A block dropping onto a spring system

[palaphys]

Homework Statement

A block dropped onto a spring system as shown in the figure. If the spring constant of the longer spring is mg/2h then find the spring constant of the larger.

Relevant Equations

Work done= change in kinetic energy

Consider this situation. A mass $$m$$ is dropped onto the spring system from rest , and it is given that the smaller spring gets compressed by a distance $$h$$

My main query in this question is regarding the use of Work Energy theorem, which states that the work done on a body by all the forces acting on it is equal to the change in its kinetic energy.

Additionally it is given that the spring constant of the longer spring is $$mg/2h$$.

Now, applying Work Energy theorem between the top point and the bottom, we get:

$$ 0= 2mgh -(mg/2h) (4h^2 /2) - k_2 h^2 /2 $$

Here the first term is the work done by gravity, second the work done by the longer spring, and third the work done by the smaller spring.

Solving, we get $$ k_2 = 4mg/h $$

However, my main problem is while trying to apply work energy theorem between the point B and A.

I decided to try a different approach to this problem and began by using Work Energy theorem between the top point and B
This gave
$$ Mv_b ^2 /2 = 2mgh- (mg/2h)(h^2 /2) $$

Solving this, I get $$KE_B= 7mgh/4 $$

Now comes the problem. Applying Work Energy theorem between B and the bottom point, I get

$$ 0- 7mgh/4 = W_{sp1} + W_{sp2} + W_g$$
$$ -7mgh/4 =-(mg/2h)h^2/2 - (k_2)( h^2/2) + mgh$$
**Now solving for k_2**,
$$k_2= 5mg/h $$

**Why is there this subtle difference?**
Has there been a misunderstanding or incorrect application of work energy theorem?
Which one is right.

And why?

 

[Orodruin]

You should be applying the work-energy theorem using the point the mass is dropped from and the point of maximal compression. Those are the points where the mass has no kinetic energy.

 

[palaphys]

You should be applying the work-energy theorem using the point the mass is dropped from and the point of maximal compression. Those are the points where the mass has no kinetic energy.

I know that but what is wrong by applying it between the two points specified int he question?

 

[Orodruin]

I know that but what is wrong by applying it between the two points specified int he question?

You would need the block’s speed at those point to apply the theorem and take that into account. The mass is not at rest at those points.

 

[kuruman]

Homework Statement: A block dropped onto a spring system as shown in the figure. If the spring constant of the longer spring is mg/2h then find the spring constant of the larger.

I find the statement of the problem confusing. Which one is the "longer" spring? In the figure, both appear to have the same relaxed length. Also, we are asked to find the spring constant of the "larger". I assume that this is the other spring, but I cannot fathom in what sense it is larger.

. . . and it is given that the smaller spring gets compressed by a distance h

Where is that given? It does not show in the problem statement. For future reference, to avoid guesswork on our part and to help us help you better, please post the problem description exactly as given to you.

 

[Orodruin]

Which one is the "longer" spring?

The one that goes from the floor to A? And the shorter the one going from the floor to B? That is how I interpreted it. The shorter one is also fatter in the image so it would also make some sort of sense to refer to it as larger. But yes, not completely clear.

 

[kuruman]

The one that goes from the floor to A? And the shorter the one going from the floor to B? That is how I interpreted it. The shorter one is also fatter in the image so it would also make some sort of sense to refer to it as larger. But yes, not completely clear.

Ah yes. I missed that one spring is inside the other and twice as long. I thought they were welded end-to-start and had equal lengths. All is clear now. Thanks.

 

[erobz]

I'm more confused than I thought...If the smaller spring (the spring the block initially contacts) is compressed $h$, then the second spring isn't applying any force? How do we find a $k$ for that spring -could be anything?

EDIT: Ok, I see that the short spring is supposed to be fixed to the long spring at $L/2$. We now return to our regularly scheduled programming...

 

[palaphys]

@oroduin diid you even read my answer?

