A 0.50 kg block sits on top of a 5.00-kg block that rests on the floor. The larger block is attached to a horizontal spring that has a force constant of 15.0 N/m. It is displaced and undergoes simple harmonic motion. What is the largest possible amplitude the 5.00-kg mass can have for the smaller mass to remain at rest relative to the larger block? The coefficient of static friction between the two blocks is 0.25. There is no friction between the larger block and the floor.(adsbygoogle = window.adsbygoogle || []).push({});

I have all the equations:

[tex] x(t) = A\cos(\omega(t) + \phi) [/tex]

[tex] v(t) = -A\omega\sin(\omega(t)+\phi) [/tex]

[tex] a(t) = -A\omega^{2}\cos(\omega(t)+\phi) [/tex]

[tex] T = \frac{1}{f} [/tex]

[tex] \omega = \sqrt{\frac{k}{m}} [/tex]

[tex] T = 2\pi\sqrt{\frac{m}{k}} [/tex]

I drew a free body diagram. I know that the velocity of the smaller block has to be 0. The force the spring exerts on the lower block has to be less than the friction force for the velocity to be 0.

Any help would be appreciated.

note: It should be [tex] \omega(t) [/tex]

Thanks

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# Homework Help: Simple Harmonic Motion of a block

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