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

[Jediknight]

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 block is removed, a block of mass 2m is hung in its place, and the experiment is repeated.

to what maximum distance does the block descend?
Express your answer in terms of π, acceleration due to gravity g, some or all of the variables m, d, and Δt.

Homework Equations

k(xeq-x0)=mg
the basics for shm
x=Asin(ωt+Φ)...a=Aω^2sin(ωt+Φ)

The Attempt at a Solution

k(xeq-x0)=mg
k(xeq2-x0)=2mg
...xeq2-x0=2(xeq-x0)
so I know xeq for 2m will be twice as far from x0 of spring as for m, but x0 is not given

and I thought I just calculated 2m would drop twice as far past xeq than m (but I can't seem to find those so they could be wrong)

I havn't got my head around this problem yet but its turning into a monster and
I'm running out of time so any guidance would be apreciated

 

[haruspex]

You have no information regarding x_eq. For each mass, you have an initial state in which the spring is partially compressed, and a later state in which the mass is at its lowest point.
Create an unknown for the initial compression and see what equations you can obtain.

 

[gneill]

Hmm. The problem says:

the block, of mass m, descends a distance d in a time interval Δt before reversing direction

You should be able to conclude some things about the amplitude of the motion, the new equilibrium point for that motion, and the period of the oscillations. An expression for the spring constant might not be out of the question...

 

[Jediknight]

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 between A and m, but I'm not seeing it yet

ok clear head, close computer, start from scratch

 

[Jediknight]

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?

 

[gneill]

i think I'm starting to get somewhere, is this problem as hard as it seems?

It's always possible to make any problem more difficult

If you consider the symmetry of a mass/spring oscillation, the mass always oscillates symmetrically about an equilibrium point. That equilibrium point is identical to the point the mass would hang at rest under the influence of gravity.

So given that the mass is released the first time and falls a total distance d from the point of release before stopping and turning around to return, you should be able to locate the new equilibrium point below the release point using that symmetry.

You should also be able to determine the period of the oscillation from that fall, which is what portion of a full cycle?

With the period and mass you should be able to find an expression for the spring constant.

 

[Jediknight]

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 it later if i get stuck when i come back.

I just feel like THIS problem has a lot of knowledge buried in it, but I am not getting something and hours of messing with equations and thinking about it aren't helping, and my test is on monday

 

[gneill]

To me it appears that conservation of energy, symmetry, Hooke's Law, and the formula for the period of a mass-spring system are put to use here. A pretty comprehensive review!

 

[Shinaolord]

And not a very hard problem if you aren't overthinking it.

Don't worry, I've been in this same situation, but with a different topic within physics. Once you figure it out you'll have that "aha!" Moment.

Anyway, if $x_{eq}$ or $x_0$ is what we're finding, at a point called $x_0$ that is a distance d from where you compressed it. that will be when the mass has a velocity of $v=0$, no?
After that, I would look at the acceleration at that point, which is equal to

$mg=k(x_{eq}-x_0)$
where $x_0$ is half the distance it has traveled since compressing and letting it go until it is at rest.
What happens if $m\rightarrow 2m$? Recall that k is a constant.

A correct analysis for several launching points for a solution, I think. I would like a confirmation though.If you're using the distance d, as it is shown in the picture, that will be twice the amplitude A. But I don't like this question. It's slightly ambiguous. Is the question asking will be the amplitude after pausing, adding the 2M, and continuing, or are you measuring two separate D's with two separate experiments?

 

[Nathanael]

at a point called $x_0$ that is a distance d from where you compressed it
...
where $x_0$ is half the distance it has traveled since compressing and letting it go until it is at rest.

I don't understand. What is x₀? You seem to have given two contradictory definitions.

If you're using the distance d, as it is shown in the picture, that will be twice the amplitude A.

Let me start by saying I haven't worked out the problem with any equations or numbers. But I think you are forgetting that it starts out compressed.
The equilibrium position, as measured from the spring's natural length, will be doubled. But since the spring does not start out at it's natural length (it starts out compressed) this will not correspond to D=2d.