 

[palaphys]

You would need the block’s speed at those point to apply the theorem and take that into account. The mass is not at rest at those points.

Please read what I have done clearly before commenting.

 

[erobz]

Please read what I have done clearly before commenting.

I don't get what you are getting for the first equation. I get $k_2 = 14 \frac{mg}{h}$ I haven't looked at the other argument.

Also, probably a good idea to tone down the attitude a tad. It makes you come off as arrogant.

 

[Orodruin]

Please read what I have done clearly before commenting.

Yes I did and you would be well adviced to lose the attitude towards people trying to help you.

The thing is you failed in that attempt as a result of needing to do the work energy theorem twice. I could point out to you what you are doing wrong, but based on your attitude I do not want to.

 

[palaphys]

Sorry for being arrogant, it's just that I have been on this problem for so long.

 

[palaphys]

Yes I did and you would be well adviced to lose the attitude towards people trying to help you.

The thing is you failed in that attempt as a result of needing to do the work energy theorem twice. I could point out to you what you are doing wrong, but based on your attitude I do not want to.

Why did I fail? Pls point out what I did wrong

 

[palaphys]

I don't get what you are getting for the first equation. I get $k_2 = 14 \frac{mg}{h}$ I haven't looked at the other argument.

Also, probably a good idea to tone down the attitude a tad. It makes you come off as arrogant.

So what I tried was that first I tried to find the velocity of the block as it compresses through the first spring. Then I considered that velocity and conserved energy for the rest of the journey.

 

[haruspex]

Now, applying Work Energy theorem between the top point and the bottom, we get:

$$ 0= 2mgh -(mg/2h) (4h^2 /2) - k_2 h^2 /2 $$

I'm lost already.
The diagram appears to show a long spring at two different lengths, y and h+y, where y is the height of B above ground and is not given but appears to be less than y; and a short spring of length y only.
So it looks as though the mass falls 2h, the long spring becomes compressed by h and the short spring not at all.

But you state that the shorter spring is compressed by h, so I'll assume the diagram is inaccurate and anyway does not show the final state. This means the total distance fallen is 3h.

According to your equation, the mass comes to a stop after falling 2h and at that point the long spring has been compressed 2h while the short spring has been compressed h.
I can think of no interpretation of the diagram which matches that.

$$ -7mgh/4 =-(mg/2h)h^2/2 - (k_2)( h^2/2) + mgh$$

I think there are a few problems on the way to that, but what I note in particular is that you need to be careful about how much energy is added to a spring that was already part compressed. If a relaxed spring is compressed by x, how much energy does it have? If by 2x how much? How much energy was added by the extra x?

 

[Orodruin]

Sorry for being arrogant, it's just that I have been on this problem for so long.

That’s better.

Your work done by the longer spring between B and the bottom is incorrect. You have computed it as if the spring was relaxed at B, but it is already stretched at that point.

Generally, of course taking the work-energy between drop point and B and then between B and the bottom must be equivalent to taking it directly between drop point and bottom. Introducing the velocity at B and doing the double application or the WE theorem just introduces the double amount of places where things can go wrong. So unless you are actually looking for the velocity at B I would strongly advice against it.

 

[Orodruin]

The diagram appears to show a long spring at two different lengths, y and h+y, where y is the height of B above ground and is not given but appears to be less than y; and a short spring of length y only.

No, based on OP’s description of the problem, those are two separate springs that differ in height by $h$ and the spring system further compresses a distance of $h$ beyond B.

 

[haruspex]

No, based on OP’s description of the problem, those are two separate springs that differ in height by $h$ and the spring system further compresses a distance of $h$ beyond B.

Isn’t that what I went on to assume in the following paragraphs? (Making the total descent of the mass 3h?)

 

[Orodruin]

Isn’t that what I went on to assume in the following paragraphs? (Making the total descent of the mass 3h?)

Yes, sorry. Your post was a bit confusing.

 

[palaphys]

That’s better.

