Re: [Phys-L] spring energy and other equation-hunting puzzles

[Moses Fayngold <moshfarlan@yahoo.com>]

On 02/09/2015 05:32 PM, Anthony Lapinski wrote:


> 3. You hang an object on a spring (with spring constant k). At the
> equilibrium position of the object, where it hangs at rest, the spring has
> stretched a distance x. You wish to find the object's mass, and ask two
> classmates for help. Equating forces, student A sets mg = kx to find that
> m = kx/g. Equating energies, student B sets mgx = 0.5kx2 to find that m =
> kx/2g -- half of student A's value. Whose method is correct and why?
> What's wrong with the other method?

At the equilibrium, the object's acceleration and therefore the net force on it is zero, 
                                                          mg - kx=0,                                         (A)
implying that m = kx/g. This method - Newton's second law used by student A is correct both - conceptually and factually.  But equating potential energies at the equilibrium point, that is setting                                                           0.5kx2-mgx=0                                  (B)is conceptually wrong since the net potential energy alone does not necessarily conserve. 
Note that by gently lowering the suspended object with negligible speed down to the equilibrium position 
you remove the important actor - kinetic energy - from the play. The resulting total mechanical energy of the object at the equilibrium position is not the same as its initial total energy, since you changed the Hamiltonian by lowering the object with support. If you want to consider only the gravitational and spring forces, then the correct use of energy conservation under the corresponding idealized conditions demands elimination of all other forces (e.g., support and air resistance). OTOH, it requires kinetic energy to stay. With the initial position (when you just suspend the object to the relaxed spring) taken as the reference point, the initial mechanical energy is zero. It must remain zero all the time after you release the object at suspension point:                                                        0.5kx2-mgx+ 0.5mv^2=0                   (C)That would be another correct equation. Unlike (A), it describes simple harmonic oscillations around equilibrium position x=x0=mg/k, which follows from (A) if we assume m known. If you solve (C) under the same assumption for a point where the velocity v=0, you will find two solutions - the initial x=0, and the lowest point  x=2x0. It is this point where you can safely use conservation law for potential energies alone. Then you can correct (B) by changing there x0 to 2x0 and get the same value as student A.  By all accounts, the approach of (A) is much more straightforward.
Moses FayngoldNJIT