A block dropping onto a spring system

[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 $\sqrt{x^2+y^2}$?

 

[Steve4Physics]

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).

Maybe a few words of explanation are in order.

The original problem-statement and diagram were poor. After ~30 posts I got the impression that there was still a degree of uncertainty/confusion.

It seemed that a clear problem-statement and diagram could be helpful to some readers (including any readers not posting replies) – even though not needed by everyone. I should have made it clear that my version was just a best guess.

I had drafted the post earlier, so by the time I actually sent it, to some extent it had been superseded by further posts.

 

[Orodruin]

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

The parallel coupling of the small spring with half of the longer spring - which is the problem interpretation @erobz was proposing. It should be clear based on that.

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:

No, this is an incorrect description of the problem @erobz was solving. Please see his posts and my replies to them for the actual interpretation he was proposing (which we have now effectively killed off as it would be impossible to get the correct maximal compression)

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 $\sqrt{x^2+y^2}$?

No, you are misrepresenting the problem interpretation @erobz was attempting to solve. The $2k_1$ is the spring constant of each half of the longer spring. One of those halves is compressed by x and the other by y so the total energy stored in that spring is $k_1 x^2$ in the first half and $k_1 y^2$ in the second. The short spring is compressed by y so the energy stored in it is $k_2 y^2/2$.

 

[Orodruin]

I should have made it clear that my version was just a best guess.

Sure, I agree that it is the most likely interpretation of the problem. It was also the assumption I made when responding in the beginning. The issue is that the fact that we cannot seem to agree on the interpretation means that it is at the very least somewhat unclear and OP should clarify by providing the original statement verbatim.

 

[erobz]

Well, there was some unclear wording in the OP and had it been @Steve4Physics diagram originally accompanying it, we would not be having this conversation (I think). I don't think the OP was confused about that, but I allowed myself to see something more exotic...and to be fair, most of the times it is the more exotic problem that sail over my head...until someone points it out. So, I thought I "caught a zebra" -turned out to be a lemon.

 

[palaphys]

All right. I've spent enough time analyzing this and I hope with this message people are clear with this question.
Sorry for the extra long answer.


So in this problem, a mass is released from a height h on top of a spring, to which a shorter spring is fixed at the bottom. The mass descends down the system and at one point attains equilibrium before maximum compression, which is given as h. As no work is done by any non Conservative forces, total mechanical energy is constant. Hence the spring constant of the shorter spring can be obtained easily, by considering the fact that the block would be at rest at maximum compression. It does NOT matter whether maximum compression occurs when the block touches the ground, as it can occur anywhere above, since it is not mentioned by the question. Secondly to determine the equilibrium position, we must consider a reference level of 0 gravitational potential, as probably the place where compression is maximum. Next, consider an additional compression x of the second spring in the downward direction. At this position, assume equilibrium. Here the compression of the longer spring would be x+h . Now, solving for x becomes simple. X turns out to be h/9. What this means in practice, is that the block attains MAXIMUM velocity at this position, as it's kinetic energy is maximum. Why is that so? Well consider the analogy of a swing. When it oscillates just past the equilibrium point, it's velocity becomes maximum as potential energy comes to a minimum. Further using energy conservation, the maximum velocity of the block can be obtained. Now coming to maximum acceleration this occurs at the maximum compression state, In an upward direction. Further analysing the motion, it can be said that ad the block passes through the equilibrium position, this maximum value of kinetic energy is converted completely to elastic potential energy, and the block begins to oscillate up and down.

And lastly, as @oro mentioned using work energy theorem between two points in between is a failed approach as it neglects the compression of the shorter force and the work done due to that.

 

[haruspex]

Unfortunately, this does nothing to clarify the question.

a mass is released from a height h on top of a spring,

On top of? That suggests being in contact. Do you mean above?

to which a shorter spring is fixed at the bottom.

Fixed to what and how?

The mass descends down the system and at one point attains equilibrium before maximum compression,

It will pass through the equilibrium point before reaching maximum compression. Is that what you mean?

which is given as h.

What is given as h? Maximum compression?

