# Homework Help: Coupled Mass System

1. Nov 4, 2015

### BOAS

Hello,

I have a problem regarding the characteristic frequencies of a coupled mass-spring system. I have made some relevant progress, but i'm unsure of where to go from here.

1. The problem statement, all variables and given/known data

Find the characteristic frequencies and the two characteristic modes of vibration if the central spring constant in the following diagram (attached) is 2k.

2. Relevant equations

3. The attempt at a solution

Using Newton's second law, I have written the following two equations for x and y.

$m \ddot{x} = - kx - 2k(y - x) = kx - 2ky$
$m \ddot{y} = - ky - 2k(y - x) = ky + 2kx$

Rearranging for accelerations;

$\ddot{x} = \frac{k}{m} x - \frac{2k}{m} y$
$\ddot{y} = \frac{2k}{m} x + \frac{k}{m} y$

I'm not sure how to do matrices properly in LaTeX, but I write the above as a matrix and solve for the Eigen values. If I call the coefficient matrix $A$ then $A_{11} = 1$, $A_{12} = -2$, $A_{21} = 2$, $A_{22} = 1$.

Using the fact that $\det (A - \mu I) = 0$ I find that $\mu = 1 \pm 2i$

Using this fact I find my eigen vectors to be $\vec e_{1} = \frac{1}{\sqrt{2}} \left(\begin{array}{c}1\\-i\end{array}\right)$ and $\vec e_{2} = \frac{1}{\sqrt{2}} \left(\begin{array}{c}1\\i\end{array}\right)$

I think the next step is for me to write out the diagonalised matrix, with the eigen values as the entries top left and bottom right.

$\left(\begin{array}{c}\ddot{X}\\\ddot{Y}\end{array}\right) = \frac{k}{m} \left(\begin{array}{c}1 - 2i & 0\\\ 0 & 1 + 2i\end{array}\right) \left(\begin{array}{c}X\\ Y\end{array}\right)$

This gives me two differential equations;

$\ddot{X} = (1-2i)X$
$\ddot{Y} = (1+2i)Y$

I'm concerned about my complex eigenvalues and think I have gone wrong somewhere, but I can't find an obvious mistake.

Please can somebody take a look and help me to understand where to go from here.

Thank you very much,

BOAS.

Edit - Figured out how to write matrices

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2. Nov 4, 2015

### PeroK

Try the special case of $y = 0$. Does it look right?

3. Nov 4, 2015

### BOAS

If $y = 0$

My initial equation is $\ddot{x} = \frac{k}{m}x$ which does not look right at all.

If I displace the first mass by a small distance x, it feels a force due to the first spring $F_{1} = -kx$. It will feel another force, whose direction depends on the position of y, due to the second spring. $F_{2} = 2k(y-x)$ and a third force due to the final spring, whose direction depends on y, $F_{3} = ky$

This gives me $\ddot{x} = -3kx + 3ky$ which makes sense in the case of y=0, because the first mass will feel a force (when displaced) due to the left and center spring in the same direction.

Thank you,

I can do the same for y and try again at this problem. :)

4. Nov 4, 2015

### PeroK

That's not correct. You were close the first time. Try a +ve x and y (with, say, x > y) and look at what the springs are doing

5. Nov 4, 2015

### BOAS

Ah, I believe I have found my mistake.

$m \ddot{x} = -3kx + 2ky$

Originally, I was forcing the direction with a minus sign, which was actually already accounted for with the (y-x) term.

6. Nov 4, 2015

### PeroK

The third spring (on the right) cannot affect the first mass, which is pushed and pulled by the first two springs only.

7. Nov 4, 2015

### BOAS

I agree. The reasoning behind my expression is as follows;

If I displace the first mass by an amount x, it will feel a force due to spring 1 of $F_{1} = -kx$. It will feel a second force due to the middle spring (which has twice the spring constant) of $F_{2} = 2k(y - x)$. This is because the extension/compression of the central spring is dependent upon the position of y. If y < x, the force pushes x back to it's equilibrium position and vice versa.

8. Nov 4, 2015

### PeroK

In the diagram you posted, all three springs have the same constant $k$.

9. Nov 4, 2015

### BOAS

True, but in the question I posted, I did specify that the central spring has twice the constant.

I apologise for causing any problems.

10. Nov 4, 2015

### PeroK

Okay. Hopefully you'll get some real eigenthings now.

11. Nov 4, 2015