Predictions for the spring system (broken?) Here is a question that I'm working on:

Let's suppose that we have an undamped spring-mass system that abides by the second order differential equation $$\frac{d^2y}{dt^2} + 4y = 3 \cos(2t).$$
This system is periodically drive by a force $$F(t) = 3 \cos (2t).$$
Explain why we should expect the spring to eventually break. How will our results differ if we add a small amount of friction into the system?

Since we are working with an undamped system, that means that our friction coefficient, call it b, will equal $0$. What's interesting is that the force $F(t)$ has the same value as the RHS of our second order differential equation. How can I show, through physical computations, that my spring will eventually break from my system. There must be a threshold of some sort (which is given by our equation describing the system). Any input for this imposed problem would be greatly appreciated.
 A: The fact that the RHS is equal to $F(t)$ is not surprising: remember that this just Newton second's law in disguise $F = m a$
$$
m \frac{{\rm d}^2 y}{{\rm d}t^2} = F_{\rm restoring} + F_{\rm drive} + F_{\rm damping} = -\omega^2_0 m y + F_{\rm drive} + F_{\rm damping} \tag{1}
$$ 
If you drop the damping force and rearrange terms you end up with the equation 
$$
\frac{{\rm d}^2 y}{{\rm d}t^2} + \omega_0^2 y = F_{\rm drive} \tag{2}
$$
To form some intuition about it, imagine $F_{\rm drive} = 0$, in this case, you have an harmonic oscillator with frequency $\omega_0$, that is, the position of mass at any time can be described as $y(t) = A_0\sin(\omega_0 t + \phi_0) $. $\omega_0$ is then the frequency at which the system spring+mass "likes" to move, that is, if there's no damping and not driving external force the mass will move at that frequency, in your case $\omega_0 = 2$.
Now, include a forcing term which periodically forces the mass with the same natural frequency $\omega_0$. What'd you expect? Since the mass likes to move at this frequency and you force at the same rate, then amplitude will increase with time until the spring reaches its plastic phase and breaks.
More formally, it is possible to show that is the forcing term is of the form $F_{\rm drive} = F_0m^{-1} \sin(\omega t)$ then the steady state solution to Eq. (1) is
$$
y(t) = \frac{F_0}{m\omega Z_m}\sin(\omega t + \phi) \tag{3}
$$
where
$$
Z_m^2 = \frac{(\omega_0 - \omega)^2}{\omega^2}
$$
You can immediately conclude that the amplitude of motion will blow up if the forcing frequency $\omega$ equals the natural frequency $\omega_0$
