## DE - mass-spring system

(I had made a thread with a problem similar to this one, but it turned a bit messy after finding out the professor made some mistakes and the wording of his problem was awkward ... however, this is it)

Problem:

Differential equation governing a forced, mass-spring system:
$$X\text{''}+4*X=0.04*\cos (\omega *t)$$

Spring constant = 4
Mass = 1Kg
Mass starts from rest, at equilibrium

Find the range of $$\omega$$ in which the system doesn't break given that the spring breaks if stretched more than 0.06 m from the equilibrium position.

Attempt:

Equilibrium position = 2.45 m

$$X(t) = \left(2.45-\frac{0.04}{4-\omega ^2}\right) \text{cos}(2*t)+\frac{0.04 *\text{cos}(t *\omega )}{4-\omega ^2}$$

So, now, how would I find all $$\omega$$ such that $$X(t) > 2.51 = 2.45 + 0.06$$ for all $$t$$?
Don't know of any exact way of calculating this. Used MATLAB to approximate the range of \omega's and it is ~ $$1.997379 < \omega < 2.303155$$

Thanks
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