Find the coefficient of stiffness of this spring (with non-negligible mass)

[Eshka]

Homework Statement

there is a spring weighing 0.2 kilograms. when not stretched, its length is 67 millimeters, and when stretched 1200 millimeters, and the spring stretches down only under its own weight, without additional load. Calculate the spring stiffness coefficient.

Relevant Equations

F = k * x maybe

So, I tried to solve this problem as follows. if the spring is stretched without load, then it is stretched by its own weight, therefore, by gravity. therefore, I decided that the elastic force is equal to gravity, but this is not the case, the teacher said that this is incorrect. Then I assume that the spring stretches unevenly (probably), that is, if it stretches under its own weight, there will be almost no tension at the beginning, and at the bottom it will stretch more. So you need to divide it into many separate sections and calculate (PROBABLY), but I don't understand what formula to use for this and if this assumption is incorrect, then I can't solve the problem at all.

 

[Orodruin]

Homework Statement: there is a spring weighing 0.2 kilograms. when not stretched, its length is 67 millimeters, and when stretched 1200 millimeters, and the spring stretches down only under its own weight, without additional load. Calculate the spring stiffness coefficient.
Relevant Equations: F = k * x maybe

So, I tried to solve this problem as follows. if the spring is stretched without load, then it is stretched by its own weight, therefore, by gravity. therefore, I decided that the elastic force is equal to gravity, but this is not the case, the teacher said that this is incorrect. Then I assume that the spring stretches unevenly (probably), that is, if it stretches under its own weight, there will be almost no tension at the beginning, and at the bottom it will stretch more. So you need to divide it into many separate sections and calculate (PROBABLY), but I don't understand what formula to use for this and if this assumption is incorrect, then I can't solve the problem at all.

How much of the spring weight does the top of the spring need to support? How much does the bottom?

Consider the top and bottom coils. How much will each coil support? How much will they stretch?

Edit: You may find parts of this Insight relevant: https://www.physicsforums.com/insights/the-slinky-drop-experiment-analysed/

 

[Eshka]

How much of the spring weight does the top of the spring need to support? How much does the bottom?

Consider the top and bottom coils. How much will each coil support? How much will they stretch?

Edit: You may find parts of this Insight relevant: https://www.physicsforums.com/insights/the-slinky-drop-experiment-analysed/

thanks for the article, I literally just came across this video in other sources :). As I understand it, the lower part of the coil spring will not stretch at all (therefore, when it falls, it hangs in the air in the video), and the beginning of the spring will be stretched as much as possible. if I say that the initial length of the spring was 0.067 meters, the final length was 1.2 meters, and the mass was 0.2 kilograms, would the correct answer be 1.77? (for g= 10 and provided that gravity is equal to the elastic force). If not, then I can assume that the elastic force is 2 times less, since it is located at the beginning of the spring, but it is absent at the end. Then the answer will be approximately 0.883.

 

[Orodruin]

A number only can never be correct. Never forget to put appropriate units! It is important!

 

[haruspex]

The first step is to stop referring to the given numbers and work algebraically. There are many advantages.
Say the spring has linear density $\rho$, unstretched length $L$ and elastic modulus $\mu$. Consider an element (unstretched) length $dx$ at (unstretched) distance $x$ from the bottom of the spring.
If $\Delta dx$ is the increase in length of element $dx$ and $x'$ is its stretched distance from the bottom, can you see that $dx+\Delta dx=dx'$?
Can you write an equation relating $g, \mu, \rho, dx, dx'$?

 

[Orodruin]

If not, then I can assume that the elastic force is 2 times less, since it is located at the beginning of the spring, but it is absent at the end.

The elastic force varies along the spring since different parts of the spring support different mass below it. However, the average elastic force is half the mass and since the elongation is linear in the load, the average elastic force will be what determines the total elongation.

That the elongation is half of what it would have been if the full mass was concentrated at the end of the spring is correct - but you will need to work on a convincing argument. Try to answer the following:

1. How much mass does the upper coil support?
2. How much mass does the coil below the upper coil support? What about the one below that and so on?
3. The elongation of each coil is linear in the mass supported below it. If half the mass is supported below it, then it will elongate half of what it would if the full mass was supported. What then is the average elongation of the coil compared to what it would have been if the full mass was supported?
4. Multiply the result of 3 with the elongation the spring would have had if the entire spring had supported the full mass. Done!

The first step is to stop referring to the given numbers and work algebraically.

Yes, that always helps.

Say the spring has linear density $\rho$, unstretched length $L$ and elastic modulus $\mu$.

However, there is no need to go into this kind of detail as the argument outlined above accomplishes the task without introducing any extra variables not explicitly mentioned by the problem.

Consider an element (unstretched) length $dx$ at (unstretched) distance $x$ from the bottom of the spring.
If $\Delta dx$ is the increase in length of element $dx$ and $x'$ is its stretched distance from the bottom, can you see that $dx+\Delta dx=dx'$?
Can you write an equation relating $g, \mu, \rho, dx, dx'$?

... and this seems to be heading into integration, which will get the same answer as above, but is shooting a mosquito with a cannon.

 

[Chestermiller]

To do this problem (which involves a non-uniform extension of the spring), you need to derive an equation which specifies the local behavior of a spring experiencing a non-uniform extension. For this purpose, let x' be the location along the deformed configuration of the spring that is at location x in the undeformed configuration. That is x'=x'(x). The usual law for a spring experiencing a uniform extension is $$F=k(L'-L)=kL\frac{L'-L}{L}$$where $\frac{L'-L}{L}$ is the linearized approximation to the extensional strain of the spring (valid for small L'/L). The extension of this relationship to a non-uniform extension is $$F(x)=kL\left(\frac{dx'}{dx}-1\right)$$However, in this problem, the extended length is much larger than the undeformed length, and the strains are high. One must thus use a measure of strain that matches the behavior of the spring more accurately both at small strains and large strains. Assume the that optimal strain measure is the logarithmic strain, we would then have $$F(x)=kL\ln{(dx'/dx)}$$

 

[Orodruin]

you need to derive an equation which specifies the local behavior of a spring experiencing a non-uniform extension

Not really in this rather simple case. See #6.

 

[Chestermiller]

The problem statement does not say that the elongation is linear in the load.

 

[Orodruin]

The problem statement does not say that the elongation is linear in the load.

That is essentially implied by asking for the coefficient of stiffness.

It is also a pretty safe assumption in introductory physics homework.