# Homework Help: Two non ideal springs hanging on a tree

1. Jun 19, 2014

### skrat

1. The problem statement, all variables and given/known data
One spring is located inside a wider spring. Both ends are welded together and hanged from a tree. both springs are $30 cm$ long when not deformed. The first spring has $30 g$ and $k=5 g/cm$ the second spring has $60 g$ and $k= 6g/cm$. How deep under the branch will be the bottom weld?

2. Relevant equations

3. The attempt at a solution

My idea was to calculate the deformation of each spring separately. Let's say that $s_1$ is the deformation of the first one and $s_2$ deformation of the second one.

Of course one will have greater deformation than the other, but when welded together, this means that one of the springs will have to carry some of the mass of the other spring with greater deformation.

For easier writing let's say that the second spring has greater deformation than the first one. So my idea was to calculate how much mass of the second spring, will the first one have to carry. I think I could do that because I know the difference between the springs deformations.

This would than mean, that the total distance of the bottom weld from the top one is a sum of deformed length of the first spring + additional deformation of spring no. 1 because it has to take some the mass of spring no 2.

Even explaining what I am trying to sounds too complex to me. So, is there an easier way? Or even better, is this even the right way?

I would love to show you my calculations yet the problem is that I have absolutely NO idea how to calculate the deformation of the hanged springs due to gravity. Whatever I do, everything seems do deduct down to zero in the first two steps :(

2. Jun 19, 2014

### BvU

OK, no calculations to show. Fair enough. Some deliberations, perhaps ?
Make things easier to start with: cut the lower weld. How far does each one stretch then ?

3. Jun 19, 2014

### CWatters

Not sure I can help but what does "The first spring has 30g" mean?

Does it mean

a) It has mass of 30g or
b) A mass of 30g is hanging on the bottom end?

4. Jun 20, 2014

### skrat

a). It has mass of 30 grams and NONE of the springs has any additional weights attached.

Yes, the problem is that I am not even sure about that part but here is how I tried without any success.

$dx$ part of the spring should stretch for $\delta x$. Meaning $k\delta x=\frac{l-x}{l}mg$. But the problem is I have no idea what to do to get integral on the right side, so I tried to use hooks law

$SE\frac{\delta x}{dx}=\frac{l-x}{l}mg$

but I have no information about $E$ .... ?

5. Jun 20, 2014

### PhanthomJay

You should first look up the deflection of a spring under its own weight, or do the calculus if you are required or so inclined to do so. Then realize that since the springs are welded together at their bases, the deflection of each must be the same.

6. Jun 21, 2014

### skrat

I want to do that. But like I said: I don't know how? I can't come up with an equation that would make sense and let me calculate the deflection of a spring under its own weight.

7. Jun 21, 2014

### PhanthomJay

The weight of the spring is distributed along its length, so calculus is required to determine the deflection of each spring if they were not welded. I have long since lost my calculus skills, that's why I had to look it up. Shouldn't be that difficult for a calculus guru, integrating the variable force overt the length using Hookes law for a spring with stiffness k. The answer is delta equals mg/3k. But that is for each spring separately. Together, you will have to find the equivalent stiffness of 2 springs in parallel, then solve some simultaneous equations for the equal deflections of each.

8. Jun 21, 2014

### BvU

Phantom has tickled me with his mg/3k. I thought it would be mg/2k (and Stockhom agrees on p5).

Maybe there's someone who can shoot a hole in my poor man's integration:

Take the 30 g 30 cm k = 500 g/m spring
I take it the second g means gram force -- a didactically rather unrecommendable formulation -- and I am going to set g = 10 m/s2 and write k = 5 N/m from now on !

Suppose it consists of 30 turns of 1 gram each. Each coil turn has a k' of 150 N/m. (This circumvents the E in Hooke's law).

Bottom coil lifts nothing. Next one up lifts 1 gram, so stretches by 0.01 N/k'. Next 0.02 etc etc up to top. So total stretch is by 29 + 28 + 27 ... + 2+ 1 = 30 * 15 times 0.01 / k', which is 0.15 times the k of the entire spring.

(Yes I cheated a little. Go to 300 or 3000 turns to soften that).

A massless spring with a separate weight of 30 g would stretch twice as much. Reasonable outcome of an integration, i would say.

9. Jun 21, 2014

### PhanthomJay

well according to the site below, you and the page 5 guy are wrong. Do the calculus.
The deflection of a hanging spring with stiffness k and uniformly distributed mass m is the same as the deflection of an ideal massless spring with a mass of m/3 hanging at the free end.

http://en.m.wikipedia.org/wiki/Effective_mass_(spring–mass_system)

10. Jun 21, 2014

### AlephZero

That is irrelevant. The result is for a different situation, and in any case it is only an approximation.

