Solving Characteristic Frequencies of Coupled Mass-Spring System

AI Thread Summary
The discussion centers on finding the characteristic frequencies and modes of a coupled mass-spring system with a central spring constant of 2k. The user has derived equations of motion using Newton's second law but is struggling with the resulting complex eigenvalues and the correct formulation of the system. After some back-and-forth, the user realizes a mistake in the force calculations affecting the first mass, particularly regarding the contributions from the springs. Clarifications are made about the forces acting on the masses, leading to a more accurate representation of the system dynamics. The user expresses gratitude for the assistance and hopes to resolve the problem effectively.
BOAS
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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. Homework Statement


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

Homework Equations

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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BOAS said:
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. Homework Statement


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

Homework Equations

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##

Try the special case of ##y = 0##. Does it look right?
 
PeroK said:
Try the special case of ##y = 0##. Does it look right?

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. :)
 
BOAS said:
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. :)

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
 
PeroK said:
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
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.
 
BOAS said:
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.

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

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.
 
BOAS said:
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.

In the diagram you posted, all three springs have the same constant ##k##.
 
PeroK said:
In the diagram you posted, all three springs have the same constant ##k##.

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

I apologise for causing any problems.

Okay. Hopefully you'll get some real eigenthings now.
 
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  • #11
PeroK said:
Okay. Hopefully you'll get some real eigenthings now.

Thank you for your help.
 
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