Simple harmonic motion and diatomic molecules

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Homework Statement


In some diatomic molecules, the force each atom exerts on the other can be approximated by F = -C/r2 + D/r3, where r is the atomic separation and C and D are positive constants. Let delta r = r - r0 be a small displacement from equilibrium, where delta r << r0. Show that for such small displacements, the motion is approximately simple harmonic, and determine the force constant. Thirdly, what is the period of such motion?

Homework Equations




The Attempt at a Solution



I have no idea where to start...
 

Answers and Replies

  • #2
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Start by finding r0.
 
  • #3
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r0 = D/C. Now what?
 
  • #4
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For simple harmonic motion, the restoring force is proportional to the displacement. Can you show that to be true in your system?
 
  • #5
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I'm not sure how to do that... How would I plug in delta r? Do I need to use the equation F = -GMm/r2?
 
  • #6
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I'm not sure how to do that... How would I plug in delta r? Do I need to use the equation F = -GMm/r2?
Uh- you can neglect gravitational interaction for molecules!

You could try a Taylor series expansion in delta r

Edit:

For simple harmonic motion you need to show that F=-k (r-r0) for small values of (r-r0). The problem then is to find k at r-r0, which you can do pretty much by direct substitution into the force equation.
 
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  • #7
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What does a Taylor series expansion mean? Not too long ago, I learned about Taylor series, and they're still a little fuzzy in my memory. I feel pretty dumb, because everything you say I seem to have questions about how to do it :frown: ...

F = -C/r2 + D/r3 = -k * (r - r0)
solving for k,
k = 1/(r - r0) * (C/r2 - D/r3)
... I think I need to do the first part... k shouldn't depend on r...
 
  • #8
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[itex]F=-C/r^2+D/r^3[/itex]

At [itex]r=r_0[/itex]

[itex]F=F(r=r_0)+\frac{dF}{dr} (r-r0)+1/2 \frac{d^2F}{dr^2} (r-r_0)^2+....[/itex]

where the derivatives are evaluated at [itex]r=r_0[/itex]
The first term should be zero at [itex]r=r_0[/itex], because you're at a minimum.

[itex]F(r=r_0)=0[/itex]

The largest term is then the second term, which is proportional to [itex]r-r_0[/itex]
 
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  • #9
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use open bracket, then "tex", then close bracket
 
  • #10
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OK, I just about got those eqns working.

You'll need to understand the Taylor series first.

This is the way I think about it.

imagine that a function f(x) can be written as a power series in x about x=0

[itex] f(x)=A_0 +A_1 x+A_2 x^2+A_3 x^3 +.....[/itex]

Then you can find the constants [itex]A_0,A_1,A_2...[/itex] from repeated differentiation.

Finding A0 is the easiest

[itex]f(x=0)=A_0[/itex]

because, when x=0 all the other terms are zero.

Next, find A1

[itex]df(x)/dx=A_1+2A_2x+3A_3x^2+...[/itex]
So, at x=0
[itex]\frac{df(x)}{dx} (x=0)=A_1[/itex]

Then find A2

[itex]\frac{d^2f(x)}{dx^2}=2A_2+6A_3x+24A_4x^2+...[/itex]

When x=0
[itex]\frac{d^2f(x)}{dx^2}(x=0)=2A_2[/itex]

And so on.

You now substitute in the values of A0,A1,A2 etc to get
[itex]f(x)=f(0)+\frac{df}{dx} x +1/2 \frac{d^2f}{dx^2} x^2+1/6 \frac{d^2 f}{dx^3} x^3+...[/itex]

where the derivatives are evaluated at (x=0)
 
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  • #11
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OK, let me know if you understand the above Taylor series expansion.
 
  • #12
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Yeah, I think so.

So do I just say that for small displacements, F is approximately equal to that first term, which is proportional to r - r0?
 
  • #13
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Yeah, I think so.

So do I just say that for small displacements, F is approximately equal to that first term, which is proportional to r - r0?
yes, that's the way I would do it.
 
  • #14
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Then F = - C4/D3 * (r - r0).
k = C4/D3
T = 2 pi * square root of (m/k), but what's m?
 
  • #15
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Then F = - C4/D3 * (r - r0).
k = C4/D3
T = 2 pi * square root of (m/k), but what's m?
You don't seem to have been provided with the mass, so just give the general expression you derived. That should be fine.
 

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