But I agree that (unless you get lucky and the errors from two incorrect assumptions cancel out)
because the "page 5 guy" doesn't include the pre-load in the springs, caused by the fact that they are welded together at the end and constrained to be the same length.

The mass distribution along the length of the two springs will be different.

Last edited by a moderator: May 6, 2017
11. Jun 22, 2014

### BvU

Hideous, hideous. I am still trying to make a start on this one and therefore still trying to work out the simple(unwelded) case for a single spring. Obviously poor man's integration and "doing the calculus" already end up different.

All this in introductory physics. I feel I should hand in my degree and start over... analyzing Slinky photographs or something.

12. Jun 22, 2014

### skrat

The same here, I can get mg/2k with a calculus that doesn't make any sense at all. That's as close as I can get to mg/3k.

Haha, I apologize for that, but moderators really like giving me warnings if I post anything in advanced physics that isn't related to quantum mechanics or similar topics.

13. Jun 22, 2014

### BvU

Stupid me. Can't find the word deflection in the slightly garbled link you give (with an ominous "last edited 2 days ago by an anonymous user"). They (?) look at an oscillating spring with a mass attached.

Raises lots of questions. I believe integrating u2 brings in a 1/3, but I hardly believe u = vy/L if the springs mass is not negligible, which is exactly what we're looking at !

Fortunately the non-anonymous Stockholm link continues for the static case with a continuously distributed mass calculation and minimizes potential energy in a (imho) sensible way and comes out with the 1/2 again.

Aren't we having fun ! Where do you think one of the two reasonings is running off the rails ?

14. Jun 22, 2014

### BvU

Yet another dead end: Den of Inquiry has a page "Hanging Slinky " where they promise a thorough (**) treatment on their website. There it says
with a time stamp March 1, 2007. Not worth holding your breath ? I might just mail the guy....

Intrigueing and encouraging for Phantom, they do reveal that
(**)
@Phantom: did you do the calculus ? What's so very wrong with the poor man's integration ?

15. Jun 22, 2014

### PhanthomJay

I am so sorry, I looked up the formula for the wrong application , as aleph zero pointed out, my bad . The deflection of a spring under its own weight, when slowly lowered to the equilibrium position, is mg/2k , which you and the 5 guy correctly stated.
But that is just the first step in solving the problem of determining the deflection of the 2 springs when welded together in parallel. The deflection of each spring must be the same, and the total deflection of the 2 spring system must be the same as the value of either spring, from a compatibility sketch. Now you should be able to calculate the equivalent stiffness of 2 springs of stiffness k1 and k2 in parallel, and then apply a load of (m1 + m2)(g/2) to the welded end of the equivalent spring, to calculate a deflection of the 2 spring system. I can't guarantee however if that result presents an exact solution, as it would if the springs were massless under the stated load applied as a point load at the free end.

16. Jun 22, 2014

### Tim Erickson

A thousand apologies to all for not having posted the Den of Inquiry solution to the hanging slinky problem. I've started the answer on my blog here.

If I'm following this thread correctly, I can certify that I'm making a whole lot of calculations that ought to be obvious but keep turning out to be wrong, leading me first to question my physics/calculus cred, but then (thankfully) to help clarify my understanding and cement my belief that there are really interesting, rich problems that lie very close to what introductory students see.

A couple of insights that might help:
• Suppose you have a spring that has a spring constant k. Now you cut it in half. What is the spring constant of each semi-spring? Answer: 2k. That is, k is NOT a property of the material.
• Consider reformulating Hooke's law so that the constant is a constant of stretchability S rather than a constant of stiffness k. Then instead of F = kx you get x = FS. This may make the calculus easier. (It did for me doing the hanging slinky.)
• There is an interesting connection between the math here and the math involving resistance in parallel and series circuits.

17. Jun 23, 2014

### dauto

They seem both correct. They are calculating different things. The 1/3m shows up when calculating the effect of the spring's mass on the oscillation period, assuming that the spring is attached to a mass M where M>>m. The 1/2m shows up when calculating the spring deflection under its own weight. There is no reason to expect those two factors would be identical. Now, I didn't double check the calculations very carefully but found no obvious mistake on either of them.

18. Jun 23, 2014

### BvU

Hello Tim, and welcome to PF -- a nice change from the math and stats world you happily got lost in !
No need to apologize. Good intentions get thwarted all the time, all over the place.

Saw your juicy data talk and liked it. Almost sorry I didn't choose for teaching .

