What Is the Maximum Amplitude of Harmonic Motion Without Slippage?

[ViewtifulBeau]

Two blocks (m = 0.468 kg and M = 2.41 kg) and a spring (k = 24.1 N/m) are arranged on a horizontal, frictionless surface. Block m is situated on top of block M. (Spring attached to block M) The coefficient of static friction between the two blocks is 0.873. What is the maximum possible amplitude of simple harmonic motion of the spring/blocks system if no slippage is to occur between the blocks?

Im not too sure what to do with this problem, I think i have to match up the normal force on the top block with the force of the spring on the bottom. but I am not sure.

 

[Astronuc]

In order for the large block (mass M) to move without slippage of the small block (mass m), the acceleration (a) of the small block must be such that the force F=ma, must not exceed \mumg.

The spring imposes a force on (M+m), and the spring force F_{spring[/sup] = kx.}

 

[ViewtifulBeau]

i did that and i got .166 meters. umg = .873*9.81*.468 = 4.008 N
F(spring) = kx
4.008 = 24.1 * x
x = .166 meters.

and this is not right. What am i doing wrong?

 

[andrevdh]

The block on top, m, are also experiencing SHM. The force that causes it to experience SHM is the frictional force. Since this force has an upper limit it puts a limit on the maximum acceleration the top block can experience. Are the acceleration of the two blocks the same during the motion?

 

[Astronuc]

F(spring) = kx
4.008 = 24.1 * x

The spring force must be applied to the combined mass (m+M) = 0.468+2.41 kg, but one needs the acceleration of the combined mass to determine F, from which one applies F = kx or x = F/k.

The problem then is - what is the acceleration of the combined blocks, knowing that the small block and large block accelerate together?