Compression in Accelerating Springs.

[jimmyw]

Suppose I have two masses with mass [tex]M[/tex] each connected together by a spring with spring constant [tex]k[/tex] like this:
[PLAIN]http://img827.imageshack.us/img827/4896/springs1.png

Then I apply a force of [tex]F[/tex] at both ends:
[PLAIN]http://img833.imageshack.us/img833/2562/spring2.png

Then using Hooke's law [tex]F=kx[/tex], the compression of the spring [tex]x[/tex] would be given by:
[tex]x=F/k[/tex]

So now if I apply the same force [tex]F[/tex] just to the right end like this:
[PLAIN]http://img810.imageshack.us/img810/4236/spring3.png

The acceleration of the system is given by Newton's 2nd Law [tex]F=ma[/tex]
[tex]F=2Ma [/tex]
[tex]a=\frac{F}{2M}[/tex]

And the acceleration of the mass on the left is the same as the entire system so the force is:
[tex]F_{left mass}=Ma [/tex]
[tex]F_{left mass}=M\frac{F}{2M}[/tex]
[tex]F_{left mass}=\frac{F}{2}[/tex]

This force must be supplied by the spring so using hookes law:
[tex]\frac{F}{2}=kx[/tex]
[tex]x=\frac{F}{2k}[/tex]

This value is half as much as before.

Here is where I'm confused can anyone explain the difference between this and the situation described in this post http://scienceblogs.com/dotphysics/2008/10/fake-vs-real-forces/" where it is said the compression is "the compression is EXACTLY the same before"

What mistakes have I made?

 

[The legend]

I am not sure ... but its wrong i believe because when you push the mass on right .. the mass on left doesn't have same acceleration until the string can't compress anymore.

Still i would prefer someone more of an expert to justify this.

 

[jimmyw]

I'm not worried about that statement.

After the spring has compressed, the acceleration of the mass on the left is the same as the entire system