19. Jun 24, 2014

### skrat

Ok se we all agree now that the deflection of single spring is

$ES\frac{\delta l}{dz}=\frac{l-z}{l}mg$
$ES\int_{0}^{\Delta l}\delta l=\frac{mg}{l}\int_{0}^{l}(l-z)dz$
$ES\Delta l=\frac{mg}{l}\frac{l^2}{2}$ which is the same as $\Delta l=\frac{mg}{2k}$ where $k=\frac{ES}{l}$.

So $\Delta l_1=\frac{m_1g}{2k_1}$ and $\Delta l_2=\frac{m_2g}{2k_2}$ threfore the difference is $\Delta l_1-\Delta l_2=\frac g 2 (\frac{m_1}{k_1}-\frac{m_2}{k_2})$.

Basically my idea was to calculate how much mass this difference represents: $m_2\frac{\Delta l_2}{l+\Delta l_2}$ meaning it should be the same as if I hanged that amount of mass on the first spring:

$F=k_1s$
$m_2g\frac{\Delta l_2}{l+\Delta l_2}=k_1s$
$s=\frac{m_2g}{k_1}\frac{\Delta l_2}{l+\Delta l_2}$

So the total distance from the top to the bottom weld wold be $\Delta l_1+s$.

This feels wrong for several reasons:
- I didn't use any integration in the second part and it feels like I should.
- I simply assumed that the second spring will stretch more than the first one, which may not be the case.

20. Jun 24, 2014

### PhanthomJay

That is an incorrect assumption. The springs are welded together at their base, so both springs work together as one , with the same deflection for each. Can you calculate the equivalent stiffness of 2 springs in parallel? It is a simple formula.

21. Jun 24, 2014

### AlephZero

The deflection is the same at the bottom of the two springs, but saying they "work as one" and calculating the equivalent stiffness is too simplistic.

Imagine you have a slinky spring, in parallel with a stiff and light elastic band. The mass of the slinky will slump towards the bottom much more than the mass of the elastic. In fact you could get a situation where the bottom of the slinky was in compression, not tension, being pushed upwards by the reaction to the tension in the elastic.

skrat has got the differential equation for the stretching of one spring. The "m/2" formula comes from the boundary condition that there is no external force applied to the bottom of the spring.

I think what you need to do is write the two DEs for each spring, and impose the boundary conditions that
1. The displacements at the bottom of each spring are equal
2. The forces applied to the bottom of each spring are equal and opposite, but not zero.

22. Jun 25, 2014

### skrat

Ammmm, on the previous page:

But... If I write the DEs for each spring, than I would have to write them with factor 1/3. Or did I just mix everything up? So my calculation in post #19 is basically useless in this case?

23. Jun 25, 2014

### BvU

It's late here and I am sleepy, but this exercise is so nice, I want to throw in a few more goodies: I really like the "so we all agree" in quote 19. Science isn't really democratic, but empiric science works a bit like that: we reason and propose some theory that we can underpin and that we agree to accept as the truth ... until someone comes with something better.

Never mind.
30 cm, 30 g spring by itself, k = 5 N/m g = 10 m/s2 , is 33 cm long
30 cm, 60 g spring by itself, k = 6 N/m g = 10 m/s2 , is 35 cm long

No compromises now (welding them together doesn't yield a 34 cm whole, although the 34 cm answer might earn you a smal credit), no more democracy, just brainwork.

Still too daft to write the DE, so I'd like to repeat the poor man's integration with a little extension: $F_{12}\$ pulling up the bottom end stiff bloke from 35 to 30 + x, and $F_{21}\$ pulling down the light one from 33 cm to 30 + x. ${\vec F}_{12} + {\vec F}_{21} = \vec 0\,$, of course (right?).

Can't use the slinky concept any more (bottom turns don't behave like F = -kΔx under compression). But otherwise I see no objections to my cheapo approach.... any takers ?

24. Jun 26, 2014

### skrat

Hmmm, is it ok to say

$m_1g\frac{l-x}{l}+m_2g\frac{l-x}{l}=S(E_1+E_2)\frac{\delta x}{dx}$ where $\delta x$ is a stretch of $dx$ and where I assumed that cross section of the springs is the same and constant.

Integrating this leaves me with:

$\delta x=\frac{g(m_1+m_2)}{2(k_1+k_2)}=4 cm$

25. Jun 26, 2014

### PhanthomJay

Well I am not sure why you need to assume the same cross sections. The stiffnesses k1 and k2 of each spring are given, and these values are based on spring material, cross section, coil spacing and thickness, unstretched spring length, etc., of each spring. It's already built in.

Anyway, I get the same result of 4.09 cm which to one sig fig is 4 cm, and which agrees with BvU's poor man's approach for at least partial credit. :tongue: