Feel free to use MATLAB for any problem larger than 2x2 and to check all solutions with MATLAB (or your calculator, abacus...).

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.

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

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 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)

For the system shown 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}=1, and k=1, find mode shapes and associated frequencies (you
can use the
code from class).
To think about: *are these approximately what you expected*?

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

a)

b)

c) Pertinent line is:

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

---------------- Eigenvectors and values v = 0.5000 -0.7071 -0.5000 -0.7071 0.0000 -0.7071 0.5000 0.7071 -0.5000 d = -3.4142 0 0 0 -2.0000 0 0 0 -0.5858 ---------------- Frequencies omega = 1.8478 1.4142 0.7654

Lowest frequency mode is smoothest. Increasing frequency with increasing number of extrema.

d) only modes one and three (with mass 1 and mass 3 in same direction) are excited.

---------------- Mode amplitudes gam = 1.0000 0.0000 -1.0000

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

This problem demonstrates that zero-input state space problems can be framed as an eigenvector problem.

Consider the state space system (with no input) given by:

Assume the solution is of the form **q**(t)=c**v**e^{st}
where c and s are scalars, and **v** is a 2x1 vector.

a) Put **q**(t) (and its derivative) into the differential
equation, and show that an eigenvalue problem results (**A**-s**I)v**=**0 **
(or** Av**=s**v**),
where **I** is the identity matrix.

b) Find the eigenvalues, s_{1} and s_{2}.

c) Find the eigenvectors, **v**_{1} and **v**_{2}.

Just as in E11, we form the generic solution as the weighted sum of these
solutions:
where the constants c_{1}
and c_{2} are determined from initial conditions. However, in
this case we easily represent the initial conditions in matrix form.

or **q**(0)=**Vc**
where **q**(0) is the (vector) initial condition, **V**
is a square matrix in which each column is an eigenvector, and **c**
is a column vector whose elements are the unknown coefficients (or weights)
of the individual solutions.

d) If , find c_{1}
and c_{2}.

e) If , solve for y(t) and plot it for 7 seconds.

f) Choose a set of initial conditions such that only the faster exponential is present in the solution.

a)

b)

c)

With MATLAB

>> A=[-5 -2; 2 0] A = -5 -2 2 0 >> [v,d]=eig(A) v = -0.8944 0.4472 0.4472 -0.8944 d = -4 0 0 -1

Eigenvalues are d(1,1) and d(2,1) or -4, -1. This
agrees with our results.

Eigvenvectors are v(:,1) and v(:,2) or [-0.9;0.45], [0.45,-0.9]. This
agrees with our results (within a multiplicative constant).

d)

>> c=inv(v)*[1; -1] c = -0.7454 0.7454

Note you may get different values for this part of the problem depending on your eigenvectors.

e)

>> % Purely numerical solution >> t=linspace(0,7); >> s=diag(d) s = -4 -1 >> % Solve for state variables >> q = c(1)*v(:,1)*exp(s(1)*t) + c(2)*v(:,2)*exp(s(2)*t); >> % Find y and plot it. >> y = [1 3/4]*q; >> plot(t,y)

>> % or symbolic solution| >> syms t >> [v,d]=eig([-5 -2;2 0]); >> s=diag(d); >> q0=[1; -1]; >> c=inv(v)*q0; >> q = c(1)*v(:,1)*exp(s(1)*t) + c(2)*v(:,2)*exp(s(2)*t) >> y=[1 3/4]*q y = (5*exp(-4*t))/12 - exp(-t)/6 >> pretty(y) 5 exp(-4 t) exp(-t) ----------- - ------- 12 6

f) The first eigenvalue represents the faster decay, so any initial conditions that is a multiple of the first eigenvector will not include the other in the solution.

>> % Check with MATLAB >> q0 = [20; -10] % This is a multiple of the eigenvector q0 = 20 -10 >> c=inv(v)*q0 c = -22.3607 0 >> % Note that c(2) (contribution from slower exponential) is zero.