2
$\begingroup$

I am trying to model a 1 DoF electromagnetic vibration sensor (geophone) analytically and with finite elements. A geophone consists of springs, a permanent magnet and coils. The coils are suspended with the springs so they induce a voltage depending on the velocity of the coils. Easy.

The coil assembly is suspended by two springs. One on the top, one on the bottomenter image description here

The mass has a weight of 11.1 g and the resonance of the spring-mass system is 4.5 Hz, thus:

\begin{align} \omega_\text{res} &= \sqrt{\frac{k}{m}} \\ (2\pi \times 4.5 \, \text{Hz})^2*11.1 \, \text{g} &= 8.8738 \, \frac{\text{N}}{\text{m}} \, . \end{align}

As the two springs are parallel we know that each spring has a stiffness of 4.4369 N/m. If the geophone is used in vertical direction we have a force $f_g$ acting on the system:

$$ f_g = 11.1 \, \text{g} * 9.81 \, \frac{\text{m}}{\text{s}^2} \, . $$ The displacement $d$ of the system is then: $$ d = \frac{f_g}{k}=12.3 \, \text{mm}. $$ And as the springs are parallel this of course means that the displacement is the same for both springs. Now comes the problem: when I measure the physical sensor I have in front of me I obtain a displacement of 6 mm which means the system has a stiffness of $$ k = \frac{11.1 \, \text{g} * 9.81\frac{\text{m}}{\text{s}^2}}{6 \, \text{mm}} = 18.1485 \, \frac{\text{N}}{\text{m}} \, . $$ This is approximately double the stiffness obtained from the resonance frequency. The springs are both placed so that in the static case they act as compression springs.

I do not understand why the stiffness of the springs is not the same, does anyone have a hint? My guesses:

  • Non-linear spring
  • (X) The pre-deformation has some influence on the springs I am not aware of

The working range of the system is +/- 2 mm. The mass of the springs is < 1 $g$ (actually my kitchen scale can't measure it), which should be neglect able compared to the 11.1 g of the mass. Here is a picture I took of the object:

enter image description here

$\endgroup$
  • $\begingroup$ I would change the title to "dynamic spring calculation" or something similar in order to get more interest. $\endgroup$ – ja72 Jun 3 '15 at 12:49
  • 2
    $\begingroup$ Carrying numbers through a calculation is a really bad habit that should be broken as early as possible. You're much more likely to get through a calculation error-free if you use symbols for everything until the very end. $\endgroup$ – DanielSank Aug 10 '15 at 16:56
  • 1
    $\begingroup$ By the way, you don't need to keep an edit history in the post. Edit history is tracked automatically. Putting the edit history in the post is therefore more of a distraction than anything else. $\endgroup$ – DanielSank Aug 10 '15 at 17:01
1
$\begingroup$

You seem to have a hidden assumption in your work, that the springs are uncompressed before you displace them and take your measurement. It is easily possible that there is something in the mechanical setup of your device which keeps the springs compressed when your geophone is in its "resting" state.

$\endgroup$
0
$\begingroup$

The first explanation that springs to mind is that the springs have a finite mass. Given the geometry, the resonant frequency will be lowered (compared to your calculation) by adding half the mass of the springs to the mass of the test object (one end of the spring doesn't move at all, the other moves at full deflection - on average they move half).

I don't know the mass of your springs, but the test mass is quite light. Could this be the reason?

Otherwise, if the springs are "predeformed" by 6 mm (what exactly do you mean by that) and you get a deflection of 6 mm, it might be that you are reaching a point where the spring becomes quite nonlinear. But this doesn't make a lot of sense to me - it seems that you would want the device to be quite linear in the range of normal operation.

$\endgroup$
  • $\begingroup$ I updated the question a bit and removed the confusing pre deformed part. I agree that the mass of the springs could take influence, but in this case the mass is much lower than the moving mass. $\endgroup$ – Tom Jan 11 '15 at 23:13
  • $\begingroup$ Are you measuring the response (frequency) while the coils are loaded? Currents in the coils will create a opposing force and make the spring constant seem larger. $\endgroup$ – Floris Jan 11 '15 at 23:17
  • $\begingroup$ The resonance frequency stays the same even when not loading the coil (open circuit). The Lorentz forces you described contribute to the damping of the system. $\endgroup$ – Tom Jan 12 '15 at 0:10
  • $\begingroup$ @Tom: What's the mass of the springs? $\endgroup$ – Qmechanic Feb 25 '17 at 13:03
0
$\begingroup$

You made a mistake in your spring calculation. From the diagram, the two springs are in series, not parallel. These add in inverse.

$$k_{effective} = (\frac{1}{k_1}+\frac{1}{k_2})^{-1} = k/2 $$

You predicted the effective spring constant to be around 8.87N/m, so you would expect the individual spring constants to be around 17.7N/m. This is much closer to your measured value of 18.1N/m.

$\endgroup$
  • 1
    $\begingroup$ Comment to the answer (v1): The springs are technically speaking connected in parallel, not in series. Adding the second spring makes it harder, not easier to move. $\endgroup$ – Qmechanic Feb 25 '17 at 12:45
0
$\begingroup$

I believe the discrepancy is due to a incorrect stiffness of 18.15 N/m in the measurement based on displacement. If I assume correctly you define displacement as the how much the mass displaces when moved to the center (by hand?) to when it moved down and only held by the springs. If so, you have to take into account that in the center point the springs could well still exert a net force on the mass so the force by gravity on the mass is not completely countered by the force to move the mass to the center point. So it is not necessarily a completely unloaded situation as David White indicated. If this net force exerted by the springs happens to be upwards you will see that the displacement when released will be less than you expect.

It is important that you define displacement here carefully.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.