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.