A block dropping onto a spring system

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I had drafted the post earlier, so by the time I actually sent it, to some extent it had been superseded by further posts.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