Call the natural length L_n and call the compressed length that it starts out at is L₀ (compressed meaning L₀ < L_n)
From what I said above (the equilibrium point as measured from L_n is twice for the mass 2m) we can write the equation:
(L₀+D/2-L_n) / (L₀+d/2-L_n) = 2
So to conclude that D=2d means to me that you assumed L₀ = L_n

 

[haruspex]

i know
d-xi=A

Where A is the subsequent amplitude? In relation to the given diagrams, where will the top of the oscillation be?

 

[Shinaolord]

I don't understand. What is x₀? You seem to have given two contradictory definitions.Let me start by saying I haven't worked out the problem with any equations or numbers. But I think you are forgetting that it starts out compressed.
The equilibrium position, as measured from the spring's natural length, will be doubled. But since the spring does not start out at it's natural length (it starts out compressed) this will not correspond to D=2d.

Call the natural length L_n and call the compressed length that it starts out at is L₀ (compressed meaning L₀ < L_n)
From what I said above (the equilibrium point as measured from L_n is twice for the mass 2m) we can write the equation:
(L₀+D/2-L_n) / (L₀+d/2-L_n) = 2
So to conclude that D=2d means to me that you assumed L₀ = L_n

Indeed that was an implicit assumption I made I should've stated that my apologies.
I was going to use that to work towards the case where $L_0 \leq L_n$ or $L_0 > L_n$. Can you see now how we could find x_0 from the information we now have?

 

[Jediknight]

where $x_0$ is half the distance it has traveled since compressing and letting it go until it is at rest.
What happens if $m\rightarrow 2m$? Recall that k is a constant.If you're using the distance d, as it is shown in the picture, that will be twice the amplitude A.But I don't like this question. It's slightly ambiguous. Is the question asking will be the amplitude after pausing, adding the 2M, and continuing, or are you measuring two separate D's with two separate experiments?

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 frequency as a horizontal, frictionless one (i knew the frequencies were)?

ok breath of fresh air, back to work, be back shortly, hopefully

 

[Jediknight]

Where A is the subsequent amplitude? In relation to the given diagrams, where will the top of the oscillation be?

yea, but I've si

I don't understand. What is x₀? You seem to have given two contradictory definitions.Let me start by saying I haven't worked out the problem with any equations or numbers. But I think you are forgetting that it starts out compressed.
The equilibrium position, as measured from the spring's natural length, will be doubled. But since the spring does not start out at it's natural length (it starts out compressed) this will not correspond to D=2d.

Call the natural length L_n and call the compressed length that it starts out at is L₀ (compressed meaning L₀ < L_n)
From what I said above (the equilibrium point as measured from L_n is twice for the mass 2m) we can write the equation:
(L₀+D/2-L_n) / (L₀+d/2-L_n) = 2
So to conclude that D=2d means to me that you assumed L₀ = L_n

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

 

[Shinaolord]

It won't be 2A, but you can get to it with the quantity 2A. Finding the difference between where it would be at equilibrium if released from equilibrium, and you can relate it to this situation.

I'm sure I'm not explaining this well, but that's basically what I was getting at. Tomorrow (I'm going to bed) I will actually do the problem and come back with my argument laid out step by step.

Good luck, see you all tomorrow.

 

[haruspex]

It won't be 2A,

Seems to me it will be 2A. Maybe we're reading the problem differently.
I see in post #9 you felt the problem was ambiguous. I think it's clear there are two separate experiments. What is a bit confusing is the reference to the system coming to rest. That appears to be irrelevant.

 

[gneill]

What is a bit confusing is the reference to the system coming to rest. That appears to be irrelevant.

I read it as being the "launch" point of the new experiment. The mass 2m is released from the equilibrium point of the previous mass m.

 

[Jediknight]

It won't be 2A, but you can get to it with the quantity 2A. Finding the difference between where it would be at equilibrium if released from equilibrium, and you can relate it to this situation.

