Simple harmonic motion and diatomic molecules

In summary, the force in some diatomic molecules can be approximated by F = -C/r2 + D/r3, where r is the atomic separation and C and D are positive constants. For small displacements from equilibrium (delta r << r0), the motion is approximately simple harmonic with a force constant of k = C4/D3. The period of such motion can be determined using the equation T = 2 pi * square root of (m/k), where m is the mass of the molecule.
  • #1
endeavor
176
0

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...
 
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  • #2
Start by finding r0.
 
  • #3
r0 = D/C. Now what?
 
  • #4
For simple harmonic motion, the restoring force is proportional to the displacement. Can you show that to be true in your system?
 
  • #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?
 
  • #6
endeavor said:
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
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
[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
use open bracket, then "tex", then close bracket
 
  • #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)
 
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  • #11
OK, let me know if you understand the above Taylor series expansion.
 
  • #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?
 
  • #13
endeavor said:
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
Then F = - C4/D3 * (r - r0).
k = C4/D3
T = 2 pi * square root of (m/k), but what's m?
 
  • #15
endeavor said:
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.
 

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which an object oscillates back and forth around an equilibrium position due to a restoring force that is directly proportional to the displacement from the equilibrium position.

2. How do diatomic molecules exhibit simple harmonic motion?

Diatomic molecules, such as H2 and O2, can exhibit simple harmonic motion because the bond between the two atoms acts as a spring, providing a restoring force that allows the atoms to oscillate back and forth.

3. What is the relationship between the potential energy and kinetic energy in simple harmonic motion of diatomic molecules?

In simple harmonic motion of diatomic molecules, the potential energy and kinetic energy are constantly changing, but the sum of these two energies remains constant. This is known as the conservation of energy.

4. What factors affect the frequency of simple harmonic motion in diatomic molecules?

The frequency of simple harmonic motion in diatomic molecules is affected by the mass of the atoms and the strength of the bond between them. As the mass increases or the bond strength decreases, the frequency decreases.

5. Can diatomic molecules exhibit more than one type of simple harmonic motion?

Yes, diatomic molecules can exhibit more than one type of simple harmonic motion depending on the direction of the oscillation. For example, they can exhibit both longitudinal and transverse simple harmonic motion.

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