# Dynamic Damping in a simple spring system

• kostoglotov
In summary: 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,.
kostoglotov

## 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?

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.

ehild said:
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.

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

ehild said:
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,.

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

kostoglotov said:
to quote the exercise, "this is an example of a Dynamic Damper"...
I meant that the damping coefficients of the springs was not mentioned.

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

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

kostoglotov said:
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?

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:

## 1. What is dynamic damping in a simple spring system?

Dynamic damping in a simple spring system refers to the process of reducing vibrations and oscillations in the system by dissipating energy. This is achieved by introducing a damping force, typically through a shock absorber or friction mechanism, which converts the kinetic energy of the system into heat.

## 2. Why is dynamic damping important in a spring system?

Dynamic damping is important in a spring system because it helps to stabilize the system and reduce the magnitude of vibrations. This is especially crucial in systems where precise control and stability are necessary, such as in suspension systems of vehicles or precision mechanical systems.

## 3. How does dynamic damping affect the behavior of a simple spring system?

Dynamic damping affects the behavior of a simple spring system by reducing the amplitude and frequency of vibrations. This results in a smoother and more controlled movement of the system, as the damping force counteracts any sudden or excessive oscillations.

## 4. What factors can affect the amount of dynamic damping in a spring system?

The amount of dynamic damping in a spring system can be affected by several factors, including the material and design of the damping mechanism, the stiffness and mass of the spring, and the amplitude and frequency of the vibrations in the system.

## 5. Can dynamic damping be adjusted or controlled in a spring system?

Yes, dynamic damping can be adjusted or controlled in a spring system by changing the properties of the damping mechanism, such as the viscosity of the fluid in a shock absorber or the amount of friction in a friction damper. This allows for fine-tuning of the damping force to suit the specific needs of the system.

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