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I'm attempting to find the equations of motion (and eventually transfer functions) for a mass-spring-damper system, but one that is slightly different from your generic damped system example.

Below I've given a picture of essentially what the system looks like.

There is a large box with mass m and spring k inside of it. This box is damped by damper c to the left-hand wall of the room. I apply a force F(t) to the side of the box, causing both the box and perhaps the mass inside to move. x is measured from the left-hand side of the room to the center of mass, so x will increase both when the box moves and when the mass inside the box moves.

What I imagine happens is that for low frequency forcing functions, the spring is essentially nonexistent and the equation of motion looks like:

$$F(t) = m\ddot{x} + c\dot{x}$$

When the forcing function frequency approaches the resonance frequency of the system, not only will the box move to the right, but the mass will vigorously experience motion inside of the box. I'm not quite sure what the equation of motion looks like for this case.

I'm looking for one differential equation that captures both of these - low frequency and high frequency forcing functions. At the end of the day I really want a transfer function to use in Simulink.

$$\frac{w_n^2}{s^2 + 2{w_n}{\zeta}s + {w_n^2}}$$

This transfer function doesn't seem to model what I want, even though it's the generic second order system transfer function.

Any help would be greatly appreciated. ^^

enter image description here

enter image description here

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    $\begingroup$ Note that unless the box has mass, or there's friction between the mass and the box, the box is irrelevant. Just think of it as applying the force to the junction of the damper and the spring. The position of this (massless) pont will always the place where the three forces on it balance to zero. This should help you set up your differential equation correctly. $\endgroup$ – Dave Tweed Feb 26 '16 at 21:08
  • $\begingroup$ Doesn't this just result in the standard second order equation of F = mx'' + cx' + kx? I understand and agree with the fact that the box is irrelevant, but I'm still confused as to how to model this system. $\endgroup$ – anonymouse Feb 26 '16 at 21:18
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    $\begingroup$ No, the usual setup for this problem is that the external force is applied directly to the mass, and that the damper and spring are in parallel, not in series. Your setup results in a very different equation. For one thing, you need to account for two different displacements, the junction of the damper and spring, and the mass itself. $\endgroup$ – Dave Tweed Feb 26 '16 at 21:21
  • $\begingroup$ I added a second picture to my original post. I think this is the correct setup, but I'm not sure how to add in the spring force. At the gray line there is a force to the left equal to cx', a force to the right equal to F(t), but... where do the parameters m and k come into the picture? $\endgroup$ – anonymouse Feb 26 '16 at 21:32
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Call the displacement of the box (where it is attached to the damper and spring) y(t).

The force of the damper is $$-c\cdot \dot{y}$$

The force of the spring is $$k\cdot(x - y)$$

y(t) is the point at which the three forces on it balance to zero:

$$F - c\cdot \dot{y} + k\cdot(x - y) = 0$$

x(t) depends only on the spring force and the mass:

$$k\cdot(x - y) + m\cdot \ddot{x} = 0$$

Combine and solve!

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  • $\begingroup$ In your equation for x(t), should the spring force have a negative sign in front of it? EDIT: Never mind, you moved it to the other side which is why it's positive. $\endgroup$ – anonymouse Feb 26 '16 at 22:03
  • $\begingroup$ It looks like the final answer ended up being F(t) = (mc/k)x''' + mx'' + cx', just in case anyone needs it in the future. $\endgroup$ – anonymouse Feb 26 '16 at 22:14

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