Completing the Square Homework Statement

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The discussion revolves around simplifying the potential energy of a charged particle in a harmonic potential influenced by an external electric field. The key method suggested is completing the square to transform the potential into a standard harmonic oscillator form. Participants express uncertainty about applying the technique to a function rather than a quadratic equation set to zero. A successful example is provided, illustrating the completion of the square process. Ultimately, the approach leads to a consistent result, confirming the validity of the method.
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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\hat{i}. The potential for this problem is simply

V(x) = 1/2 mw^{2}x^{2} - 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)^{2} = a^{2}x^{2} - 2abx + b^{2}

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)^{2} I believe 1/2 mw^{2} is a^{2} and -2ab is -qε but I'm unsure.

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

V(x) = 1/2 mw^{2}x^{2} - qεx + q^{2}ε^{2}/2mw^{2} - q^{2}ε^{2}/2mw^{2} = (\sqrt{1/2 mw^{2}}x - qε/\sqrt{2mw^2})^2 - q^{2}ε^{2}/2mw^{2}
 
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denjay said:

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\hat{i}. The potential for this problem is simply

V(x) = 1/2 mw^{2}x^{2} - 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)^{2} = a^{2}x^{2} - 2abx + b^{2}

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)^{2} I believe 1/2 mw^{2} is a^{2} 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.
 
Cool yeah, I did it the way you suggested and got the same answer as I did in my original post. Thanks!
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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