Vertical Spring-mass system.

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Homework Statement:
A mass is attached to an un-stretched spring, vertically, what displacement does it make finding displacement using forces (F_g=F_s) gives half the answer from energy method (U_g=U_s), what mistake am I making?
Relevant Equations:
mg=kx
mgx=0.5kx^2
The spring is first relaxed at its lower end h_1, thats when we attached mass m to itand let it fall freely and come to rest. Now by Force analysis we can show that in final position of mass the forces of gravity and restoring force from spring are balanced. i.e. F_g=F_s which yields
x=-mg/k.
Solving the same problem with energy argument; initial potential energy in mass in mgh_1 and final would be mgh_2, meanwhile the initial potential energy in spring is zero and finally it would be0.5kx^2.
Solving by enrgy, K_i + U_g_i + U_s_i = K_f + U_g_f + U_s_f
=> 0 + mgh_1 + 0 = 0 + mgh_2 + 0.5k(h_1-h_2)^2
=> mg(h_1-h_2) = 0.5k(h_1-h_2)^2
=> mgx = 0.5kx^2
=> x = 2mg/k

which is double the displacement we found from force method... what is going wrong here
 

Answers and Replies

  • #2
gneill
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Problem Statement: A mass is attached to an un-stretched spring, vertically, what displacement does it make finding displacement using forces (F_g=F_s) gives half the answer from energy method (U_g=U_s), what mistake am I making?
Relevant Equations: mg=kx
mgx=0.5kx^2

The spring is first relaxed at its lower end h_1, thats when we attached mass m to itand let it fall freely and come to rest.
Why would it come to rest? Would it not oscillate about some median point? The so-called equilibrium point?

Letting the mass of a mass-spring system drop freely and carefully lowering it to the equilibrium point via some external agency (such as your hand) are quite different situations.
 
  • #3
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Why would it come to rest? Would it not oscillate about some median point? The so-called equilibrium point?

Letting the mass of a mass-spring system drop freely and carefully lowering it to the equilibrium point via some external agency (such as your hand) are quite different situations.
No external force is applied. The mass was left to fall freely, hanging by the spring. The analysis is made when system has now come to rest. That's when we can take F_g=F_s and also complete energy transfer from gravitational to elastic.
 
  • #4
gneill
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No external force is applied. The mass was left to fall freely, hanging by the spring. The analysis is made when system has now come to rest. That's when we can take F_g=F_s and also complete energy transfer from gravitational to elastic.
There is no mention of damping that would allow the oscillations to die out. The mass will pass through the equilibrium point where gravitational force equals spring restoring force, and proceed until its progress is finally arrested by the restoring force whereupon it will return upwards, again passing through the equilibrium point. In other words it will oscillate about the equilibrium point. The greatest displacement will not be the equilibrium point.
 
  • #5
haruspex
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and come to rest
Solving the same problem with energy argument
To put @gneill's answer the other way around, if SHM comes to rest then energy is not conserved, so you cannot use the energy argument.
To make energy a valid method, you would need it to come to rest by some means other than dissipation, e.g. a ratchet that holds it at max extension.
 

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