For the circuit shown L_{1}=1, L_{2}=2, C=1.

a) Determine the circuit equation in the form:

b) Find frequencies of vibrations, and mode shapes (do by hand, you can check with Matlab).

c) Draw the force-current analog of the system.

a)

b)

Solve for ω^{2}.

Find **v**_{1}.

Find **v**_{2}.

Check with Matlab

>> [v,d]=eig([-3/2 1/2; 1/2 -3/2]) v = 0.7071 0.7071 -0.7071 0.7071 d = -2 0 0 -1

c)

In class I used code (code from class) to calculate modes and frequencies of vibration, and displayed mode shapes and mass trajectories. Add comments to the code between the line starting with %% Begin Comments here and ending at the line starting with %% Stop Comments here. The comments should be written for yourself, yesterday. In other words, make the comments such that if you had read them yesterday you could understand how the code is used to solve the problem and display the results.

If you are using an older version of MATLAB that generates an error, you can comment out the lines that say set(gca,'ColorOrderIndex',1);

The system shown is a very crude model of a building with no damping. The masses can only move in a horizontal direction.

a) draw free body diagrams

b) write equations of the form

c) If m_{1}=m_{2}=m_{3}=m_{4}=1, and k=1, find mode shapes and associated frequencies (you
can use the code from class). Note: these mode shapes are similar to those
of the air pressure in wind instruments.

d) Examine the mode shapes and associated frequencies. Is the lowest frequency associated with the expected mode shape? Why or why not?

Note: you have just developed the solution to an 8^{th} order problem.

a)

b)

c) Pertinent equation is:

A=[-2 1 0 0; 1 -2 1 0; 0 1 -2 1; 0 0 1 -1];

---------------- Eigenvectors and values v = -0.4285 -0.6565 0.5774 0.2280 0.6565 0.2280 0.5774 0.4285 -0.5774 0.5774 -0.0000 0.5774 0.2280 -0.4285 -0.5774 0.6565 d = -3.5321 0 0 0 0 -2.3473 0 0 0 0 -1.0000 0 0 0 0 -0.1206 ---------------- Frequencies omega = 1.8794 1.5321 1.0000 0.3473

d) The lowest frequency is associated with the smoothest mode shape - the building moves almost as one piece. As frequencies get higher, there are more maxima and minima in the mode shape and consecutive floors can exhibit displacements in opposite directions.

The system shown is a very crude model of a building with no damping. The masses can only move in a horizontal direction. The ground under the building has moved suddenly to the left at t=0, stretching the bottom spring. If we call the ground position "zero" this corresponds to the initial conditions:

x_{1}(0)= x_{2}(0)= x_{3}(0)= x_{4}(0)=1.

a) Which mode do you think will have the largest magnitude? Why?

b) Find the mode weights (i.e., the γ's) and verify your answer.

c) simulate the system and plot the trajectories of the masses. (you can use the code from class)

a) These initial conditions most resemble the lowest frequency mode, so we expect that mode magnitude to be largest.

b) Choose as initial conditions a multiple of the eigenvector associated with the lowest frequency (in this case the 4th eigenvector). We can choose this eigenvector in Matlab

x0 = [1 1 1 1]' %Initial condition (size must match that of "A" array).

The output shows that the lowest frequency mode has the largest amplitude (1.89)

Mode amplitudes

gam =

-0.1213

-0.2797

0.5774

1.8904

c)

For the previous problem

a) find initial conditions such that only the lowest frequency mode is excited.

b) simulate the system and plot the trajectories of the masses. (you can use the code from class)

a) Choose as initial conditions a multiple of the eigenvector associated with the lowest frequency (in this case the 4th eigenvector). We can choose this eigenvector in Matlab

x0=v(:,4) %Initial condition (size must match that of "A" array).

b)

Note: we can get animation to work properly with the following lines (in place of the original lines in code).

% Animate (animates objects as beads on a string. . . . yRope=[0 yBeads]; % The y deflection of the rope (0 at left end) xRope=[0 xBeads]; . . . set(ropeLine,'Ydata',[0 x(:,i)']); % Change y values of rope

For the system shown

As you are doing this problem, think about the how the system would behave as the middle mass gets to be much heavier than the other two. This should allow you to figure out in advance approximately what the mode shapes and frequencies should be.

a)
If m_{1}=m_{3}=1, m_{2}=10, and k=1, find mode shapes and associated frequencies (you
can use the
code from class).
Try to figure out approximately what the mode shapes and frequencies will be
before solving. Recall that for a mass-spring system the natural
frequency is (k/m)^{½}.

b) Find the weights (i.e., the "γ" terms) associated with each eigenvector and plot trajectories for the initial conditions .

a) Because the middle mass is so heavy, it won't generally move much.

There will be two modes with similar frequencies in which the outer masses are
in and out of phase with each other, but m2 barely moving. These frequencies will be close.
Theses are modes 2 and 3 in the solution below. Since the middle
mass is not moving it is as if m1 and m3 have an effective spring constant
of 2*k attached (this is because there are two springs attached, and m3 is
practically fixed). The frequency will be (2*k/m1)^{½}=1.41.

The other mode has m2 moving, and m1 and m3 just going
along without affecting the dynamics of the problem very much.
Therefore we expect m1 and m3 to be in phase with m2, but with an amplitude
about half as large. If we ignore m1 and m3, then the mass m2 has 2
springs attached each of magnitude k_{eff}=k/2 (this is the
equivalent spring constant of two springs in series between the mass and the
wall). So the total spring constant is approximately k, and the
frequency should be (2*k_{eff}/m2)^{½}=(1/10)^{½}=0.32

Pertinent line is:

A=[-2 1 0; 0.1 -0.2 0.1; 0 1 -2];

---------------- Eigenvectors and values v = -0.4215 0.7071 0.7052 -0.8029 0.0000 -0.0740 -0.4215 -0.7071 0.7052 d = -0.0950 0 0 0 -2.0000 0 0 0 -2.1050 ---------------- Frequencies omega = 0.3082 1.4142 1.4509

b)

---------------- Initial Conditions x0 = 1 0 1 ---------------- Mode amplitudes gam = -0.1239 -0.0000 1.3440