Please confirm or correct one of the clearly spelt out interpretations to be found in this thread, such as post #23.

to determine the equilibrium position, we must consider a reference level of 0 gravitational potential, as probably the place where compression is maximum.

No idea what you mean by that. Where you set the PE reference level is irrelevant so long as you are consistent.

Next, consider an additional compression x of the second spring in the downward direction. At this position, assume equilibrium.
Here the compression of the longer spring would be x+h . Now, solving for x becomes simple. X turns out to be h/9. What this means in practice, is that the block attains MAXIMUM velocity at this position,

Yes, since equilibrium is where there is no net force on the mass, so no acceleration.

 

[Orodruin]

And lastly, as @oro mentioned using work energy theorem between two points in between is a failed approach as it neglects the compression of the shorter force and the work done due to that.

I am pretty sure I never said this. I am uncertain what statement you would interpret like that.

 

[SammyS]

All right. I've spent enough time analyzing this and I hope with this message people are clear with this question.
Sorry for the extra long answer.


So in this problem, a mass is released from a height h on top of a spring, to which a shorter spring is fixed at the bottom. The mass descends down the system and at one point attains equilibrium before maximum compression, which is given as h. As no work is done by any non Conservative forces, total mechanical energy is constant. Hence the spring constant of the shorter spring can be obtained easily, by considering the fact that the block would be at rest at maximum compression. It does NOT matter whether maximum compression occurs when the block touches the ground, as it can occur anywhere above, since it is not mentioned by the question. Secondly to determine the equilibrium position, we must consider a reference level of 0 gravitational potential, as probably the place where compression is maximum. Next, consider an additional compression x of the second spring in the downward direction. At this position, assume equilibrium. Here the compression of the longer spring would be x+h . Now, solving for x becomes simple. X turns out to be h/9. What this means in practice, is that the block attains MAXIMUM velocity at this position, as it's kinetic energy is maximum. Why is that so? Well consider the analogy of a swing. When it oscillates just past the equilibrium point, it's velocity becomes maximum as potential energy comes to a minimum. Further using energy conservation, the maximum velocity of the block can be obtained. Now coming to maximum acceleration this occurs at the maximum compression state, In an upward direction. Further analysing the motion, it can be said that ad the block passes through the equilibrium position, this maximum value of kinetic energy is converted completely to elastic potential energy, and the block begins to oscillate up and down.

And lastly, as @oro mentioned using work energy theorem between two points in between is a failed approach as it neglects the compression of the shorter force and the work done due to that.

The description here differs from the description in the OP (Originating Post) But neither is particular clearer than the other.

A main problem is ambiguity and last of consistency in terminology a in referring to objects and their configurations. It's also very difficult to distinguish which statements are part of the given problem information from which statements are part of your analysis / solution.

The following examples would have improved clarity greatly:

... a mass is released from a height $h$ on above the top of the long spring ... (Referring to the figure in the OP)

Rather than ...The mass descends down the system and at one point attains equilibrium before maximum compression, which is given as h. ...
You might rather say something like:
As the mass descends, it passes a point at which it is in equilibrium. The decent finally ends as the short spring is compressed by an amount $h$.

Most of the rest of your post explains your thoughts about the problem, your analysis and your results.

Finally, I draw attention to :

Secondly to determine the equilibrium position, we must consider a reference level of 0 gravitational potential, as probably the place where compression is maximum. Next, consider an additional compression x of the second spring in the downward direction. At this position, assume equilibrium. Here the compression of the longer spring would be x+h . Now, solving for x becomes simple. X turns out to be h/9.

The choice of the location for zero potential is unimportant. It seems that here you're saying that the equilibrium position is where compression is maximum . We know that you know better than that.

However, it was interesting to me to observe that plugging in values of $\dfrac h 9$ and $\dfrac {10h} 9$ respectively for compressions of the short and long springs, the resultant force was $mg$, using @Steve4Physics's $k_2$ value.

 

[SammyS]

And lastly, as @oro (@Orodruin) mentioned using work energy theorem between two points in between is a failed approach as it neglects the compression of the shorter force and the work done due to that.

Here is @Orodruin's answer regarding how to use the Work-Energy Theorem when the velocity is non-zero at the chosen points.

From Post #4 .

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