# M vs. 2m shm verticle spring

1. Feb 21, 2015

### Jediknight

1. The problem statement, all variables and given/known data

The hand in (Figure 1) (a) holds a vertical block-spring system so that the spring is compressed. When the hand lets 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.
2. Relevant equations
k(xeq-x0)=mg
the basics for shm
x=Asin(ωt+Φ)...a=Aω^2sin(ωt+Φ)

3. 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

2. Feb 21, 2015

### haruspex

You have no information regarding xeq. 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.

3. Feb 21, 2015

### Staff: Mentor

Hmm. The problem says:
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...

4. Feb 21, 2015

### Jediknight

ok ill be back in a minute, getting out a fresh peice 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

5. Feb 21, 2015

### 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?

6. Feb 21, 2015

### Staff: Mentor

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.

7. Feb 21, 2015

### 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 thats helping me, might post it later if i get stuck when i come back.

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

8. Feb 21, 2015

### Staff: Mentor

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!

9. Feb 21, 2015

### 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?

Last edited: Feb 21, 2015
10. Feb 21, 2015

### Nathanael

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

11. Feb 21, 2015

### haruspex

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

12. Feb 21, 2015

### Shinaolord

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?

Last edited: Feb 21, 2015
13. Feb 21, 2015

### Jediknight

OK thats 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

14. Feb 21, 2015

### Jediknight

yea, but ive 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

15. Feb 21, 2015

### 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.

16. Feb 21, 2015

### haruspex

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.

17. Feb 21, 2015

### Staff: Mentor

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.

18. Feb 21, 2015

### Jediknight

me too, im loving and hating this problem, and yea i agree wont be 2A, cause theres 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,

19. Feb 21, 2015

### haruspex

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.

20. Feb 21, 2015

### haruspex

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).