Re: Why can potential energy due to gravity be ignored on a vertical spring

Dear Matt,

Let me try to guess what the original problem looks like.  If I
misstate the problem, please send me email and I'll try again.

First, let's look at the horizontal problem as a warm-up exercise.
There is a mass "m" attached to a rigid wall by a spring of force
constant "k", and the mass slides along a horizontal frictionless
surface.  You displace the mass a distance "x" away from its
equilibrium position and release it.  Then you ask for the speed of
the mass as it flies through its equilibrium position.

The principle of Conservation of Mechanical Energy says that the sum
of kinetic energy and potential energy does not change with time.  For
the initial (i) time choose the moment when the mass is released, and
for the final (f) time choose the instant when the mass passes through
equilibrium.

KE(i) +  PE(i)  =  KE(f)  + PE(f) 

         kx^2       mv^2
  0   +  -----  =  -----  +  0
           2         2   

where the initial kinetic energy is zero because the mass is released
with zero speed, and the final potential energy is zero because at the
equilibrium position, the spring is neither stretched nor compressed so
no energy is stored in the spring.

Solving this equation for the speed of the mass at equilibrium we get

v = x sqrt[k/m]

Notice that this differs from the speed in your question by a factor
"x", the maximum displacement of the mass away from equilibrium.
_______________________________________________________________________

Now we will tackle the vertical problem.  The same mass "m" hangs from
the same spring which is attached to the ceiling.  This time, the
equilibrium position (the place where the net force is zero) will be
where the spring is stretched a distance mg/k.

To see this, recall that the Hooke's Law force has magnitude k times
the displacement from the rest length of the spring, which we will
call y to remind us that the displacement is vertical.  The direction
of this spring force will be up.  The only other force acting on the
mass is the gravitational force which has magnitude mg and direction
down.  Since the net force is zero at equilibrium,

ky - mg = 0     or     y = mg/k

Now we will pull the mass a distance "x" away from its new equilibrium
position and release it.  Then we will find the speed of the mass as
it flies through its new equilibrium position using Conservation of
Mechanical Energy.  Just as previously, for the initial (i) time
choose the moment when the mass is released, and for the final (f)
time choose the instant when the mass passes through its new equilibrium.

The difference is that now we must account for two potential energies,
one from the spring and the other gravitational.  We also need to 
choose the location where the gravitational potential energy is zero.
This choice is complete arbitrary and does not affect the final answer.
Another way to say this is that the potential energy is only defined
up to a constant.  I will take the gravitational potential energy to
be zero at the initial position of the mass.  Try to work the problem
with a different choice and you will see that it does not matter.


KE(i) +          PE(i)     =  KE(f)  +       PE(f) 

         k(x+y)^2              mv^2     ky^2  
  0   +  -----     +  0    =  -----  +  -----  + mgx
           2                    2         2   


Some terms deserve further explanation.  The initial potential energy
stored in the spring is k/2 times (the total distance that the spring
has been stretched)^2.  The spring was stretched a distance y=mg/k by
gravity, then a further distance x by hand.  In the final spring
potential energy, the spring is still stretched a distance y by
gravity.  The initial gravitational potential energy is zero by
choice, and the final gravitational potential energy is the mass times
the gravitational constant "g" times the distance above the zero
level, which is "x" for us.

Next, substitute mg/k for y in the Conservation of Mechanical Energy
equation above to get

1/2 k (x + mg/k)^2  = 1/2 mv^2 + 1/2 k(mg/k)^2 + mgx

and expand the left hand side

1/2 kx^2 + kxmg/k + 1/2 k(mg/k)^2 = 1/2 mv^2 + 1/2 k(mg/k)^2 + mgx

Now notice that the third term on the left is identical to the second
term on the right, and that the second term on the left (after
cancelling the k's) is the same as the third term on the right.  This
leaves

 kx^2       mv^2 
 -----  =  ----- 
   2         2   

which is exactly the equation we solved in the horizontal case above.
The speed of the mass as it passes through its new vertical
equilibrium is

v = x sqrt[k/m]

the same speed as in the horizontal case.


Here is one of many web sites that illustrates vertical springs 
http://stravinsky.ucsc.edu/~josh/5A/book/harmonic/node13.html

--Dr. Randall J. Scalise    http://www.phys.psu.edu/~scalise/