2
$\begingroup$

I would like to set up a simple laser interferometer to measure the compression of a material experiencing forces around $8 \times 10^{-4}$ N. But I need to find the spring constant to calculate the force, since the interferometer only gives me distance. I could use a photodiode to count the number of waves moving across the projection for a few known masses as small as 1 gram and calculate its constant, but I'm not sure if materials behave the same under microscopic compression. Will I be able to accurately find the force applied that is changing the distance? The material to be compressed is still to be determined.

$\endgroup$
3
  • $\begingroup$ It occurs to me that there are digital scales that could easily handle 2 milligrams, but why do it the easy way when I could use lasers? $\endgroup$
    – griffin175
    Aug 29 '15 at 22:10
  • $\begingroup$ You'll want to search for "elastic modulus". $\endgroup$ Aug 29 '15 at 22:39
  • $\begingroup$ Do you want to measure the flexing (bending) of material, or do you want to measure its compression? If you want to measure how much a bar bends, you can use the flexural modulus. If compression is what you're after, use Young's modulus. These are the ratio of stress to strain. There are tables of these ratios for different materials. $\endgroup$
    – Ernie
    Aug 30 '15 at 1:51
1
$\begingroup$

I am not sure if I understand exactly what you want to do, but let me assume and reply accordingly: I imagine you have a Michelson type interferometer with a little block of material at the end of one of the interferometer arms off which the laser light reflects of. Now if a force acts on the front face of the block it changes the optical path in that arm and you'll see a change in your fringe pattern on the interferometer output.

Like commenters have said: What determines the volume change (compression) under pressure is the bulk modulus of the material. For the interferometer more important would be Young's modulus, that determines with how much linear strain your material responds to a certain tension (Force per cross section area). For an isotropic solid the two are linked by $E = 3K \cdot (1-2\nu)$, where $E$ is Young's modulus, $K$ the bulk modulus, and $\nu$ Poisson's ratio.

If you want to make sure to get large displacement for small forces, you want to pay attention to use a sample with a large aspect ratio: The overall arm-length change $\Delta L$ is proportional to the absolute length $L$ of the sample: $\Delta L = \epsilon L$, where $\epsilon$ is the strain.

The strain is linked to the force by Hooke's law (linear response): $F/A = E \cdot \epsilon$, with $F$ the force and $A$ the cross section area of the sample.

As @tom has already pointed out: Hooke's law is best fulfilled the smaller the strain $\epsilon = \Delta L/L$, so you shouldn't be worried at the low end of the strain. That has to do with that when the elastic sample just sits there and does nothing, the atoms it consists of sit in equilibrium positions, where there is a minimum of the potential energy of the chemical binding to its neighbouring atoms. When you think of the Taylor expansion of this potential landscape with respect to the atoms relative coordinates (strain), any such potential minimum will be well approximated by a constant term plus a parabola (to lowest order). The linear term vanishes by definition, as its a local minimum. When you compute the force $F = - \frac{\partial V}{\partial x}$ for such a potential $V = c + k \cdot x^2$, you'll find it's proportional to the displacement $x$, so you find again Hooke's law $F = -k \cdot x$, force proportional to displacement.

$\endgroup$
0
$\begingroup$

I remember hearing on the radio sometime ago of an experiment where a very small mass was placed on a large steel girder sticking out of a wall and a microscopic displacement was observed - I think this was in Cranfield University in the UK, but can't remember any more details about it.

I would be very suprised if Hooke's Law did not work for microscopic forces because it is linear and will normally break down for large forces.

$\endgroup$
2
  • $\begingroup$ linearity should break down since matter is ultimately discrete just as does Ohm's law $\endgroup$
    – hyportnex
    Aug 30 '15 at 0:36
  • $\begingroup$ @user31748 - I disagree, but I may be wrong - I disagree because current flows with discrete charges, but the force on, for example, a crystal lattice can vary smoothly from zero to a large value with atoms pushed slightly closer together all the time and a small shift in equilibrium geometry due to the force applied... $\endgroup$
    – tom
    Aug 30 '15 at 18:02

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.