Your work done by the longer spring between B and the bottom is incorrect. You have computed it as if the spring was relaxed at B, but it is already stretched at that point.

Generally, of course taking the work-energy between drop point and B and then between B and the bottom must be equivalent to taking it directly between drop point and bottom. Introducing the velocity at B and doing the double application or the WE theorem just introduces the double amount of places where things can go wrong. So unless you are actually looking for the velocity at B I would strongly advice against

Thank you so much. I cemetery neglected the fact that b also stretches. Now I get why I was incorrect.

 

[palaphys]

I'm lost already.
The diagram appears to show a long spring at two different lengths, y and h+y, where y is the height of B above ground and is not given but appears to be less than y; and a short spring of length y only.
So it looks as though the mass falls 2h, the long spring becomes compressed by h and the short spring not at all.

But you state that the shorter spring is compressed by h, so I'll assume the diagram is inaccurate and anyway does not show the final state. This means the total distance fallen is 3h.

According to your equation, the mass comes to a stop after falling 2h and at that point the long spring has been compressed 2h while the short spring has been compressed h.
I can think of no interpretation of the diagram which matches that.


I think there are a few problems on the way to that, but what I note in particular is that you need to be careful about how much energy is added to a spring that was already part compressed. If a relaxed spring is compressed by x, how much energy does it have? If by 2x how much? How much energy was added by the extra x?

Hey one thing I gotta tell you, the maximum compression of the spring NEED NOT be in such a manner that the smaller spring touches the ground. There is nothing about that in the question, and you can set ANY reference point above the ground to solve this problem. Your answer will NOT change.

 

[haruspex]

Hey one thing I gotta tell you, the maximum compression of the spring NEED NOT be in such a manner that the smaller spring touches the ground. There is nothing about that in the question, and you can set ANY reference point above the ground to solve this problem. Your answer will NOT change.

Ok, but I still am unclear on the set up.
In my interpretation,

1. the mass free falls distance h
2. it hits the taller spring and compresses it by h. The mass has now descended 2h. At this point it contacts the shorter spring.
3. the mass descends a further h before coming to an instantaneous stop. At this point, the mass has descended 3h, the tall spring has compressed by 2h, and the shorter by h.

Is this correct? If not, please state it in similar detail.

 

[erobz]

Seems like we all aren't on the same page.

I'm thinking that based on the diagram the short spring is fixed to the long spring at its midpoint making it a "spring system" where if the long spring compressed by $h$ by the falling mass the short spring compresses by $\frac{h}{2}$.

If the springs were independent and the long spring compresses by $h$ then the other short spring appears to do nothing if the diagram is correct. So I'm not leaning that way.

 

[Orodruin]

Seems like we all aren't on the same page.

I'm thinking that based on the diagram the short spring is fixed to the long spring at its midpoint making it a "spring system" where if the long spring compressed by $h$ by the falling mass the short spring compresses by $\frac{h}{2}$.

This is assuming that the shorter spring is attached to the midpoint of the longer one, information that we are not given (the distance to the floor is not available in the figure).

If the springs were independent and the long spring compresses by $h$ then the other short spring appears to do nothing if the diagram is correct. So I'm not leaning that way.

The object is said to fall $h$ below the end point of the thicker spring (i.e., B). This is also what the OP is seemingly attempting to use for the solution. In this case, the elongation of the longer spring is $2h$ and that of the thicker spring is $h$ at the point of maximal compression so the thicker spring definitely does something.

That said, the problem could be clearer formulated.


Edit: The issue of the OP's first solution is that they consider the fall distance to be $2h$ and not $3h$ as they should. They are doing the compressions of the springs as if the fall distance were $3h$ so that the longer spring is compressed by $2h$ and the shorter by $h$. The issue of the second solution is that they are miscalculating the work done by the longer spring from point B to the lowest point. Both therefore result in the wrong answer.

 

[erobz]

That said, the problem could be clearer formulated.

