Physics Help Forum Angular frequency

 Mar 8th 2014, 09:58 AM #1 Member   Join Date: Feb 2013 Location: Greater St. Louis area Posts: 43 Angular frequency 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.
 Mar 10th 2014, 07:46 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,324 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 by ChipB; Mar 11th 2014 at 07:26 AM.
 Mar 10th 2014, 05:28 PM #3 Member   Join Date: Feb 2013 Location: Greater St. Louis area Posts: 43 Thank you Thank you so much Chip. Yes, I can solve this now.
Mar 11th 2014, 07:26 AM   #4
Physics Team

Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,324
Upon review, I'm afraid this part is incorrect:

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

Sorry for this.

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