What's Wrong with Energy Argument in Vertical Spring-Mass System?

  • Thread starter Thread starter waqar
  • Start date Start date
  • Tags Tags
    System Vertical
AI Thread Summary
The discussion centers on the discrepancy between the displacement calculated using force analysis and energy methods in a vertical spring-mass system. The force method yields a displacement of x = -mg/k, while the energy method results in x = 2mg/k, indicating a mistake in the energy approach. Participants highlight that the mass, when dropped freely, will oscillate around an equilibrium point rather than come to rest immediately, which complicates energy conservation. It is emphasized that energy conservation cannot be applied if the system is allowed to oscillate without external damping. The conclusion is that the energy argument fails due to the assumption of the mass coming to rest without accounting for oscillations.
waqar
Messages
2
Reaction score
0
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, that's 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
 
Physics news on Phys.org
waqar said:
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, that's 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.
 
gneill said:
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.
 
waqar said:
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.
 
  • Like
Likes waqar
waqar said:
and come to rest
waqar said:
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.
 
  • Like
Likes waqar
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top