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transmini

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## Homework Statement

A damped harmonic oscillator is driven by an external force of the form $$F_{ext}=F_0sin(\omega t)$$

Show that the steady state solution is given by $$x(t)=A(\omega)sin(\omega t-\phi)$$

where $$ A(\omega)=\frac{F_0/m}{[(\omega_0^2-\omega^2)^2+4\gamma^2\omega^2]^{1/2}} $$

and $$tan\phi=\frac{2\gamma\omega}{\omega_0^2-\omega^2}$$

## Homework Equations

Equation of motion: $$\ddot{x}+2\gamma\dot{x}+\omega_0^2x=\frac{F_0}{m}sin(\omega t)$$

## The Attempt at a Solution

Using a characteristic equation ##\lambda^2+2\gamma\lambda+\omega_0^2## I found the solution to the equation of motion to be $$x(t)=Acos(\omega t)+Bsin(\omega t)$$ with another term that goes to zero at steady state, which we are considering only steady state. ##A## and ##B## are both arbitrary constants.

Then using the steady state equation found, I plugged this back into the equation of motion to find the coefficients.

$$\dot{x} = -A\omega sin(\omega t)+B\omega cos(\omega t)$$

$$\ddot{x} = -A\omega^2cos(\omega t)-B\omega^2sin(\omega t)$$

so

$$-A\omega^2cos(\omega t)-B\omega^2sin(\omega t)-2\gamma A\omega sin(\omega t)+2\gamma B\omega cos(\omega t)+\omega_0^2Acos(\omega t)+\omega_0^2Bsin(\omega t) = \frac{F_0}{m}sin(\omega t)$$

grouping sines and cosines:

$$(-A\omega^2+2\gamma B\omega+\omega_0^2)cos(\omega t)+(-B\omega^2-2\gamma A\omega+\omega_0^2B)sin(\omega t) = \frac{F_0}{m}sin(\omega t)$$

Since the coefficients for cosines on the left have to match the ones on the right, similarly for the sines, I got

$$-A\omega^2+2\gamma B\omega+\omega_0^2A = 0$$

$$-B\omega^2-2\gamma A\omega+\omega_0^2B=\frac{F_0}{m}$$

Then solving for the coefficients A and B, I got

$$A=\frac{-2F_0\gamma\omega}{m(\omega_0^2-\omega^2)^2+4m\gamma^2\omega^2}$$

$$B=\frac{F_0(\omega_0^2-\omega^2)}{m(\omega_0^2-\omega^2)^2+4m\gamma^2\omega^2}$$

Which when combining the sine and cosine term into ##A(\omega)cos(\omega t-\phi)## where ##A(\omega)=\sqrt{A^2+B^2}## and ##tan(\phi)=\frac{B}{A}##,

I receive exactly what I was supposed to get with the exception that I get $$tan(\phi)=\frac{-2\gamma\omega}{\omega_0^2-\omega^2}$$ rather than the positive version of the argument in the trig function.

I got told that "the method is wrong" and the " 'A' you received is not the same as the one we want". I don't see how the

*method*is wrong, seeing as how it does in fact give

*a*steady-state solution to the equation of motion (I did verify this to be sure). When using complex roots you get the correct signs on all the terms, but where did I go wrong with solving the differential equation using this method? If this method actually does not work, why does it not work when it was how we were taught in an Elementary Differential Equations course?

I can post the finding of the coefficients if need be, but since it was a few fairly ugly matrices, I left it out of the initial post.

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