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Elwin.Martin
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Homework Statement
"A large block of mass m1 executes horizontal SHM as it slides across a friction-less surface under the action of a spring with spring constant k. A block of mass m2 rests upon m1. The coefficient of friction between the two blocks is [itex]\mu[/itex]. Assume that m2 does not slip relative to m1.
(c) what is the angular frequency of oscillations [itex] \omega [/itex] ?
(d) What is the maximum amplitude, A, that the system can have if m2 does not slip relative to m1? (express in terms of [itex] \omega [/itex]
Homework Equations
F=-kx Hooke's Law
[itex]\omega = \sqrt{\frac{k}{m}}[/itex]
The Attempt at a Solution
I'm reviewing for a transfer exam and I'm going through their old finals problems. This came up and I'm hoping I'm not seriously oversimplifying here.
c. [itex]\omega = \sqrt{\frac{k}{m_1+m_2}}[/itex]
d. [itex]k(A)= \mu N [/itex] where N is the normal force from the contact of the blocks
[itex]kA= \mu m_2 g [/itex]
[itex]A= \frac{\mu m_2 g}{k} [/itex]
[itex]A= \frac{\mu m_2 g}{k} [/itex]
From [itex]\omega = \sqrt{\frac{k}{m_1+m_2}}[/itex] , [itex] \frac{m_2}{k} = \frac{m_2}{\left(m_1+m_2\right)\omega^2}[/itex]
[itex]A= \frac{\mu m_2 g}{\left(m_1+m_2\right)\omega^2} [/itex]
...This comes across as odd to me, if I had a normal force which result from both masses, I'd be considerably happier. I see no reason to express the answer in terms of [itex] \omega[/itex] when it comes out kind of ugly. The only explanation I have is that there is some contribution to the normal force I'm neglecting or I made some careless mistake elsewhere.
Any and all help would be great! Just trying to re-learn basic mechanics ><
Edited to add an m_2 where it needed to be.
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