Recent content by 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...

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

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...

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

xi=the value given to the starting point based on the choice of origin, thinking about it your way, which I'm thinking is best but still struggling, xi will be the same value for both set ups, but xi≠A in this frame xi-xeq=A , and xeq2=2xeq1 [(also this means x at time t or x≠Acos(ω(t), but...

calling A (d/2) gave me some contradictions but maybe I was just getting jammed up, starting over with xrelaxed as the origin, that makes xi the same value in both problems, but that also means xi≠A, but as you said that doesn't help us anyway -Starting with a cos function this time, as it...

but that would make xi=A, but that would make the energy at the tp greater than the energy at the bottom (due to gravitational potential energy), when I look at the ΣF at each point your statement makes sense, I'm not saying your wrong just that I'm not getting the concept yet, at least when I...

i feel like xi is a fixed distance from a stationary reference point (such as the ceiling) for both experiments (but will be a different value for each expiriment based on me choosing xeq as the origin) at xi U=(1/2)mω^2xi^2+mgd=E xi=d-A sin(Φi)= xi/A ∑F=-kxi-mg at xΔt U=(1/2)/ω^2A^2=E...

me too, I am loving and hating this problem, and yea i agree won't be 2A, cause there's gravitational and elastic potential energy at xi, and only elastic at xΔt I'm ... going to sleep now, time to hit it hard tomorrow, got to get down the last bit of shm, jump back to relativity, then review...

yea, but I've si so are you saying d doesn't equal 2A, I was really starting to think it would, some thinking is in order i wish I didn't have to learn this this fast

OK that's what was ment by symetry, for some reason I was thinking it would bounce back up higher than that!, ok that makes sense, gravitys pulling it down harder but it also pulls it down with the same acceleration on the way up, does that mean a verticle spring has the same amplitude and...

AHHHH I guess I should switch gears do some re-redimg and come back to this, idk what "bumping" a thread is but if anyone on here already could keep tabs on it itd be cool I got d in terms of k m and g with an energy, acceleration, position system but i don't think that's helping me, might post...

taking a minute to think I know my previous statement of ω being constant is false,since ω^2=k/m i know d-xi=A xi=A sin(ω(0)+Φi) ... 1+(d/A)=sin(Φi) ΣFi=-kxi-mg=-mAω^2sin(ω(0)+Φi) k(A+d)+mg=mAω^2(1+(d/A)) i think I'm starting to get somewhere, is this problem as hard as it seems?

ok ill be back in a minute, getting out a fresh piece of paper i know same spring means k is constant I know ω is independant of mass, so its constant i figure Φi is not constant because xeq and A will be different I know there must be a simple relationship between A and F and there fore...

Homework Statement The hand in (Figure 1) (a) holds a vertical block-spring system so that the spring is compressed. When the hand let's go, the block, of mass m, descends a distance d in a time interval Δt before reversing direction in (Figure 1) (b). Once the system comes to rest, the...