# 2 Masses attached by a spring in the vertical axis

• Sam Jelly
Sam Jelly
Homework Statement
Two identical masses m are attached by a spring of constant k in the vertical axis.
Determine the minimum compression of the spring so that when it is released the bottom mass is slightly off the ground.
Relevant Equations
MAcm = Fexternal
I am trying to solve this problem without using energy conservation. How can I solve this using the 2 body system and center of mass.
I know that in order for mass m at the bottom to be slightly off the ground the reaction force must be equal to zero. I don’t know if the reaction force before releasing the spring is equal to after it is released.

Sam Jelly said:
Homework Statement: Two identical masses m are attached by a spring of constant k in the vertical axis.
Determine the minimum compression of the spring so that when it is released the bottom mass is slightly off the ground.
Relevant Equations: MAcm = Fexternal

I am trying to solve this problem without using energy conservation. How can I solve this using the 2 body system and center of mass.
I know that in order for mass m at the bottom to be slightly off the ground the reaction force must be equal to zero. I don’t know if the reaction force before releasing the spring is equal to after it is released.View attachment 342587
The normal force is decreasing as the spring elongates. The tension force in the spring changes direction once the free length of the spring is passed. ( i.e. the spring goes from compression to tension), just make sure your math makes sense in that regard.

The problem statement seems to be erroneous.

The bottom mass is not supposed to be slightly off the ground at the moment of release. It is supposed to be slightly off the ground after the top mass has sprung into the air, when it is exerting maximum tension before coming back down.

erobz
We’ll, at least the intent of this one is clear.

haruspex said:
Not sure that would help much anyway.
Using energy conservation must be the simplest approach.

erobz and haruspex
PeroK said:
Using energy conservation must be the simplest approach.
Unless I am missing something, the motion after release will be simple harmonic motion. This allows one to almost read off the answer without invoking energy conservation.

jbriggs444 said:
Unless I am missing something, the motion after release will be simple harmonic motion. This allows one to almost read off the answer without invoking energy conservation.
Yeah, after trying to set it up I came to a simple answer too.

jbriggs444 said:
Unless I am missing something, the motion after release will be simple harmonic motion. This allows one to almost read off the answer without invoking energy conservation.
I must admit I don't see it myself. That said, I automatically did the problem with different masses, because that's the way I tend to do things!

jbriggs444 said:
Unless I am missing something, the motion after release will be simple harmonic motion. This allows one to almost read off the answer without invoking energy conservation.
erobz said:
Yeah, after trying to set it up I came to a simple answer too.
Me too three. One has to realize that, when the top mass is released from rest at distance ##A## below the equilibrium position, it will execute simple harmonic motion according to ##y=-A\cos\omega t## and that the normal force exerted by the support on the bottom mass is ##N=2mg## when the top mass is at the equilibrium position, whether moving or not.

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