I'm sure I'm not explaining this well, but that's basically what I was getting at. Tomorrow (I'm going to bed) I will actually do the problem and come back with my argument laid out step by step.

Good luck, see you all tomorrow.

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 those two and gravity and make a notecard,

 

[haruspex]

i agree won't be 2A, cause there's gravitational and elastic potential energy at xi, and only elastic at xΔt

I don't see why that stops d being twice the amplitude. The relevant fact is that the KE is zero at both positions. Since there's no loss of energy (we assume), it must keep returning to its launch position.

 

[haruspex]

I read it as being the "launch" point of the new experiment. The mass 2m is released from the equilibrium point of the previous mass m.

Hmm... maybe, but I still read "experiment is repeated" as meaning it is raised as before. Otherwise, you could even read it as the 2m block being released from its own equilibrium position (leading to a rather trivial answer).

 

[gneill]

Hmm... maybe, but I still read "experiment is repeated" as meaning it is raised as before.

That would make mentioning that the block comes to rest for replacing the block entirely pointless. I submit that it has a meaning here.

I think that repeating the experiment means dropping the new mass from the current position, which if you think about it, is above the equilibrium position as was the first mass.

Otherwise, you could even read it as the 2m block being released from its own equilibrium position (leading to a rather trivial answer).

No, I don't see how that interpretation could be justified by the wording of the problem: "Once the system comes to rest, the block is removed, a block of mass 2m is hung in its place". Sounds like a swap out at the current equilibrium point to me.

But I'd be happy to go with whatever scenario is agreed upon.

 

[Shinaolord]

That would make mentioning that the block comes to rest for replacing the block entirely pointless. I submit that it has a meaning here.

I think that repeating the experiment means dropping the new mass from the current position, which if you think about it, is above the equilibrium position as was the first mass.

No, I don't see how that interpretation could be justified by the wording of the problem: "Once the system comes to rest, the block is removed, a block of mass 2m is hung in its place". Sounds like a swap out at the current equilibrium point to me.

But I'd be happy to go with whatever scenario is agreed upon.

That's why my original post was as vague as possible, because I didn't understand the true nature of what the question is attempting to ask.

 

[Jediknight]

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
sin(ω(Δt)+Φi)=-1...ω(Δt)+Φi=3π/2
∑F=-kx-mg (since x is negative at Δt force points in positive direction, which I've chosen to be up, but mg still points down)

at xeq U=mgA+(1/2)mveq^2
|veq|=|vmax|=|A|ω
ω(teq)+Φi=kπ
∑F=mg

theres ways to compare xeq in expiriment 1 to xeq in expiriment 2 because there both proportional to xrelaxed of spring and there weight, but I'd need to
eliminate xrelaxed which seems like it may be doable in one step but I didn't earlier so its probably not that easy

Ive combined and substituted and divided the these equations til my eyes went cross, now its keeping me up, damn you physics how do you do it, need to sleep, start fresh in morning (at that time ill check new messages that i just seen come in, if I can wait that long, knowing me ill lay awake and be back before too long)

 

[Nathanael]

I think that repeating the experiment means dropping the new mass from the current position, which if you think about it, is above the equilibrium position as was the first mass.

Ah, I assumed it meant they dropped the second mass from the initial spot, but I like your understanding better. I think that's the way they meant it.

so are you saying d doesn't equal 2A, I was really starting to think it would,

d is 2A regardless. Like gneill pointed out, there's a certain symmetry with SHM: the oscillation is around the equilibrium point.
This means it goes as far below equilibrium as it goes above, which means the equilibrium point will always be half way between it's highest and lowest points (assuming energy is conserved).

there weight, but I'd need to eliminate xrelaxed which seems like it may be doable in one step but I didn't earlier so its probably not that easy

Take haruspex's advice from post #2 and use a variable for the initial compression. If you use x_relaxed, you will have to bring in another variable which tells the initial length. This is redundant because only the difference between the relaxed and initial lengths is important, so you can just make it a single unknown; the initial compression.