Well, the diagram is horrible then, because compressing $h$ below B puts the mass below ground and our spring system have dematerialized.

 

[Orodruin]

Well, the diagram is horrible then, because compressing $h$ below B puts the mass below ground and our spring system have dematerialized.

Never assume diagrams are to scale unless explicitly stated. This issue exists also in your interpretation as the statement says that the smaller spring is compressed by $h$.

given that the smaller spring gets compressed by a distance $$h$$

(my emphasis)

 

[erobz]

Never assume diagrams are to scale unless explicitly stated. This issue exists also in your interpretation as the statement says that the smaller spring is compressed by $h$.

(my emphasis)

But I thought we decided that the smaller spring was the longer spring! In the sense that it is not as "girthy", lower spring constant. I'm so turned around...Yikes.

 

[Orodruin]

But I thought we decided that the smaller spring was the longer spring! In the sense that it is not as "girthy", lower spring constant. I'm so turned around...Yikes.

I don't think anyone decided that. Everything mentioned in the OP suggests that the "smaller spring" is the same as the "shorter string", including when the OP states this (for example):

$$ 0= 2mgh -(mg/2h) (4h^2 /2) - k_2 h^2 /2 $$

Here the first term is the work done by gravity, second the work done by the longer spring, and third the work done by the smaller spring.

The "smaller spring" here clearly refers to the spring with unknown spring constant $k_2$.

 

[erobz]

I don't think anyone decided that. Everything mentioned in the OP suggests that the "smaller spring" is the same as the "shorter string", including when the OP states this (for example):

The "smaller spring" here clearly refers to the spring with unknown spring constant $k_2$.

The problem statement says find the spring constant of the larger ("larger" $k$ in my estimation)...or a typo.

Homework Statement: A block dropped onto a spring system as shown in the figure. If the spring constant of the longer spring is mg/2h then find the spring constant of the larger.

emphasis added

including when the OP states this (for example):

I argue that the OP's attempt is extraneous to the problem statement and should not be considered as evidence of the correct interpretation. At the very least it should be given less weight than the accompanying diagram.

 

[Orodruin]

The problem statement says find the spring constant of the larger ("larger" $k$ in my estimation)...or a typo.

The shorter spring is larger in the transversal direction so it could be referring to that. It also makes no sense for the question to ask for the spring constant that was given as part of the problem statement (again, my emphasis):

Homework Statement: A block dropped onto a spring system as shown in the figure. If the spring constant of the longer spring is mg/2h then find the spring constant of the larger.

Additionally it is given that the spring constant of the longer spring is $$mg/2h$$

The first of those quotes structurally would read "find X given X" if "larger spring" is to be interpreted as the longer spring. All in all, I think it is safe to conclude that it is the spring constant of the shorter spring that is sought.

Edit: Typos.

Edit 2: It is also clear that what we are given is not the actual problem statement, but the OP's interpretation of the problem statement. Showing again the importance about giving the actual problem statement.

 

[erobz]

The shorter spring is larger in the transversal direction so it could be referring to that. It also makes no sense for the question to ask for the spring constant that was given as part of the problem statement (again, my emphasis):


The first of those quotes structurally would read "find X given X" if "larger spring" is to be interpreted as the longer spring. All in all, I think it is safe to conclude that it is the spring constant of the shorter spring that is sought.

Edit: Typos.

I agree that its the spring constant of the shorter (free length) spring that is sought.

What I'm not sure about is that they have referred to the shorter spring as the larger, then in the next statement said about the smaller spring compressing $h$, which is not inconsistent with the springs being fixed to each other in my estimation. Also, the diagram does not explicitly show a compression of $h$ below B. I think it's not even implied.

 

[erobz]

Edit 2: It is also clear that what we are given is not the actual problem statement, but the OP's interpretation of the problem statement. Showing again the importance about giving the actual problem statement.

That very well could be the problem.

 

[kuruman]

I think that we are owed a complete statement of the problem exactly as given to @palaphys. Let's see what we have and see what we can deduce from what is "shown".

