# Homework Help: Compression in Accelerating Springs.

1. Sep 26, 2010

### jimmyw

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

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

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

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

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

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

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

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/" [Broken] where it is said the compression is "the compression is EXACTLY the same before"

Last edited by a moderator: May 4, 2017
2. Sep 27, 2010

### The legend

3. Sep 27, 2010

### 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