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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?

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  • $\begingroup$ I haven't checked your whole derivation (it's rather a tedious bit of ODE math) but the amplitude goes to $\infty$ when $\omega_0=\omega$ for a driven oscillator, so I'm guessing that's where your mistake lies, somewhere. $\endgroup$
    – Gert
    Apr 17 at 13:21
  • $\begingroup$ I believe I have found the error, which I haven't checked yet. I't involves me assuming the standard steady state solution for harmonic osscillators is valid for resonance. The formula actually breaks down for resonance as normal substitution into the differential equation actually yields no solution for the special case where omega is resonance. Which I was not aware of at the time. $\endgroup$ Apr 17 at 13:23
  • $\begingroup$ Link to previous question? $\endgroup$
    – Qmechanic
    Apr 17 at 13:27
  • $\begingroup$ physics.stackexchange.com/q/704112 sals answer is what cleared my doubt. Didn't read it at the time. Very cool though that at resonance even the steady state solution needs modifying. $\endgroup$ Apr 17 at 13:29
  • $\begingroup$ If you are at resonance and you do not have damping, then you are constantly pumping energy into the system, at all times, and your solutions need to account for this. If you do not want to include a damping term (which is the only way to have a steady state driven solution at resonance) then you are missing a solution of the inhomogeneous problem, which is of the form $t\cos(\omega t) $. $\endgroup$ Apr 17 at 15:05

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