In post #1 there is no mention of height $h$ by which the inner spring is compressed, it is just an assertion by the OP.

See the detail on the right of the posted picture. It shows a long spring inside a shorter of spring of larger diameter. There is a thicker line inside the long coil that is is the only part of the "system" touching the floor. Clearly, someone went through the trouble to draw that thicker line. What is it trying to "show"?

• It could be a third spring, but I doubt it because that would introduce a third spring constant.
• I could be the continuation of the longer string that goes through the shorter spring. The two springs are not connected in any way; the shorter spring just rests on the floor although it is not shown doing that. The bottom spring is compressed only after the mass is at height $h$ above the floor. The mass descends to some height, yet to be specified but lower than $h$, from the floor.
• It could be an attempt to show that the longer spring is welded to the shorter spring along the entire length of the latter. This means that both springs are compressed right from the start until the mass descends to some height, yet to be specified, from the floor.

Personally, I don't like guessing; I would like to see a complete, exact statement of the problem before investing more time here.

 

[Orodruin]

That very well could be the problem.

Ot is certainly confusing and the OP should clarify, but let me also add another reason I don’t really buy your interpretation:

If we assume that your interpretation is correct that the shorter spring is connected to the middle of the larger spring, then it is no longer that simple to relate the spring constants and compression of the springs. Half the long spring would be in series with the parallel of its other half and the shorter spring. You therefore cannot simply state that the long spring compression is h and the short spring compression is h/2 - that will give you an incorrect answer. I think that is too complicated for the level of the OP.

 

[erobz]

If we assume that your interpretation is correct that the shorter spring is connected to the middle of the larger spring, then it is no longer that simple to relate the spring constants and compression of the springs. Half the long spring would be in series with the parallel of its other half and the shorter spring. You therefore cannot simply state that the long spring compression is h and the short spring compression is h/2 - that will give you an incorrect answer. I think that is too complicated for the level of the OP.

That does sound more complicated that the assumption of proportional deflections. Out of curiosity, how bad do you think that is. Is it a "well...technically", or "holy crap, that's a bad assumption almost always"

EDIT: Ok, I decided to check that myself. I see there is some dependency on the ratio $\frac{k_2}{2k_1}$. Thanks for pointing that out. I obviously didn't think about it. So my needle has shifted a bit towards the simpler of the two interpretations, but still not certain (because I think both could have a claim to solution). I admittedly haven't gone all the way, so a claim to solution for my interpretation, (your corrections to it) could get foiled in the ensuing algebra to isolate $k_2$.

 

[Orodruin]

That does sound more complicated that the assumption of proportional deflections. Out of curiosity, how bad do you think that is. Is it a "well...technically", or "holy crap, that's a bad assumption almost always"

Somewhere in between. It is certainly going to matter for the result, but I would probably not have any problems solving it (assuming the problem is well defined enough). I am not an average high school student though.

 

[Orodruin]

As it turns out after actually having a look at it, there is no solution to that problem. The issue is the following:

Assume a long spring of whatever length and spring constant mg/2h and a short spring of spring constant k = zmg/h where z is a dimensionless constant to be determined. The short spring has half the length of the long one and is connected in parallel to half the long one.

The spring constant of half the long spring is mg/h and after some algebra the effective spring constant of the full system is$$
K = \frac{mg}{h}\frac{1+z}{2+z} = \frac{mg}{h} f(z)$$
The factor $f(z)$ varies between 1/2 ($z=0$) and 1 ($z\to \infty$). This is reasonable. If z=0 we essentially only have the full long spring and recover its spring constant. In the $z\to \infty$ limit the now very strong short spring keeps that half immovable and only the spring constant of the other half of the long spring matters. All good.

Now, here is the thing. If the mass is dropped from height $h$ above the end of the spring system, then the displacement $d$ from that point must satisfy$$
mg(h+d) = \frac{mg}{2h} f(z) d^2
$$
This solves to (for the relevant root)$$
d = \frac{h}{f(z)} \left( \frac 12 + \sqrt{ \frac 14 + f(z)}\right) > h $$
for all $z$.

