Dynamic Damping in a simple spring system

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1. Feb 3, 2016

kostoglotov

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

2. Relevant equations

3. The attempt at a solution

From the FBD it is apparent that there is a constraint

$$-k_1x_1 + k_2(x_2-x_1) + 5\cos{10t} = 0$$

If you combine this with

$${x}''_1 = -(k_1+k_2)x_1 + k_2x_2 + 5\cos{10t}$$

and

$$m{x}''_2 = -k_2(x_2 - x_1)$$

then by subbing this last equation immediately above into the constraint

$$-k_1x_1-m{x}''_2 + 5\cos{10t} = 0$$

I produce a set of equations

$${x}''_1 = -(k_1+k_2)x_1 + k_2x_2 + 5\cos{10t}$$

and

$${x}''_2 = \frac{k_1}{m}x_1 - \frac{5}{m}\cos{10t}$$

giving the system

$$\vec{{x}''}=\begin{bmatrix} -60 & 10\\ \frac{50}{m} & 0 \end{bmatrix}\vec{x} + \begin{bmatrix} 5\\ -\frac{5}{m} \end{bmatrix}\cos{10t}$$

Now, without the constraint $-k_1x_1 + k_2(x_2-x_1) + 5\cos{10t} = 0$ the system would be

$$\vec{{x}''}=\begin{bmatrix} -60 & 10\\ \frac{10}{m} & -\frac{10}{m} \end{bmatrix}\vec{x} + \begin{bmatrix} 5\\ 0 \end{bmatrix}\cos{10t}$$

I don't think those two systems are equivalent, but I'm not sure. I'll get back to that soon, but moving ahead...

From that first system with the main matrix A where

$$A = \begin{bmatrix} -60 & 10\\ \frac{50}{m} & 0 \end{bmatrix}$$

A very nice characteristic equation for the eigenvalues pops out (the characteristic equation for the eigenvalues of the second unconstrained system is very awkward and messy)

$$\lambda = -30 \pm \sqrt{900+\frac{500}{m}}$$

It is here that I start to become uncertain...I want to assert that the frequency of the second mass should be the same as that of the driving force acting on the first mass. The only thing that seems really convincing to me is that the constraint $-k_1x_1 + k_2(x_2-x_1) + 5\cos{10t} = 0$ implies linear dependence of $x_1,x_2$ and $\cos{10t}$, which means that the sinusoids of the position functions of the masses must have the same frequency as the driving force, ie, 10 rad/s.

Clearly the frequency of the first mass becomes zero in the steady state, but the system of equations I've used to get this far do not allow for any eigenvalue to equal zero, and it does not allow for repeated eigenvalues.

If I run with my idea that lambda must equal -100, ($\lambda = -\omega^2 = -100$), and I solve for the mass, I get a nice number, 0.125 kg. And then if I put that mass back into the first matrix system that got me here I get eigenvalues +40 and -100...that +40 leads to one solution having simply $e^{2\sqrt{10}t}$ in it, rather than e to the power of a complex number, which would result in sinusoids...however, those terms with $e^{2\sqrt{10}t}$ would definitely need to cancel, as they are unbounded.

Feeding m = 0.125 kg into the second unconstrained system produces eigenvalues -100 and -40, leading to two different frequencies, and I can't see how the sinusoidal terms with $\omega = 2\sqrt{10}$ would cancel...

I'm also having trouble figuring out what form the particular solution should take, but for the moment, could someone let me know if I'm on the right track, or any flaws there may be in my reasoning, or otherwise steer me in a better/correct direction?

2. Feb 4, 2016

ehild

The problem says: "find m2 so that in the resulting steady periodic oscillations the mass m1 will remain in rest". It is forced oscillation. Damping is not mentioned, but you can take that there is damping and all masses move with the frequency of the force after long time, when the own oscillations of the system decay. Just substitute ω=10 s-1.
It is not clear what you call x1 and x2. If they are the displacements from equilibrium when F = 0 and the springs are relaxed, $\ddot x_1 = -ω^2 x_1$ and $\ddot x_2 = -ω^2 x_2$. Solve the problem for x1=0.

3. Feb 4, 2016

kostoglotov

What I call x_1 and x_2 is what the question and diagram provided call x_1 and x_2...

4. Feb 4, 2016

kostoglotov

to quote the exercise, "this is an example of a Dynamic Damper"...

5. Feb 4, 2016

ehild

I meant that the damping coefficients of the springs was not mentioned.

6. Feb 4, 2016

kostoglotov

I also never used them in my model or any systems of equations.

7. Feb 4, 2016

ehild

You do not need the eigenvalues of the homogeneous system. At the steady vibration, it is only the particular solution, with ω=10s-1.
The general solution of the system is a linear combination of three SHM-s, two with the eigenfrequencies of the system and one with ω=10. You can reach no motion of m1 only assuming zero amplitudes for the vibrations with the eigenfrequencies.

Last edited: Feb 4, 2016