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Consider a mass-spring system with $m = 1$, no damping, and $k = 4$. Find the general form of the mass’ motion if there is no forcing. Now suppose that we apply an external force $F(t) = 3\cos^3(\omega t)$ for a constant $\omega$. Show that there are two values of $\omega$ at which resonance occurs, and find both.

$m = 1$

$k = 4$

$my'' + ky = 0$

$y'' +4y = 0$

$y(t) = c_1\cos (2t) + c_2\sin (2t)$

Now I'm unsure how to show/find the two values for resonance. I know the equation is

$y'' + 4y = 3\cos^3(\omega t)$

but what does that tell me?

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Hint: $\cos^3(\omega t) = \cos(3 \omega t)/4 + 3\cos(\omega t)/4$. You get resonance in $m y'' + k y = \cos(r t)$ when $\cos(r t)$ is a solution of the homogeneous equation.

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  • $\begingroup$ So resonance will occur at 1/4 and 3/4? $\endgroup$ – user1038665 Oct 31 '12 at 5:22
  • $\begingroup$ No. The important thing is the frequency, not the amplitude. $\endgroup$ – Robert Israel Oct 31 '12 at 6:54
  • $\begingroup$ So is this when $3\omega = 2$, or $\omega = \frac{2}{3}$, and when $\omega = 2$? $\endgroup$ – user1038665 Nov 1 '12 at 4:04
  • $\begingroup$ Yes, that's right. $\endgroup$ – Robert Israel Nov 1 '12 at 7:01

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