="width:95%" |
 * }
 * }

$$x $$     represents the displacement of the elastic bar

$$E $$     is a constant that is a property of the material

$$A $$     is the cross-sectional area of the bar

$$L $$     is the initial length of the bar

We can compare this to the equation for a spring:


 * {| Italic textstyle="width:100%" border="0"

$$ F= kx $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

$$x $$ represents the displacement of the spring

$$k $$ is a spring constant that is a property of the spring

Thus, we can think of $$\frac{EA}{L} $$ as being a spring constant $$k $$.

With this assumption that the elastic bars act as springs, we can analyze the system as if the two bars are two springs in series.


 * {| Italic textstyle="width:100%" border="0"

$$ \frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} $$
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 * style="width:95%" |
 * }
 * }

Applying this equation to the system:


 * {| Italic textstyle="width:100%" border="0"

$$ \frac{1}{k_{eq}} = \frac{L_1}{E_1 A_1} + \frac{L_2}{E_2 A_2} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solving:


 * {| Italic textstyle="width:100%" border="0"

$$ k_{eq} = \frac{1}{\frac{L_1}{E_1 A_1} + \frac{L_2}{E_2 A_2}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The constants for each elastic rod are then plugged in:


 * {| Italic textstyle="width:100%" border="0"

$$ E_1 = 2, \ A_1 = 1 / 2 , \ L_1 = 5, E_2 = 2, \ A_2 = 1 / 2 , \ L_2 = 4 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| Italic textstyle="width:100%" border="0"

$$ k_{eq} = \frac{1}{\frac{5}{(2)(0.5)} + \frac{4}{(2)(0.5)}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| Italic textstyle="width:100%" border="0"
 * {| Italic textstyle="width:100%" border="0"

$$ k_{eq} = \frac{1}{9} $$
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 * style="width:95%" |
 * }
 * }

Now, apply the same method followed for elements 1 and 2, sum the forces and plug in the constants. The only difference is that there is an extra force in the equation due to the two elastic rods.


 * {| Italic textstyle="width:100%" border="0"

$$ f_C^{(3)} = k_3 ( -d_D + d_C ) + c_3 ( -d'_D + d'_C ) +k_{eq}(-d_D + d_C)= 5 ( -d_D + d_C ) + 2 ( -d'_D + d'_C ) + \frac{1}{9} (-d_D + d_C) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| Italic textstyle="width:100%" border="0"

$$ f_D^{(3)} = k_3 ( -d_C + d_D ) + c_3 ( -d'_C + d'_D ) + k_{eq}(-d_C + d_D)= 5 ( -d_C + d_D ) + 2 ( -d'_C + d'_D) + \frac{1}{9}(-d_C + d_D) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} f_C^{(3)} \\ f_D^{(3)} \end{Bmatrix} = \begin{bmatrix}   k_3 & - k_3 \\ -k_3 &   k_3 \end{bmatrix} \begin{Bmatrix} d_C \\ d_D \end{Bmatrix} + \begin{bmatrix}  c_3 & - c_3 \\ -c_3 & c_3 \end{bmatrix} \begin{Bmatrix} d'_C \\ d'_D \end{Bmatrix} + \begin{bmatrix}  k_{eq} & - k_{eq} \\ -k_{eq} &   k_{eq} \end{bmatrix} \begin{Bmatrix} d_C \\ d_D \end{Bmatrix} = \begin{bmatrix}   5 & - 5 \\ -5 &   5 \end{bmatrix} \begin{Bmatrix} d_C \\ d_D \end{Bmatrix} + \begin{bmatrix}  2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_C \\ d'_D \end{Bmatrix} + \begin{bmatrix}  \frac{1}{9} & - \frac{1}{9} \\ -\frac{1}{9} &   \frac{1}{9} \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Mass B
Begin with a free body diagram for mass B.



Determine the forces acting on mass B:

$$ \sum_{} f = F_B(t)- m_Bd''-f_B^{(1)}-f_B^{(2)} = 0 $$


 * {| style="width:100%" border="0"

$$ f_B^{(1)} = \begin{bmatrix} -k_1d_A + k_1d_B \end{bmatrix} + \begin{bmatrix} -c_1d'_A + c_1d'_B \end{bmatrix} = \begin{bmatrix} -5d_A + 5d_B \end{bmatrix} + \begin{bmatrix} -2d'_A + 2d'_B \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_B^{(2)} = \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} = \begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Sum the forces acting on mass B and plug in the constants into the equation of equilibrium:


 * {| style="width:100%" border="0"

$$ F_B(t) = m_Bd''_B + (\begin{bmatrix} -k_1d_A + k_1d_B \end{bmatrix} + \begin{bmatrix} -c_1d'_A + c_1d'_B \end{bmatrix}) + (\begin{bmatrix} k_2d_B - k_2d_C \end{bmatrix} + \begin{bmatrix} c_2d'_B - c_2d'_C \end{bmatrix}) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ = 1.3d''_B + (\begin{bmatrix} -5d_A + 5d_B \end{bmatrix} + \begin{bmatrix} -2d'_A + 2d'_B \end{bmatrix}) + (\begin{bmatrix} 5d_B - 5d_C \end{bmatrix} + \begin{bmatrix} 2d'_B - 2d'_C \end{bmatrix}) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Mass C
Begin with a free body diagram for mass C.



