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For an assignment in one of my maths units at uni, I've been asked to derive and solve the differential equation of motion for a forced harmonic oscillator, with the forcing function having the form $F_0sin(\omega t)$, with some given properties (mass = 1kg, spring constant = 400N/m, amplitude of the sinusoidal driving force = 20N).

I am reasonably confident of the solution I found;

$$x(t) = -0.0245\cos(20t) + \frac{\omega}{\omega^2 - 400}\sin(20t) + \frac{20}{400 - \omega^2}\sin(wt) + 0.0245$$

Next the question tells us that the spring will fail if its extension exceeds 1 metre, and asks us what forcing frequencies will allow for safe oscillations.

I formulated this condition as;

$$1 \geq |-0.0245\cos(20t) + \frac{\omega}{\omega^2 - 400}\sin(20t) + \frac{20}{400 - \omega^2}\sin(wt)+0.0245|$$

Obviously this condition needs to hold for all t, so it seems to me that I need only look at the sum of the amplitudes, however $\cos(20t)$ and $\sin(20t)$ will never superimpose entirely constructively, and I cannot find an adequate way of expressing them as one trig function.

How should I approach this?

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Hint: $cos(\theta) = sin(\theta+\frac{\pi}{2})$ and there are identities for $sin(\theta) \pm sin(\phi)$ – FrankH Oct 16 '11 at 6:04
Unfortunately combining all of the trig terms into one does not yield anything that I can use. – Daniel Blay Oct 16 '11 at 8:24
combine the two "20" terms, and then use the fact that $\omega$ and 20 are not going to stay in phase, so that you can just add the amplitudes of these two terms to get the maximum amplitude of oscillation. – Ron Maimon Oct 16 '11 at 11:40

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