Therefore, no matter how large the spring constant of the small spring becomes, the mass will not turn around after only compressing the spring system by $h$.

This further reinforces my view that this is not the intended problem.

 

[erobz]

Are you saying this cannot have a solution where the system deflects some distance $h$ before instantaneously coming to rest?

$$ 2 mgh = \frac{1}{2}(k_{\text{eff}}) y^2 + \frac{1}{2}( 2k_1) x^2 $$

Where:

$$ k_{\text{eff}} = 2k_1+k_2 $$

$$ y = \frac{h}{2+\frac{k_2}{2k_1}}$$

$$ x = h - y $$

And that would presumably be because of this definition for $k_1 = \frac{mg}{2h}$? I'll try to see if I can carry this out to failure. Or maybe something wrong with applying work-energy the way I have?

 

[Orodruin]

View attachment 357311

Are you saying this cannot have a solution where the system deflects some distance $h$ before instantaneously coming to rest?

$$ 2 mgh = \frac{1}{2}(k_{\text{eff}}) y^2 + \frac{1}{2}( 2k_1) x^2 $$

Where:

$$ k_{\text{eff}} = 2k_1+k_2 $$

$$ y = \frac{h}{2+\frac{k_2}{2k_1}}$$

$$ x = h - y $$

And that would presumably be because of this definition for $k_1 = \frac{mg}{2h}$? I'll try to see if I can carry this out to failure. Or maybe something wrong with applying work-energy the way I have?

You are missing that x and y are restricted to return the same force in the respective strings (they must convey the same for e due to force balance on all parts of the ideal massless springs).

You are better served finding the effective spring constant of the full system.

 

[erobz]

You are missing that x and y are restricted to return the same force in the respective strings (they must convey the same for e due to force balance on all parts of the ideal massless springs).

I feel like I used that to arrive at the equation for $y$.

I say:

$$ y \left( 2k_1 + k_2\right) = 2 k_1 x $$

Thats just saying the red forces in the diagram are the same magnitude

You are better served finding the effective spring constant of the full system.

perhaps, but it's not obvious that it should matter.

 

[Steve4Physics]

I hope this helps.
________________________
Problem statement
[Edit. Posted in case it's useful to anyone - it's one possible interpretation - an expansion of a previous interpretation, e.g. see @haruspex's Post #23.]

Spring-1 (S1), shown blue, has a spring constant $k_1 = \frac {mg}{2h}$.
Spring-2 (S2), shown red, has a spring constant $k_2$.

S1’s natural length is greater than S2’s by an amount $h$. S2 has a larger diameter than S1 so that S1 fits freely inside S1 as shown.

A mass $m$ is released from height $h$ above S1 and comes to instantaneous rest after descending a total distance $3h$, compressing both springs.

Express $k_2$ in terms of $m, g$ and $h$
_________________
Solution (Edit - solution removed.)

 

[erobz]

So, I do end up with a negative $k_2$ after I solve after combining into entire effective spring constant:

$$ 2mgh = \frac{1}{2} \frac{4k_1^2 + 2 k_1 k_2}{4 k_1 + k_2} h^2 $$

$$ k_2 = -\frac{7}{3}\frac{mg}{h} $$



Surely this is tied to the definition of $k_1$ being dependent on $\frac{1}{h}$ - and that is what @Orodruin has eluded to. if $k_1$ were a constant I don't see this problem could exist.

 

[Orodruin]

Surely this is tied to the definition of $k_1$ being dependent on $\frac{1}{h}$ - and that is what @Orodruin has eluded to. if $k_1$ were a constant I don't see this problem could exist.

You would have to specify a larger maximal compression somewhere in between the values it would take if the short spring had spring constant 0 and if it had a spring constant approaching infinity.

