October 5th, 2017, 12:44 PM  #1 
Senior Member Joined: Nov 2015 From: United States of America Posts: 181 Thanks: 21 Math Focus: Calculus and Physics  Finding frequency of oscillation
I am a little stuck and would love a hint or two if anyone has got any for me. There's a particle with mass m that is trapped in a potential. $$U(x) = \frac{U_o}{(\frac{a}{a})^2 + 1}$$ where $$U_o > 1$$ and $$a>0$$. Assuming the amplitude is small, what would be the frequency of oscillation? My plan of attack was to use Taylor Expansion for $U(x)$ considering the particle remains close to equilibrium (x=a). $$U(x) = U(a) + U'(a)(xa)+ \frac{1}{2}U''(a)(xa)^2 + ... \approx U_o + \frac{1}{2}k(xa)^2$$ My problem now is trying to figure out frequency from this information. Perhaps there is some relation with Simple Harmonic Oscillation and a potential function. Thanks for any tips! 
October 5th, 2017, 12:56 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond 
Looks like there is a typo. You've got a over a squared plus one in the denominator of the righthand side.

October 5th, 2017, 05:09 PM  #3 
Senior Member Joined: Nov 2015 From: United States of America Posts: 181 Thanks: 21 Math Focus: Calculus and Physics  
October 5th, 2017, 06:16 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,266 Thanks: 1198  Quote:
$\dfrac{dU}{dx} \approx \dfrac{2 U_0 x}{a^2}$ and thus the restoring force is given by $\kappa = \dfrac{2U_0}{a^2}$ should be cookie cutter from here  
October 6th, 2017, 10:27 AM  #5  
Senior Member Joined: Nov 2015 From: United States of America Posts: 181 Thanks: 21 Math Focus: Calculus and Physics  Quote:
Wouldn't the restoring force be equal to the derivative of the potential function? What happened to the x in the in $\kappa = \dfrac{2U_0}{a^2}$ ? compared to $\dfrac{dU}{dx} \approx \dfrac{2 U_0 x}{a^2}$ where there is the x. Do we remove the x since we are assuming the amplitude is small?  
October 6th, 2017, 10:36 AM  #6  
Senior Member Joined: Sep 2015 From: USA Posts: 2,266 Thanks: 1198  Quote:
Yes, the force is the derivative of the potential. The "spring constant", $\kappa$ is what I gave you. With that constant you can determine the period of oscillation from the commonly known formula for simple harmonic oscillators with no damping.  
October 6th, 2017, 10:48 AM  #7  
Senior Member Joined: Nov 2015 From: United States of America Posts: 181 Thanks: 21 Math Focus: Calculus and Physics  Quote:
$F = \frac{2U_o}{a^2} = kx = ma$ Then considering... $\sqrt{\frac{k}{m}} = 2 \pi f$ $\Rightarrow \frac{k}{4\pi^2 f} = m$ I then multiply both sides by $a$ $\frac{ak}{4\pi^2 f} = ma = F$ Then plug back in and solve for f $ f = \sqrt{\frac{a^3 k}{8U_o \pi^2}}$ However, I have a constant k that is now in the formula that I did not have originally.  
October 6th, 2017, 11:05 AM  #8  
Senior Member Joined: Sep 2015 From: USA Posts: 2,266 Thanks: 1198  Quote:
and $k = \dfrac{2U_o}{a^2}$ so $f = \sqrt{\dfrac{a}{4\pi^2m}}$ now you are back to your original constants.... (you dropped the $m$ in the original) Last edited by romsek; October 6th, 2017 at 11:08 AM.  
October 7th, 2017, 11:30 AM  #9 
Senior Member Joined: Nov 2015 From: United States of America Posts: 181 Thanks: 21 Math Focus: Calculus and Physics  

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finding, frequency, oscillation 
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