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As per a previous question:

Transient behavour For a driven harmonic oscillator:

I was trying to show an exponential increase in amplitude using the transient solution, however I still got the sake result that the amplitude is infinitely large, even at the start. Can anyone figure out where I went wrong?

$$m(\ddot{x_{1}}+\omega_0^2x_{1})=F_0\cos(\omega t) \tag{1}$$

$$m(\ddot{x_{2}}+\omega_0^2x_{2})= 0 \tag{2}$$

(2) Is the homogenous version of the differential equation for a driven harmonic osscillator.

Any solution to (1) can be added with a solution to (2) to give a new solution to (2)

Since (1) + (2) is in the same form as (1) with the solution being $(x_{1} + x_{2})$

Solving (2)

Using the trial function

$x= Ae^{i(bt+\phi)}$

$\frac{d^2x}{dt^2} =-b^2 Ae^{i(bt+\phi)}$

Substituting into (2):

$m(-b^2 Ae^{i(bt+\phi)} +\omega_0^2 Ae^{i(bt+\phi)})= 0 $

$(A)(-b^2 +\omega_0^2 ) = 0 $

$b = \omega_0$

$x= Ae^{i(\omega_0 t+\phi)}$

We only want the real part, to be a physical solution

$x= Acos(\omega_0 t +\phi)$

Added with the inhomogenous solution

Gets us:

$$ x= Acos(\omega_0 t +\phi) + \frac{F_0} {m(\omega_0^2-\omega^2)}\cos(\omega t) $$

Breaking the first half into

$Acos(\omega_0 t )cos(\phi)-Asin(\omega_{0} t )sin(\phi)$

And substituting:

$ x = Acos(\omega_0 t )cos(\phi)-Asin(\omega_{0} t )sin(\phi) + \frac{F_0} {m(\omega_0^2-\omega^2)}\cos(\omega t)$

Factoring $cos(\omega t)$

$(Acos(\phi) + \frac{F_0} {m(\omega_0^2-\omega^2)})cos(\omega t) -Asin(\omega_{0} t ) $

We need to simplify A,$\phi$ in terms of my initial displacement $x_{0}$ and initial velocity $v_{0}$

Look back at x, substitute $x(0)= x_{0}$

We see that (a)

$(Acos(\phi) + \frac{F_0} {m(\omega_0^2-\omega^2)}) = x_{0}$

Substituting this into our factored formula gives

$x = x_{0}cos(\omega t ) -Asin(\omega_{0} t )$

Substituting $\frac{dx}{dt}(0)= v_{0}$

We see that (b)

$-Asin(\phi)\omega_0 = v_{0}$

Solving this system of equations of (a) and (b) i find that:

A is infinity.

Where in my derivation have I gone wrong? A is not a function of time, and thus there is no increase in amplitude as time goes on?

Qmechanic
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jensen paull
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