Simple Harmonic Motion, masses, and springs.

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Homework Statement


Two blocks (m = 1.22 kg and M = 8.73 kg) and a spring (k = 344 N/m) are arranged on a horizontal, frictionless surface, with m resting on top of M and the spring attached to a wall and M. The coefficient of static friction between the blocks is 0.42. Find the maximum possible amplitude of the simple harmonic motion is no slippage is to occur between the blocks.


Homework Equations





The Attempt at a Solution


So, after drawing free body diagrams for both blocks, I get:

m: [tex]\sum F = -\mu_s m g = m a[/tex] or just [tex]a = -\mu_s g[/tex]

M: [tex]-k x = (M + m) a[/tex] or just [tex]a = -\frac{M + m}{k x}[/tex]

Setting these two equations equal, I finally get:

[tex]x = \frac{M + m}{k \mu_S g}[/tex]

However, when I plug in the values, I get the wrong answer. I think something may be wrong with my FBD's?

Thanks in advance.
 
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The acceleration is a function of time. Diffentiate the position function twice with respect to time and find the maximum acceleration. This expression will give the acceleration as a function of omega and amplitude.
 
Wouldn't the acceleration be the greatest at the maximum displacement? That's why I thought it would be find to only consider the acceleration at the point of maximum amplitude, x.

Also, I have two equations for the acceleration? which would I differentiate? Or does it matter?
 
Your method is fine; you just made a simple error.
csnsc14320 said:
M: [tex]-k x = (M + m) a[/tex]
OK.
or just [tex]a = -\frac{M + m}{k x}[/tex]
Oops... redo this step. (You would have caught this if you had checked units.)
 
Doc Al said:
Your method is fine; you just made a simple error.

OK.

Oops... redo this step. (You would have caught this if you had checked units.)

Oops! I need to slow down a little bit when doing my simple algebra :-P

I re-did it, checked my units (came out to meters),plugged it, and got the right answer.

Thanks