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Simple Harmonic Motion--Quadratic Perturbation


Given a simple harmonic oscillator with a quadratic perturbation, write the perturbation term in the form alphaepsilonx^2,

 x^..+omega_0^2x-alphaepsilonx^2=0,
(1)

find the first-order solution using a perturbation method. Write

 x=x_0+epsilonx_1+...,
(2)

and plug back into (1) and group powers to obtain

 (x^.._0+omega_0^2x_0)+(x^.._1+omega_0^2x_1-alphax_0^2)epsilon+2alphax_0epsilon^2+....
(3)

To solve this equation, keep terms only to order epsilon^2 and note that, because this equation must hold for all powers of epsilon, we can separate it into the two simultaneous differential equations

x^.._0+omega_0^2x_0=0
(4)
x^.._1+omega_0^2x_1=alphax_0^2.
(5)

Setting our clock so that x_0(0)=0, the solution to (4) is then

 x_0=Acos(omega_0t).
(6)

Plugging this solution back into (5) then gives

 x^.._1+omega_0^2x_1=alphaA^2cos^2(omega_0t).
(7)

The equation can be solved to give

 x_1=(alphaA^2)/(6omega_0^2)[3-cos(2omega_0t)]+C_1cos(omega_0t)+C_2sin(omega_0t),
(8)

Combining x_0 and x_1 then gives

x(t)=x_0+epsilonx_1
(9)
=Acos(omega_0t)-(alphaA^2)/(6omega_0^2)epsilon[cos(2omega_0t)-3],
(10)

where the sinusoidal and cosinusoidal terms of order epsilon (from the x_1) have been ignored in comparison with the larger terms from x_0.

SHOPerturbed

As can be seen in the top figure above, this solution approximates x(t) only for epsilon<<1. As the lower figure shows, the differences from the unperturbed oscillator grow stronger over time for even relatively small values of epsilon.


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Cite this as:

Weisstein, Eric W. "Simple Harmonic Motion--Quadratic Perturbation." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/SimpleHarmonicMotionQuadraticPerturbation.html

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