When a spring is cut in half

[silverice]

Okay, I get why the spring constant doubles when the spring is cut in half according to the F=-kx formula.

But when I think about it, it doesn't really make sense to me. The spring constant is depends on the material make up of the spring, while not any of that has changed by breaking the spring in half, how does the spring constant change?

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[Hemorrage]

From my understanding, the spring constant k is defined by the equilibrium position of the spring, therefore if you cut the length of the spring in half, k would have to double to compensate. I'm not 100% sure though

 

[Laureatebarrel]

Can you post the whole question? I don't think k is gonna double since like you said it is an intrinsic quality.

 

[silverice]

Can you post the whole question? I don't think k is gonna double since like you said it is an intrinsic quality.

sure, below it's the question from EK physics Q 251

A spring (k=400N/m) is cut in half to make two new springs. what is teh spring constant of each of the new springs?
A 100 N/m
B 200 N/m
C 400 N/m
D 800 N/m
Answer is D 800 N/m

 

[silverice]

From my understanding, the spring constant k is defined by the equilibrium position of the spring, therefore if you cut the length of the spring in half, k would have to double to compensate. I'm not 100% sure though

Yeah, you are right, that's exactly the logic that was provided by the answer. I do get why the spring constant is double from this perspective.

But i'm just saying, it doesn't make sense if i think of it the other way, as in the spring constant is a property of the make up of the spring material (the stiffness), how does cutting it in half make a difference.

 

[cheesier]

You can solve it via conservation of energy.

The potential energy of the big spring is 1/2kx^2

When you cut it in half it becomes 1/2k(new)(x/2)^2 = 1/2k(new)x^2 * 1/4 (you have two smaller springs that have this as their potential energy)

If you add both of the potential energies of the smaller springs together, it has to be equal to the bigger spring.

1/2k(new)x^2*1/4 + 1/2k(new)x^2*1/4 = k(new)x^2*1/4

k(new)x^2*1/4 = 1/2k(old)x^2

k(new) * 1/4 = 1/2k(old) ==> k(new) = 2k(old)

 

[sazerac]

I think silverice is looking for a more intuitive explanation.

You are correct that cutting the spring in half does not change any intrinsic property of the underlying metal. Your fundamental hang-up is that X, the spring displacement, is measured not in terms of multiples of the length of the original spring, but rater in terms of absolute displacement.

Let's say you have a spring, and it is 12 coils. You pull on it with 50lbs of force. It stretches one foot. Each individual coil in the spring stretches one inch, contributing 1/12 of the total force.

Now cut the spring in half, so it is only 6 coils long. Pull on it with 50lbs of force again. Each coil will stretch one inch, and the total displacement will be 0.5 feet. How can you get the half-size spring to stretch a full foot? Pull twice as hard, with 100lbs. Each coil will now stretch 2 inches.

By cutting the original spring in half, the remaining coils will pull twice as hard when stretched to the original displacement. Working twice as hard means K doubled.

 

[contraband]

Think of k, the spring constant, like R, resistance in a circuit. R = resistivity * (length/area). Resistivity is intrinsic to the material of the resistor - it would be the same for a long, skinny copper wire (with high resistance) and a short, fat copper wire (with low resistance). Thus, the dimensions/configuration of the resistor along with the type of material (resistivity) will determine the overall resistance. By changing the dimension/configuration of the spring, you're changing its overall resistance to a change in shape (restoring force) without changing the type of material the spring is made of.

 

[silverice]

Thank you everyone for all the interesting ways of explaining this problem.

sazerac you knew exactly of what I was looking for! Thx a bunch!

 

[anon4895]

Okay, I get why the spring constant doubles when the spring is cut in half according to the F=-kx formula.

But when I think about it, it doesn't really make sense to me. The spring constant is depends on the material make up of the spring, while not any of that has changed by breaking the spring in half, how does the spring constant change?

The spring constant depends on what the spring is made out of as well as the geometry of the spring.

You can calculate it based on the cross sectional area, length, and youngs modulus of the material. Youngs modulus depends only on what the material is.