VERTICAL SPRING AND MASS OSCILLATORS,
THE GRAVITY OF OUR NEGLECT
When we discussed the oscillations of a mass hung from a spring, we only
talked about the force and energy due to the spring; and I told you we could
neglect gravity.
So how, exactly is this possible?
Let’s do the full treatment, with gravity and spring and see how it all comes
out. Let’s call the energy of the system at the unstretched length of the
spring zero.
At the equilibrium point, we know that the force of gravity equals the force
of the spring to give a net force of zero:
Fnet = kΔL – mg = 0 so mg = kΔL
The potential energy will be different from the original unstretched
position. The gravitational energy will be less by mgΔL. The spring potential
energy is greater by ½ k(ΔL)2. This gives the energy at equilibrium as
Eeq = ½ k(ΔL)2 – mgΔL.
So how about some arbitrary position a distance x above equilibrium?
Then the gravitational energy is – mgΔL +mgx due to the change in
height. The spring potential energy is not so simple. Since it depends on the
square of the stretch, we need to do calculate the new stretch first, then
square it. The new stretch will be ΔL -x , less by the increased height.
This gives a spring potential energy of ½ k(ΔL-x)2 and a total potential
energy of
Ex = ½ k(ΔL-x)2 - mgΔL +mgx
Now for the algebraic onslaught! The general road map is to take the total
potential energy at height x above equilibrium, expand the square of the
distance and subtract the potential energy at equilibrium to see what’s left.
Then we will turn our attention to the force acting at each point and
subtract see how this works.
Energy for x above equilibrium:
(ΔL-x)2 = ΔL2 - 2xΔL +x2 so
Ex = ½ k(ΔL-x)2 - mgΔL +mgx
Ex = ½ k(ΔL2 - 2xΔL +x2) - mgΔL +mgx
Subtracting the potential energy at the equilibrium point gives:
-
Ex = ½ k(ΔL2 - 2xΔL +x2) - mgΔL +mgx
Eeq = ½ k(ΔL)2
– mgΔL
2
ΔE = -kxΔL + ½ kx + mgx
= ½ kx2 + mgx - kxΔL
= ½ kx2 + (mg – kΔL)x
From the force balance at the equilibrium point, mg = kΔL
So,
mg – kΔL = 0 and
ΔE = ½ kx2 + (mg – kΔL)x
= ½ kx2 and we can use this because we can set the energy under any
particular circumstance equal to our zero level.
How about that force?
If we want to treat the equilibrium position as our zero, then the net force
is already set for us; it is zero there already. What about the force on the
mass at position x above the equilibrium position?
There we have F(x) = -mg + k(ΔL-x) since the stretch is less by x.
Expanding this gives:
F(x) = -mg + kΔL- kx
= (kΔL-mg) - kx
= 0
- kx using the force balance at equilibrium.
Notice that, if the mass is below the equilibrium position a distance –x,
everything still works; the energy is still ½ kx2 and the force is +kx.
So you see our net force is just that of a spring compressed from
equilibrium by x, F= –kx. And, our potential energy is just that of a spring
compressed by x : E = ½ kx2 , so we can ignore gravity and just measure
everything from the equilibrium position!