Simple harmonic motion and diatomic molecules

[endeavor]

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...

 

[christianjb]

Start by finding r0.

 

[endeavor]

r₀ = D/C. Now what?

 

[christianjb]

For simple harmonic motion, the restoring force is proportional to the displacement. Can you show that to be true in your system?

 

[endeavor]

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?

 

[christianjb]

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.

 

[endeavor]

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 ...

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...

 

[christianjb]

F=-C/r^2+D/r^3

At r=r_0

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

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

F(r=r_0)=0

The largest term is then the second term, which is proportional to r-r_0

 

[endeavor]

use open bracket, then "tex", then close bracket

 

[christianjb]

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

f(x)=A_0 +A_1 x+A_2 x^2+A_3 x^3 +...

Then you can find the constants A_0,A_1,A_2... from repeated differentiation.

Finding A0 is the easiest

f(x=0)=A_0

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

Next, find A1

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

Then find A2

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

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

And so on.

You now substitute in the values of A0,A1,A2 etc to get
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+...

where the derivatives are evaluated at (x=0)

 

[christianjb]

OK, let me know if you understand the above Taylor series expansion.

 

[endeavor]

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?

 

[christianjb]

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.

 

[endeavor]

Then F = - C⁴/D³ * (r - r₀).
k = C⁴/D³
T = 2 pi * square root of (m/k), but what's m?

 

[christianjb]

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.