Determine the forces acting on mass C:

$$ \sum_{} f = 0 = F_C(t)- m_Cd''_C- f_C^{(2)} - f_C^{(3)} $$


 * {| style="width:100%" border="0"

$$ f_C^{(2)} = \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} = \begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_C^{(3)} = \begin{bmatrix} k_3d_C - k_3d_D \end{bmatrix} + \begin{bmatrix} c_3d'_C - c_3d'_D \end{bmatrix} + \begin{bmatrix} k_{eq}d'_C - k_{eq}d'_D \end{bmatrix} = \begin{bmatrix} 5d_C - 5d_D \end{bmatrix} + \begin{bmatrix} 2d'_C - 2d'_D \end{bmatrix} + \begin{bmatrix} \frac{1}{9}d'_C - \frac{1}{9}d'_D \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Sum the forces acting on mass C and plug in the constants into the equation of equilibrium:


 * {| style="width:100%" border="0"

$$ F_C(t) = m_Cd''_C + (\begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix}) + (\begin{bmatrix} k_3d_C - k_3d_D \end{bmatrix} + \begin{bmatrix} c_3d'_C - c_3d'_D \end{bmatrix} + \begin{bmatrix} k_{eq}d'_C - k_{eq}d'_D \end{bmatrix}) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">


 * }
 * {| style="width:100%" border="0"

$$ = 1.3d''_C + (\begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix}) + (\begin{bmatrix} 5d_C - 5d_D \end{bmatrix} + \begin{bmatrix} 2d'_C - 2d'_D \end{bmatrix} + \begin{bmatrix} \frac{1}{9}d'_C - \frac{1}{9}d'_D \end{bmatrix}) ) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">


 * }

Apply the boundary conditions to both equations of motion (this will eliminate some terms):


 * {| style="width:100%" border="0"

$$ d_A(t) = d'_A(t) = 0 $$
 * style="width:95%" |
 * style="width:95%" |


 * }
 * }


 * {| style="width:100%" border="0"

$$ d_D(t) = d'_D(t) = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Equations of motion after substituting in boundary conditions:


 * {| style="width:100%" border="0"

$$ F_B(t) = m_Bd_B + k_1d_B + c_1d'_b + \begin{bmatrix} k_2d_B - k_2d_C \end{bmatrix} + \begin{bmatrix} c_2d'_b - c_2d'_C \end{bmatrix} = 1.3d_B + 5d_B + 2d'_b + \begin{bmatrix} 5d_B - 5d_C \end{bmatrix} + \begin{bmatrix} 2d'_b - 2d'_C \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$ F_C(t) = m_Cd_C + \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} + k_3d_C + c_3d'_C + k_{eq}d_C= 1.3d_C + \begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} + 5d_C + 2d'_C + \frac{1}{9}d_C $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now, assembling these final two equations into matrix form:


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} F_B(t) \\ F_C(t) \end{Bmatrix} = \begin{bmatrix}   m_B & 0 \\ 0 &  m_C \end{bmatrix} \begin{bmatrix} d_B \\ d_C \end{bmatrix} + \begin{bmatrix}  c_1 + c_2 & - c_2 \\ -c_2 & c_2 + c_3 \end{bmatrix} \begin{bmatrix} d'_B \\ d'_C \end{bmatrix} + \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 + k_3 \end{bmatrix} \begin{bmatrix} d_B \\ d_C \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & k_{eq} \end{bmatrix} \begin{bmatrix} d_B \\ d_C \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} F_B(t) \\ F_C(t) \end{Bmatrix} = \begin{bmatrix}   m_B & 0 \\ 0 &  m_C \end{bmatrix} \begin{bmatrix} d_B \\ d_C \end{bmatrix} + \begin{bmatrix}  c_1 + c_2 & - c_2 \\ -c_2 & c_2 + c_3 \end{bmatrix} \begin{bmatrix} d'_B \\ d'_C \end{bmatrix} + \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 + k_3+k_{eq} \end{bmatrix} \begin{bmatrix} d_B \\ d_C \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Substitute the values into the matrix:


 * {| style="width:100%" border="0"

$$ k_1 = k_2 = k_3 = 5 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ m_1 = m_2 = 1.3 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ c_1 = c_2 = c_3 = 2 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ m_B = m_1 $$
 * style="width:95%" |
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 * }
 * }


 * {| style="width:100%" border="0"

$$ m_C = m_2 $$
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 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} F_B(t) \\ F_C(t) \end{Bmatrix} = \begin{bmatrix}   1.3 & 0 \\ 0 &  1.3 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} + \begin{bmatrix}  4 & -2 \\ -2 & 4 \end{bmatrix} \begin{bmatrix} d'_1 \\ d'_2 \end{bmatrix} + \begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & \frac{1}{9} \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The final answer is:


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$$ \begin{Bmatrix} F_B(t) \\ F_C(t) \end{Bmatrix} = \begin{bmatrix}   1.3 & 0 \\ 0 &  1.3 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} + \begin{bmatrix}  4 & -2 \\ -2 & 4 \end{bmatrix} \begin{bmatrix} d'_1 \\ d'_2 \end{bmatrix} + \begin{bmatrix} 10 & -5 \\ -5 & 10.111 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} $$
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