My Math Forum Finding frequency of oscillation

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 October 5th, 2017, 11:44 AM #1 Senior Member     Joined: Nov 2015 From: United States of America Posts: 176 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). $$U(x) = U(a) + U'(a)(x-a)+ \frac{1}{2}U''(a)(x-a)^2 + ... \approx U_o + \frac{1}{2}k(x-a)^2$$ 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, 11:56 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 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 right-hand side. Thanks from SenatorArmstrong
October 5th, 2017, 04:09 PM   #3
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Quote:
 Originally Posted by greg1313 Looks like there is a typo. You've got a over a squared plus one in the denominator of the right-hand side.
Embarrassing... Thanks for telling me!

Should be...

$$U(x) = \frac{-U_o}{(\frac{x}{a})^2 + 1}$$

October 5th, 2017, 05:16 PM   #4
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Quote:
 Originally Posted by SenatorArmstrong Embarrassing... Thanks for telling me! Should be... $$U(x) = \frac{-U_o}{(\frac{x}{a})^2 + 1}$$
$U(x) \approx -U_0+\dfrac{U_0 x^2}{a^2}$

$\dfrac{dU}{dx} \approx \dfrac{2 U_0 x}{a^2}$

and thus the restoring force is given by

$\kappa = \dfrac{2U_0}{a^2}$

should be cookie cutter from here

October 6th, 2017, 09:27 AM   #5
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Quote:
 Originally Posted by romsek $U(x) \approx -U_0+\dfrac{U_0 x^2}{a^2}$ $\dfrac{dU}{dx} \approx \dfrac{2 U_0 x}{a^2}$ and thus the restoring force is given by $\kappa = \dfrac{2U_0}{a^2}$ should be cookie cutter from here
Thanks for the help romsek.

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, 09:36 AM   #6
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Quote:
 Originally Posted by SenatorArmstrong Thanks for the help romsek. 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?
I'm sorry I guess I wasn't clear.

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, 09:48 AM   #7
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Quote:
 Originally Posted by romsek $U(x) \approx -U_0+\dfrac{U_0 x^2}{a^2}$ $\dfrac{dU}{dx} \approx \dfrac{2 U_0 x}{a^2}$ and thus the restoring force is given by $\kappa = \dfrac{2U_0}{a^2}$ should be cookie cutter from here
Do you think this is an acceptable way to relate frequency and restoring force?

$F = \frac{2U_o}{a^2} = -kx = ma$

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, 10:05 AM   #8
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Quote:
 Originally Posted by SenatorArmstrong Do you think this is an acceptable way to relate frequency and restoring force? $F = \frac{2U_o}{a^2} = -kx = ma$ 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.
...

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 10:08 AM.

October 7th, 2017, 10:30 AM   #9
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Quote:
 Originally Posted by romsek ... 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)

Crystal clear. Thank you for your help as always!

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