I would write an equation that describes the two equilibrium positions in terms of d, D, and the initial compression. Then you can combine them because you know x_eq.2=2x_eq.1

Then you just need to eliminate the unknown for the initial compression, which is where the given Δt comes into play. What is the period in terms of Δt? What is the period of an arbitrary mass on a spring?

 

[Jediknight]

d is 2A regardless. Like gneill pointed out, there's a certain symmetry with SHM: the oscillation is around the equilibrium point.
This means it goes as far below equilibrium as it goes above, which means the equilibrium point will always be half way between it's highest and lowest points (assuming energy is conserved).?

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 look at energy.



Still waking up, drinking coffee trying to get folks to take the kids, I feel like your right cause as said there is no kinetic energy at that point so without dissipating any energy it has to bounce back to that spot again with zero kinetic energy, so xi has to be the highest it goes above xeq, so xi has to equal A, so d has to equal 2A, but where does that gravitational potential energy go?

I guess I should just run with that, since it must be true but I can't stand that I don't get it. Maybe I should picture it all horizontally to get my head around things.

starting to think A might be constant with xeq being the only thing that changes, but that can't be true, need to get into this with paper and pencil for a minute...tick tock

be back soon as possible, will re-read problem and posts and if necessary the end of chapter 15, see if I get anywhere

 

[Nathanael]

x_i is the starting point? Measured from the equilibrium position? This means x_i = d/2 which gives us nothing new... Have you considered taking the spring's relaxed length as the origin?

but that would make xi=A, but that would make the energy at the tp greater than the energy at the bottom

I think you're not seeing that the object does not rise as far above the relaxed length as it falls below it. You can't even write any equations to check the energy without introducing another unknown (the initial compression).

but where does that gravitational potential energy go?

You have to remember, the spring-potential-energy depends on the spring's relaxed length. But the block does not oscillate about the spring's relaxed length... It oscillates about the equilibrium position which is below the relaxed length. So the system has more spring-potential-energy at the bottom than at the top, and it has more gravitational-potential-energy at the top than at the bottom.

Anyway, I don't think you need to use energy to solve this. (But it is good to understand that energy is conserved and how.)

need to get into this with paper and pencil for a minute...

Use the pencil to draw a picture of the situation. Use the relaxed length as the origin. Include the initial compression as an unknown. Sometimes equations can make things a little foggy, it is important when solving a problem to have a clear idea as to what you're trying to do and (ideally) what's stopping you.

 

[Jediknight]

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 seems to start at the amplitude (i.e. xi-xeq=A)

 

[Nathanael]

starting over with xrelaxed as the origin, that makes xi the same value in both problems,

What does x_i represent? Please say explicitly.

Also, are you working under gneill's understanding of the problem statement, in which the second block is released from the bottom point of the first block's motion?

 

[haruspex]

but that would make x_i=A

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.

 

[Jediknight]

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 x-xeq does) (but ill have to be careful of my signs during using these as choosing "up" as my positive axis makes xeq a negative value)]


I think the problems saying the point of release (xi) will be a fixed distance from a fixed point (say the ceiling) I think they would've drawn 2 pictures if that wasn't the case, I think there reason for saying experiment was repeated is just so they can avoid saying "Identical" so many times to describe the spring and elevation of experiment.

I know the "hung in its place" is a little vague, but honestly I don't care about getting the problem r"ight" is long as I figure out the correct answer with respect to the interpretation being followed, if both versions of the problem are possible I'd like to go at the harder one. I just know as hard as this seems for me my test tomorrow will have one 20X harder with something I couldn't expect or predict thrown in (they always do) I'm just trying to get an understanding, and if I understood this stuff I'd understand how to solve the problem (with either interpretation I do believe)

I think d is the distance from that fixed point of release (xi) the object travels before turning around, and I think the time it takes for object 2 to do it will be Δt√(2) due to the fact ω^2=k/m and ω(t)=π when object is at its lowest position

 

[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