(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

"A large block of mass m_{1}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 m_{2}rests upon m_{1}. The coefficient of friction between the two blocks is [itex]\mu[/itex]. Assume that m_{2}does not slip relative to m_{1}.

(c) what is the angular frequency of oscillations [itex] \omega [/itex] ?

(d) What is the maximum amplitude, A, that the system can have if m_{2}does not slip relative to m_{1}? (express in terms of [itex] \omega [/itex]

2. Relevant equations

F=-kx Hooke's Law

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

3. 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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# Homework Help: Review problems in Mechanics I

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