Help with the phase of the solution for a driven oscillator

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SUMMARY

The discussion focuses on the solution phase of the undamped driven harmonic oscillator, specifically analyzing the equation of motion $$\ddot{x} + {\omega_{0}} ^{2}x = \frac{F_{0}}{m} \cos{\omega t}$$. The complementary solution is established as $$x = A\cos{(\omega_{0}t + \phi)}$$, while the particular solution can be derived using either $$x = B\cos{\omega t}$$ or $$x = B\cos{(\omega t + \psi)}$$. Both approaches yield the same result, demonstrating that the phase shift can be ignored in the absence of damping, as it does not affect the outcome of the solution.

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etotheipi
Homework Statement
Obtain the general solution for an undamped harmonic oscillator with a forcing term proportional to a cosine wave, with a maximum of force at t = 0.
Relevant Equations
N/A
My question also applies to the damped driven oscillator, however for simplicity I will first consider an undamped oscillator.

The equation of motion is $$-kx + F_{0} \cos{\omega t} = m \ddot{x}$$ or in a more convenient form $$\ddot{x} + {\omega_{0}} ^{2}x = \frac{F_{0}}{m} \cos{\omega t}$$The auxiliary equation is {\lambda}^{2} + {\omega_{0}} ^{2} =0, which has solutions \lambda = \pm i \omega_{0}. So the complementary solution is $$x = A\cos{(\omega_{0}t + \phi)}$$Now for the particular solution. For a non-homogenous part relating to sine or cosine, the ansatz is of the form of a sine or cosine of the same argument plus a phase. However, most texts I've read just choose x = B\cos{\omega t}, which of course cancels out more nicely.

If we just use x = B\cos{\omega t}, the particular solution comes out to be x = \frac{F_{0} \cos{\omega t}} {m(\omega_{0} ^{2} - \omega ^{2})}.

However, if we choose the ansatz x = B\cos{(\omega t + \psi)} which is equivalent to x = Re(Be^{i(\omega t + \psi)}), we get $$- B \omega ^{2} e^{i \psi} + B \omega_{0} ^{2} e^{i \psi} = \frac{F_{0}}{m}$$ and thus B = \frac{F_{0} e^{-i \psi}}{m(\omega_{0} ^{2} - \omega ^{2})}. When we put this together, the e^{-i \psi} will cancel out and we end up with the same particular solution as before when we take the real part.

My question is why are we allowed to ignore the phase of the guess for the particular solution and still obtain the correct answer?
 
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etotheipi said:
My question also applies to the damped driven oscillator, however for simplicity I will first consider an undamped oscillator.

The equation of motion is

−kx+F0cosωt=m¨x​
This is the equation of motion for the undamped driven harmonic oscillator.

Note that the particular solution should have frequency ##\omega##, not ##\omega_0## and that there is no phase shift in absence of a damping term (if you input a phase shift it will come out to be zero).
 
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Orodruin said:
This is the equation of motion for the undamped driven harmonic oscillator.

Note that the particular solution should have frequency ##\omega##, not ##\omega_0## and that there is no phase shift in absence of a damping term (if you input a phase shift it will come out to be zero).

Yes sorry about the ##\omega_0##, it was a typo.

That makes perfect sense. Thanks!
 

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