Vertical Spring-mass system.

In summary: Solving the same problem with energy argumentTo 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.
  • #1
waqar
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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, 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
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
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  • #5
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.
 
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Related to Vertical Spring-mass system.

1. What is a vertical spring-mass system?

A vertical spring-mass system is a mechanical system consisting of a mass attached to a vertical spring, which is fixed at one end and free to move at the other end. When the mass is disturbed from its equilibrium position, the spring will stretch or compress, resulting in an oscillatory motion of the mass.

2. What is the equation of motion for a vertical spring-mass system?

The equation of motion for a vertical spring-mass system is given by F = -kx - mg, where F is the net force acting on the mass, k is the spring constant, x is the displacement of the mass from its equilibrium position, and g is the acceleration due to gravity.

3. How does the spring constant affect the motion of a vertical spring-mass system?

The spring constant, k, determines the stiffness of the spring and thus affects the frequency of oscillation of the mass. A higher spring constant will result in a higher frequency and a shorter period of oscillation, while a lower spring constant will result in a lower frequency and a longer period of oscillation.

4. What factors affect the amplitude of oscillation in a vertical spring-mass system?

The amplitude of oscillation in a vertical spring-mass system is affected by the initial displacement of the mass, the mass of the object, the spring constant, and the damping force (if present). A larger initial displacement or a smaller mass will result in a larger amplitude, while a larger spring constant or a stronger damping force will result in a smaller amplitude.

5. What is the relationship between the period of oscillation and the mass in a vertical spring-mass system?

The period of oscillation, T, in a vertical spring-mass system is directly proportional to the square root of the mass, m. This means that as the mass increases, the period of oscillation will also increase, and vice versa. This relationship is given by T = 2π√(m/k).

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