Solving the Spring Paradox: Uncovering the Error

  • #1
New user has been reminded to use the Homework Help Template when starting schoolwork threads at the PF
Good evening.

The problem states: Spring paradox. What is wrong with the following argument? Consider a mass m held at rest at y = 0, the end of an unstretched spring hanging vertically. The mass is now attached to the spring, which will be stretched because of the gravitational force mg on the mass. \Vhen the mass has lost gravitational potential energy mgy and the spring has gained the same amount of potential energy so that mgy = ~ Cy²/2, the mass will come to equilibrium. Therefore the position of equilibrium is given b y=2mg/C.

Working out by the principle of forces, where the weight, when equated to the elastic force shows,
Fg= -Fk or Fg + Fk = 0 (in the equilibrium point).
-mg = -Cy, and y = mg/C. (What is different from what would be obtained from the spring paradox).
But i simply don't know how to show that the spring argument is wrong, as well.

Please, help!

Answers and Replies

  • #2
A very interesting problem. A couple of observations solve this paradox, so that it is no longer a paradox. When the mass is released, simple harmonic motion occurs where the total energy=kinetic energy plus potential energy is constant. To lead you to the solution to this paradox, what is the kinetic energy the instant the mass is released? Thereby is not the total energy equal to the potential energy? Isn't there another place in the simple harmonic motion where the kinetic energy is zero? What is the potential energy there? Answering these questions should help resolve the paradox. ## \\ ## Meanwhile, along the path of simple harmonic motion, when the kinetic energy is non-zero, is there anywhere that the potential energy is equal to the initial potential energy? ## \\ ## If the spring system is slightly damped by air resistance, where will the equilibrium position be where the mass on the spring finally stops? (Hint: It's the place in the simple harmonic motion where the acceleration is zero).
  • #3
At the point described in the bold text the PE lost might be equal to the KE gained but is the system really in equilibrium?
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Likes Charles Link
  • #4
At the point described in the bold text the PE lost might be equal to the KE gained but is the system really in equilibrium?
To add further to @CWatters input, is the acceleration equal to zero either at the position ## y=0 ## or at the position where ## mgy=-\frac{1}{2}Cy^2 ##? What is the velocity at both of these points, and also what is the kinetic energy? ## \\ ## Is the system at equilibrium if the velocity is zero if it is accelerating? ## \\ ## An additional question that I think you @LukasMont can answer is what is the acceleration at ## y=0##? Without computing it, what do you expect the acceleration to be at the point where ## mgy=-\frac{1}{2}Cy^2##? And then go and compute it, to see if everything is as expected.
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  • #5
It is an interesting question as to how exactly equilibrium is defined in physics. Certainly it includes that forces and torques balance, so no acceleration; but does it also require zero velocity? None of the definitions I have found online specify that.

Anyway, while the above replies are correct, I would simply observe that there is no basis for claiming that equilibrium is when the GPE lost equals the EPE gained. This circumvents the question of how the mass descends, e.g. it might be lowered gradually.

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