Angular frequency

Feb 2013
43
0
Greater St. Louis area
A mass moves along the x axis with potential energy
U(x)= - U0 a^2 / (a^2 + x^2). What is the angular frequency assuming the oscillation is small enough to be harmonic?

w^2 = k/m with w as the angular frequency

F= -kx = -(gradient) U

Since this is one-dimensional we take the derivative of U with respect to x.

I get -(gradient) U = -2 U0 a^2 x / (a^2 + x^2)^2

Therefore k= 2 U0 a^2 / (a^2 + x^2)^2

The correct answer does not have an x term in it.

Is there a binomial expansion that would essentially eliminate the x term in the denominator?

Thanks for any help.
 

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
If the motion is harmonic then x is a function of time:

x = A sin(wt)

where w = rotational velocity in radians/second. The equation of motion from conservation of energy is:

m(d^2x/dt^2)+U = 0. [**EDIT - this is wrong - see correction in post #4 **]

So:

-mAw^2 sin(wt)+2Asin(wt)cos(wt)U_0a^2/(a^2+A^2sin^2(wt))^2 = 0

Can you take it from here? You can apply a couple of trig identities for cos(2x) and sin(2x) to simplify things.
 
Last edited:
Feb 2013
43
0
Greater St. Louis area
Thank you

Thank you so much Chip.

Yes, I can solve this now.
 

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
Upon review, I'm afraid this part is incorrect:

m(d^2x/dt^2)+U = 0.
This should be: m(d^2x/dt^2) + dU/dx = 0.

Sorry for this.