Maximum height reached by object released from a vertical spring

  • Thread starter Happiness
  • Start date
  • #1
665
26
Mentor Note: thread moved, therefore no template
A mass m is placed on a vertical spring and allowed to reach equilibrium, whose level is ##e_0## below the original level of the spring before placing the mass. It is then pushed downwards such that it is now a distance ##x_0## below the equilibrium level. What is the maximum height d from the level of release it can reach?

Method 1
By conservation of energy, loss in elastic potential energy (EPE) = gain in gravitational potential energy (GPE).
$$\frac{1}{2}k(e_0+x_0)^2=mgd$$
$$d=\frac{k(e_0+x_0)^2}{2mg}$$

Method 2
After release, the mass moves upwards and loses contact with the spring at a level distance x above the equilibrium level. At the point where contact is lost, ##a=g=\omega^2x##, where ##\omega=\sqrt{\frac{k}{m}}##, since it is in simple harmonic motion.
And velocity $$v=\omega\sqrt{x_0^2-x^2}$$
By conservation of energy, loss in kinetic energy (KE) = gain in gravitational potential energy.
##\frac{1}{2}mv^2=mgh##, where h is the distance from the point where contact is lost to the maximum point. Making h the subject and substituting v from the equation above and ##x=\frac{g}{\omega^2}##, $$h=\frac{\omega^2x_0^2}{2g}-\frac{g}{2\omega^2}$$
$$d=h+x+x_0=\frac{\omega^2x_0^2}{2g}+\frac{g}{2\omega^2}+x_0=\frac{kx_0^2}{2mg}+\frac{mg}{2k}+x_0$$

It can be shown that the two answers are the same. My question is just before contact is lost, the mass has EPE, KE and GPE, but just after contact is lost, it only has KE and GPE, so where does the EPE go to? EPE isn't included in the calculation of h in method 2 (in the step where conservation of energy is used). So it seems like the mass loses some energy (equal to the EPE) after it loses contact. So I would expect method 2's answer to be smaller than method 1's, where no such loss in energy is used.
 
Last edited by a moderator:

Answers and Replies

  • #2
jbriggs444
Science Advisor
Homework Helper
2019 Award
9,071
3,798
In taking the zero point for the spring potential energy at x0, you are implicitly removing gravity from the situation. Instead of considering a mass at equilibrium with a stressed spring plus gravity, you are considering an equivalent situation with a mass at rest at x0 connected to an unstressed spring with no gravity. That is an excellent simplification.

Solution 1 complicates the simplification by re-injecting gravity. That's erroneous.
 
  • #3
665
26
In taking the zero point for the spring potential energy at x0, you are implicitly removing gravity from the situation. Instead of considering a mass at equilibrium with a stressed spring plus gravity, you are considering an equivalent situation with a mass at rest at x0 connected to an unstressed spring with no gravity. That is an excellent simplification.

Solution 1 complicates the simplification by re-injecting gravity. That's erroneous.
I have edited solution 1. Is it now correct?

I added a question in the last paragraph. How could the two method give the same answer when one involves a loss of EPE and the other doesn't?
 
Last edited:
  • #4
jbriggs444
Science Advisor
Homework Helper
2019 Award
9,071
3,798
You have modified the original post. Let me see if I can agree with the updated equations.
A mass m is placed on a vertical spring and allowed to reach equilibrium, whose level is ##e_0## below the original level of the spring before placing the mass. It is then pushed downwards such that it is now a distance ##x_0## below the equilibrium level. What is the maximum height d from the level of release it can reach?

Method 1
By conservation of energy, loss in elastic potential energy (EPE) = gain in gravitational potential energy (GPE).
$$\frac{1}{2}k(e_0+x_0)^2=mgd$$
$$d=\frac{k(e_0+x_0)^2}{2mg}$$
Slow down and back up. What is the left hand side of that first equation supposed to represent? What is the right hand side of the first equation supposed to represent? What justification do you have for assuming conservation of energy between the two represented states?
 
  • #5
665
26
You have modified the original post. Let me see if I can agree with the updated equations.

Slow down and back up. What is the left hand side of that first equation supposed to represent? What is the right hand side of the first equation supposed to represent? What justification do you have for assuming conservation of energy between the two represented states?
Since there is only EPE at first (just before release) and only GPE in the end (at the maximum height), the left hand side is loss in EPE and right hand side is gain in GPE. I assume the spring is massless, so it cannot have any energy at the final state.
 
  • #6
665
26
I think I have found my misconception. There is no loss of EPE in method 2 too.
 
  • #7
jbriggs444
Science Advisor
Homework Helper
2019 Award
9,071
3,798
Since there is only EPE at first (just before release) and only GPE in the end (at the maximum height), the left hand side is loss in EPE and right hand side is gain in GPE. I assume the spring is massless, so it cannot have any energy at the final state.
That does not answer the first two questions I asked. What two states are you comparing and asserting have equal energies?

As for the third (is energy conserved), look at the problem statement:
It is then pushed downwards
Does that not pose a problem for a conservation of energy analysis?
 

Related Threads on Maximum height reached by object released from a vertical spring

Replies
2
Views
2K
Replies
7
Views
55K
Replies
10
Views
5K
Replies
3
Views
5K
  • Last Post
Replies
3
Views
6K
  • Last Post
Replies
5
Views
4K
Replies
9
Views
25K
Replies
7
Views
4K
  • Last Post
Replies
7
Views
9K
Replies
2
Views
21K
Top