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Simple harmonic motion and diatomic molecules

  1. Apr 7, 2007 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations


    3. The attempt at a solution

    I have no idea where to start...
     
  2. jcsd
  3. Apr 7, 2007 #2
    Start by finding r0.
     
  4. Apr 7, 2007 #3
    r0 = D/C. Now what?
     
  5. Apr 7, 2007 #4
    For simple harmonic motion, the restoring force is proportional to the displacement. Can you show that to be true in your system?
     
  6. Apr 8, 2007 #5
    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?
     
  7. Apr 8, 2007 #6
    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.
     
    Last edited: Apr 8, 2007
  8. Apr 8, 2007 #7
    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...
     
  9. Apr 8, 2007 #8
    [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]
     
    Last edited: Apr 8, 2007
  10. Apr 8, 2007 #9
    use open bracket, then "tex", then close bracket
     
  11. Apr 8, 2007 #10
    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)
     
    Last edited: Apr 8, 2007
  12. Apr 8, 2007 #11
    OK, let me know if you understand the above Taylor series expansion.
     
  13. Apr 8, 2007 #12
    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?
     
  14. Apr 8, 2007 #13
    yes, that's the way I would do it.
     
  15. Apr 9, 2007 #14
    Then F = - C4/D3 * (r - r0).
    k = C4/D3
    T = 2 pi * square root of (m/k), but what's m?
     
  16. Apr 9, 2007 #15
    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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