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

[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.