Spring Constant Dilemma: Understanding Discrepancies in Solving for k

[argentum108]

Consider the following question: a woman of mass m steps onto a vertical spring and compresses it a distance d. What is the value of the spring constant k?

The dilemma is that I found what I believe are two valid ways of solving this problem, but the two results for k are not equal!

If someone can explain where the discrepancy comes in, I'd much appreciate it.

Method I
For a spring, the force of compression is F=kd.
In this case, F also equals mg.
Hence, k=mg/d

Method II
ΔPE woman equals mgd
ΔPE spring equals kx²/2, which in this case is kd²/2
Hence, k=2mg/d

 

[AlephZero]

If depends what you think "compresses the spring a distance d" means.

If the woman suddenly steps onto the spring, the spring will move in simple harmoic motion, and in Method II you took d = the maximum displacement of the spring, when the KE of the woman is 0. That is twice the displacement to the equilibrium position of the spring.

If d is the equilibrium position of the spring, then Method I is correct.

 

[Emilyjoint]

The energy stored (PE) is 0.5Fd (average force x distance) so PE = 0.5 x mg x d.
The definition of spring constant is force/extension so k = mg/d.
This means mg = kd, this gives the PE equation PE = 0.5kd^2

 

[argentum108]

The energy stored (PE) is 0.5Fd (average force x distance) so PE = 0.5 x mg x d.
The definition of spring constant is force/extension so k = mg/d.
This means mg = kd, this gives the PE equation PE = 0.5kd^2

Well, I was looking for and explanation of how two different ways of deriving k gave two different answers, not a derivation of the PE of a spring. You still have not explained why k derived through PE is not the same as k derived by force.

If depends what you think "compresses the spring a distance d" means.

If the woman suddenly steps onto the spring, the spring will move in simple harmoic motion, and in Method II you took d = the maximum displacement of the spring, when the KE of the woman is 0. That is twice the displacement to the equilibrium position of the spring.

If d is the equilibrium position of the spring, then Method I is correct.

AlephZero, would it be correct then to say that Method II ignores the fact that the woman's loss of gravitational potential energy is not just converted into spring potential energy but also kinetic energy of the woman as she lowers?

 

[Studiot]

What makes you think either method is correct?

It is impossible to apply a force mg to the unstrained spring, since F_s=k(0) = 0
Where F_s is the instantaneous force in the spring.

What actually happens is that the mass is accelerated by mg downwards and this acceleration is resisted by an increasing spring force F_s upwards.

At the point where the spring force up equals mg down the mass has traveled down a distance d and is traveling at velocity v.

The potential energy mgd has been converted into the strain energy in the spring plus the kinetic energy of motion at displacement = d

The mass continues downwards meeting increasing resistance from the spring until the upwards force from the spring decelerates the mass to zero velocity.
The force from the spring is then greater than than mg so the mass is then accelerated upwards to again pass through the point with displacement d below the unstrained position, this time moving upwards with velocity v.

Yes, in the absence of dissipative mechanisms such as heating in the spring, air resistance, noise etc this simple harmonic motion witll continue indefinitely, swopping bewteen potential strain energy in the spring, potential gravitational energy of the mass m and kinetic energy of the mass m.

It is not hard to establish the apportionment of these energies can you do this?

You should note that you cannot apply an equilibrium analysis to this system since the system is never in equilibrium.

 

[Nugatory]

What makes you think either method is correct?
...
You should note that you cannot apply an equilibrium analysis to this system since the system is never in equilibrium.

You are being too pessimistic about approach number 1, I think.

Whether the system is in equilibrium or not, we have mg=kd at the point where the acceleration is zero, so approach 1 is a valid way of measuring the spring constant. It just so happens that it is a damn sight easier to find the point where the acceleration is zero when the system is in equilibrium... and the easiest way of doing that is to transfer the weight to the scale slowly enough that the oscillations can be ignored.

In my experience, non-physicists step onto bathroom scales slowly, not because they understand any of this stuff, but because they don't want to wait longer than necessary for the scale to settle down... And every bathroom scale that I've encountered has a fairly high damping factor, perhaps because the designer does understand this stuff...

 

[Studiot]

Whether the system is in equilibrium or not, we have mg=kd at the point where the acceleration is zero,

Yes this may be true, depending upon the actual system and the circumstances.

However the rest does not follow.

Since the strain energy in the spring is exactly half that of the loss of PE of the mass, one of three things have happened, and since energy must be conserved in the universe.

1) The remainder of the energy remains in the system but not as strain energy.
2) The remainder of the energy is dissipated by/ transferred to an external agent.
3) The remainder of the energy is dissipated by the process of applying mg slowly to the spring.

Expanding upon these three possibilities, note they are not mutually exclusive.

1) The system is in oscillation as described.

2) The unaccounted for half of the energy has been dissipated in the damping mechanism.

3) The mass m has not acted on the spring for the full distance d. ie it has been applied slowly. The only way this can occur is if some of the mass is supported on another structure during transfer. For example the person standing on one leg and stepping gently onto the scale with the other, slowly transferring weight as indicated. In this instance the force mg has not been applied constantly to the spring through its compression. The remaining energy is dissipated within the mass and its other supports, perhaps rotational work is done, perhaps some is converted to heat etc.

So the actual situation is complicated (which is why it is usually avoided in texts).

Well done to argentum for noticing the apparent discrepency.