 

[haruspex]

View attachment 357311

Are you saying this cannot have a solution where the system deflects some distance $h$ before instantaneously coming to rest?

$$ 2 mgh = \frac{1}{2}(k_{\text{eff}}) y^2 + \frac{1}{2}( 2k_1) x^2 $$
Where:
$$ k_{\text{eff}} = 2k_1+k_2 $$

You are making the same mistake as the OP that I noted in post #16.
The energy stored in compressing a spring an extra $y$ is not $\frac{1}{2}k y^2$; it is $\frac{1}{2}k ((x+y)^2-x^2)=\frac{1}{2}k (2xy+y^2)$

 

[Orodruin]

I hope this helps.
________________________
Problem statement:

View attachment 357315
Spring-1 (S1), shown blue, has a spring constant $k_1 = \frac {mg}{2h}$.
Spring-2 (S2), shown red, has a spring constant $k_2$.

S1’s natural length is greater than S2’s by an amount $h$. S2 has a larger diameter than S1 so that S1 fits freely inside S1 as shown.

A mass $m$ is released from height $h$ above S1 and comes to instantaneous rest after descending a total distance $3h$, compressing both springs.

Express $k_2$ in terms of $m, g$ and $h$
_________________
Solution (click spoiler):

The mass descends a distance $3h$. S1 is compressed by $2h$. S2 is compressed by $h$.

GPE lost by mass = $3mgh$.
EPE gained by S1 $= \frac 12 (\frac {mg}{2h}) (2h)^2 = mgh$.
EPE gained by S2 $= \frac 12 k_2 h^2$

$3mgh = mgh + \frac 12 k_2 h^2$x ..some simple algebra.... $k_2 = \frac {4mg} h$

I don’t think anyone was doubting the solution to that interpretation of the problem except the OP. The discussion has evolved into arguing if that is the correct interpretation of the problem or not (me seeing it as the natural interpretation and @erobz arguing for a different one).

 

[Orodruin]

You are making the same mistake as the OP that I noted in post #16.
The energy stored in compressing a spring an extra $y$ is not $\frac{1}{2}k y^2$; it is $\frac{1}{2}k ((x+y)^2-x^2)=\frac{1}{2}k (2xy+y^2)$

No he is not. He is defining x as the compression of the first half of the long spring and y as the compression of the parallel of the second part of the long spring and the short spring.

 

[Orodruin]

I feel like I used that to arrive at the equation for $y$.

I say:

$$ y \left( 2k_1 + k_2\right) = 2 k_1 x $$

Thats just saying the red forces in the diagram are the same magnitude

Yes but I don’t see that equation in your original post.

perhaps, but it's not obvious that it should matter.

If you take it into account properly, you will essentially just derive the expression for the full effective spring constant along the way. Easier just use it if you know it. Of course it will not matter for the end result.

 

[erobz]

I don’t think anyone was doubting the solution to that interpretation of the problem except the OP. The discussion has evolved into arguing if that is the correct interpretation of the problem or not (me seeing it as the natural interpretation and @erobz arguing for a different one).

Well, at least that interpretation is now sacked...What am I going to do

 

[haruspex]

the compression of the parallel of the second part of the long spring

Sorry, I don’t know what that is saying. How do you compress a parallel of something?

The diagram in post #39 appears to show one spring being compressed by x, then that spring being compressed a further y as the short spring is simultaneously compressed by y.
Regardless of that, look at the equations:

$$ 2 mgh = \frac{1}{2}(k_{\text{eff}}) y^2 + \frac{1}{2}( 2k_1) x^2 $$

Where:

$$ k_{\text{eff}} = 2k_1+k_2 $$

I.e. $$ 2 mgh = \frac{1}{2}(2k_1+k_2) y^2 + \frac{1}{2}( 2k_1) x^2 $$
$$=\frac{1}{2}(k_2) y^2 + \frac{1}{2}( 2k_1)( x^2+y^2)$$
So did something get compressed by $\sqrt{x^2+y^2}$?