Motion equation of moving particle

[qoqosz]

Hi,
assume that motion equation of moving particle are given by:

$$\ddot{x} + \omega^2 x = \frac{f}{m} \cos ( \gamma t + \beta )$$​

the solution is of course $$x = a \cos (\omega t + \alpha) + \tfrac{f}{m(\omega^2 - \gamma^2)} \cos ( \gamma t + \beta)$$.

My question is - what exactly happens when $$\omega = \gamma$$ ?

As $$\gamma \to \omega$$ but $$\omega \not= \gamma$$ the amplitude of oscillations grows rapidly, right?
We might guess that when $$\omega = \gamma$$ aplitude is infinite! But... let's right $$x$$ as:

$$x = b \cos (\omega t + \alpha) + \frac{f}{m(\omega^2 - \gamma^2)} \left\{ \cos ( \gamma t + \beta) - \cos (\omega t + \beta) \right\}$$​

.
(b is a new const). In the second term of the sum we have $$0/0$$ symbol and using de L'Hospital rule we obtain

$$x = b \cos (\omega t + \alpha) + \frac{f}{2m \omega} t \sin (\omega t + \beta)$$​

This function grows to infinity as $$t \to \infty$$ but not so fast and it of course has a finite amplitude.

So again - what exactly happens when $$\omega = \gamma$$ ?

 

[Dale]

Why don't you try to set them equal in the original equation and solve that.

 

[qoqosz]

I did that and got solution of the form $$x = b \cos (\omega t + \alpha) + \frac{f}{2m \omega} t \sin (\omega t + \beta)$$ again.
But I still wonder why when $$\gamma \to \omega$$ the amplitude can be very very large and when $$\gamma = \omega$$ it is totally different.