What is the maximum distance traveled by the block in a vertical spring system?

[Jediknight]

A is a distance, x_i is a position. Whether they're equal depends on your reference point for positions, which you've not defined.

at that time I was calling xeq the origin, but I think calling xrelaxed of spring without masses (which I would now like to denote as x0 If its ok with everybody) the origin will give me some more insight to this problem, as was suggested

 

[haruspex]

xeq2=2xeq1

Again, that is not meaningful unless you state where your origin is. For that specific statement to be true, your origin would have to be the bottom of the relaxed spring. I feel that this confusion between absolute and relative positions is a key part of your difficulties.

Ok, that post crossed with yours, defining the origin as the bottom of the relaxed spring.

 

[haruspex]

at that time I was calling xeq the origin, but I think calling xrelaxed of spring without masses (which I would now like to denote as x0 If its ok with everybody) the origin will give me some more insight to this problem, as was suggested

Ok, with that definition, all of your post #30 looks right. So, how about some equations?

 

[Jediknight]

I just wanted to say real quick THANK YOU ALL FOR PUTTING UP WITH ME SO FAR I should've defined variables better and specified reference frames better from the start, I'm new here and will do better next time and I think just thinking about that is an improvement to my problem solving skills SORRY I DIDN'T DEFINE THINGS BETTER SOONER

now ill check that last alert, but I probably won't respond til I go get my head around all this again, again thank you guys, I'm really trying here

I'm also happy to "give up" and get there correct answer if it will help in understanding how to get to it/defining what the problem was actually asking (my HW is a very low percentage of my grade, all its really therefor is understanding the material)

 

[Jediknight]

ok that didn't work, be back more organized without typo's don't have time to figure out how to fix that now, it looked good before I hit post

 

[Jediknight]

I keep running into contradictions, like 8mg=1 and others, I spent the better part of yesterday and all day today, I am going to go get the answer...

d2 = d+(2gΔt^2)/(π^2) which is d+g(ω2)^-2

so the extra acceleration is g (that makes sense kinda they pulled out and eliminated an m from ((2m)g-(m)g) somewhere

and that over ω2^2 gives you A2 but why would A2 simply be added on to d, idk I'm going to switch to relativity or damping constants, this was the last question in non-dissipative shm, if someone feels like laying out for me how you get there itd really help me, I know coming to it myself is better but with the pace of this class the times just not there :(

 

[haruspex]

I keep running into contradictions, like 8mg=1 and others, I spent the better part of yesterday and all day today, I am going to go get the answer...

d2 = d+(2gΔt^2)/(π^2) which is d+g(ω2)^-2

so the extra acceleration is g (that makes sense kinda they pulled out and eliminated an m from ((2m)g-(m)g) somewhere

and that over ω2^2 gives you A2 but why would A2 simply be added on to d, idk I'm going to switch to relativity or damping constants, this was the last question in non-dissipative shm, if someone feels like laying out for me how you get there itd really help me, I know coming to it myself is better but with the pace of this class the times just not there :(

That answer confirms the interpretation that the two masses are released from the same position.
Pick a reference point for gravitational PE.
In terms of the mass m, acceleration g, the relaxed length of the spring, the initial compression, and the spring constant k, write an expression for the initial PE.
Do the same for the PE at lowest point.
What can you deduce?
You can express k in terms of m and $\Delta t$. From all that, find the initial compression. Then move on to the 2m mass.