- #1

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 [itex]{\lambda}^{2} + {\omega_{0}} ^{2} =0[/itex], which has solutions [itex]\lambda = \pm i \omega_{0}[/itex]. 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

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

However, if we choose the ansatz [itex]x = B\cos{(\omega t + \psi)}[/itex] which is equivalent to [itex]x = Re(Be^{i(\omega t + \psi)})[/itex], we get $$- B \omega ^{2} e^{i \psi} + B \omega_{0} ^{2} e^{i \psi} = \frac{F_{0}}{m}$$ and thus [itex] B = \frac{F_{0} e^{-i \psi}}{m(\omega_{0} ^{2} - \omega ^{2})}[/itex]. When we put this together, the [itex]e^{-i \psi}[/itex] 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?

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 [itex]{\lambda}^{2} + {\omega_{0}} ^{2} =0[/itex], which has solutions [itex]\lambda = \pm i \omega_{0}[/itex]. 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 [itex]x = B\cos{\omega t}[/itex], which of course cancels out more nicely.If we just use [itex]x = B\cos{\omega t}[/itex], the particular solution comes out to be [itex]x = \frac{F_{0} \cos{\omega t}} {m(\omega_{0} ^{2} - \omega ^{2})}[/itex].

However, if we choose the ansatz [itex]x = B\cos{(\omega t + \psi)}[/itex] which is equivalent to [itex]x = Re(Be^{i(\omega t + \psi)})[/itex], we get $$- B \omega ^{2} e^{i \psi} + B \omega_{0} ^{2} e^{i \psi} = \frac{F_{0}}{m}$$ and thus [itex] B = \frac{F_{0} e^{-i \psi}}{m(\omega_{0} ^{2} - \omega ^{2})}[/itex]. When we put this together, the [itex]e^{-i \psi}[/itex] 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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