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Completing the Square

  • Thread starter denjay
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Homework Statement



Consider a charged particle of mass m in a harmonic potential and in the presence also of an
external electric field E = E[itex]\hat{i}[/itex]. The potential for this problem is simply

V(x) = 1/2 mw[itex]^{2}[/itex]x[itex]^{2}[/itex] - qεx

where q is the charge of the particle.

1) Show that a simple change of variables turns this problem into one of a particle under
only a harmonic oscillator potential. (Hint: Complete the square.)

Homework Equations



(ax-b)[itex]^{2}[/itex] = a[itex]^{2}[/itex]x[itex]^{2}[/itex] - 2abx + b[itex]^{2}[/itex]

The Attempt at a Solution



So I know the way to simplify the potential is by completing the square. I only know the way of completing the square when a quadratic equation is equal to 0 but in this case it's a function. So with that formula for (ax-b)[itex]^{2}[/itex] I believe 1/2 mw[itex]^{2}[/itex] is a[itex]^{2}[/itex] and -2ab is -qε but I'm unsure.

So what I got was that b = qε/(2mw[itex]^{2}[/itex]) so the equation is

V(x) = 1/2 mw[itex]^{2}[/itex]x[itex]^{2}[/itex] - qεx + q[itex]^{2}[/itex]ε[itex]^{2}[/itex]/2m[itex]w^{2}[/itex] - q[itex]^{2}[/itex]ε[itex]^{2}[/itex]/2m[itex]w^{2}[/itex] = ([itex]\sqrt{1/2 mw^{2}}x[/itex] - qε/[itex]\sqrt{2mw^2}[/itex])^2 - q[itex]^{2}[/itex]ε[itex]^{2}[/itex]/2m[itex]w^{2}[/itex]
 
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Answers and Replies

  • #2
LCKurtz
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Homework Statement



Consider a charged particle of mass m in a harmonic potential and in the presence also of an
external electric field E = E[itex]\hat{i}[/itex]. The potential for this problem is simply

V(x) = 1/2 mw[itex]^{2}[/itex]x[itex]^{2}[/itex] - qεx

where q is the charge of the particle.

1) Show that a simple change of variables turns this problem into one of a particle under
only a harmonic oscillator potential. (Hint: Complete the square.)

Homework Equations



(ax-b)[itex]^{2}[/itex] = a[itex]^{2}[/itex]x[itex]^{2}[/itex] - 2abx + b[itex]^{2}[/itex]

The Attempt at a Solution



So I know the way to simplify the potential is by completing the square. I only know the way of completing the square when a quadratic equation is equal to 0 but in this case it's a function. So with that formula for (ax-b)[itex]^{2}[/itex] I believe 1/2 mw[itex]^{2}[/itex] is a[itex]^{2}[/itex] and -2ab is -qε but I'm unsure.
You don't have to have an equation. Here's an example, complete the square on: ##2x^2-12x##. You factor out the ##2## getting ##2(x^2-6x)##. Now inside the parentheses you need a ##9## so add it and subtract it: ##2(x^2-6x + 9 - 9)## which is the same as ##2(x-3)^2 - 18##. Try something like that.
 
  • #3
77
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Cool yeah, I did it the way you suggested and got the same answer as I did in my original post. Thanks!
 

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