What Range of Frequencies Prevents a Mass-Spring System from Breaking?

[DualCortex]

(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:
[tex]X\text{''}+4*X=0.04*\cos (\omega *t)[/tex]

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

Find the range of [tex]\omega[/tex] 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

[tex]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}[/tex]

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

Thanks

 

[mighty2000]

for sharing your problem and solution attempt. It's always interesting to see how different people approach and solve problems.

To find the range of \omega in which the system doesn't break, we need to consider the maximum displacement of the mass from the equilibrium position. As you mentioned, the spring breaks if the displacement is more than 0.06 m from the equilibrium position.

From the equation for X(t), we can see that the maximum displacement occurs when the term \frac{0.04}{4-\omega ^2} is at its maximum value. This happens when \omega = 4, as the denominator becomes 0 and the term becomes undefined.

So, we can say that the system will not break if \omega < 4, as the maximum displacement will always be less than 0.06 m. However, if we consider the term \frac{0.04}{4-\omega ^2} to be approximately equal to 0.06, we can solve for \omega to get an approximate range.

0.06 = \frac{0.04}{4-\omega ^2}
0.06(4-\omega ^2) = 0.04
0.24 - 0.06\omega ^2 = 0.04
0.06\omega ^2 = 0.2
\omega ^2 = \frac{0.2}{0.06}
\omega ^2 = \frac{10}{3}
\omega = \sqrt{\frac{10}{3}}
\omega \approx 1.82574

So, we can say that the system will not break if 1.82574 < \omega < 4.

However, this is just an approximation and using MATLAB to get a more accurate range is a good idea. It's always important to double check our calculations and solutions.

Hope this helps! Let me know if you have any further questions or if I can assist in any other way.