If I have the system that could be observed in the next Image:

Mechanical system

I want to know the transfer function, where the external force $f$ is the entry and $x_1$ is the output. The direction and positive sense of movement and force is right to left, as the image shows.

Assume that the mass $m_x$ is zero, the initial conditions are zero, and the effects of gravity are null.
Moreover, the two springs are taken as a single equivalent ($k=k_1+k_2$) and the same with the shock absorbers ($b=b_1+b_2$), since it is assumed that each end of them suffers the same displacement, speed and acceleration.

My transfer function is:


It contains a null damping factor, that is, it is pure oscillatory. However, I know that I have a mass that acquires kinetic energy, a spring that acquires potential energy, and there will be an exchange of energy between them during the oscillations, but I also have a damper dissipating energy, which I don't see in my transfer function. How can I interpret this?

Attached is my development.

Development to get the transfer function


2 Answers 2


but I also have a damper dissipating energy, which I don't see in my transfer function

So, this is a fun one.

You have a damper that dissipates energy when the external force is applied. However, your $m_x = 0$; your connecting bar thingie on the input is massless, and you are (properly) modeling your dampers as massless, too.

As a consequence, the end bar isn't anchored to anything -- it "sees" an infinite mechanical compliance.

From an intuitive perspective, this means that with no force applied, whatever your main mass does, your connecting bar thingie will do the exact same thing, only with whatever offset happens to be present in the dampers. This is what leads to your mathematical result that shows the dampers having no effect.


That looks correct. The issue is that the normal way to analyze the mass-spring-damper system that gets a non-zero damping term has the spring and the damper in parallel.

This answer on the Physics SE site comes to a similar conclusion to you (though uses different notation).

They end up with

$$m\,\ddot x_2=b\,(\dot x_1-\dot x_2)=-k\,x_1$$

where I've changed their $\sigma$ to your $b$.


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