Class content > Oscillations and Waves > Harmonic Oscillation > Mass on a spring
Prerequisites:
Though our analysis of the oscillation of an idealized horizontally moving mass on a spring was fairly straightforward (since there was only a single horizontal force), it's not very easy to set up. Getting the friction down to a negligible point for horizontal motion requires either very good wheels (ball bearings) or an air track, and most springs don't obey Hooke's law (see Realistic Springs) under compression. A cleaner and easier to set up experiment is to hang a mass from a spring, let it settle down, and take that as our reference point. A realistic experimental setup is sketched at the right.


If we let the mass hang from the spring and settle down until it comes to rest, the spring will be stretched until the upward force of the spring exactly balanced the gravitational pull of the earth  the mass's weight. This is displayed in the freebody diagram for the mass shown at the right.


Suppose the rest length of the spring (with nothing hanging from it) is L_{0} and that when the mass is on it, the spring stretches to a length L. If the spring constant of the spring is k, then the force balance at the equilibrium point will be
k(LL_{0}) = mg.
Let's choose the vertical coordinate of our mass to be y with the positive direction up and the origin chosen to be at the position when the mass is at equilibrium. If we now let the mass move up and down, the actual length of the spring will be L  L_{0}  y. (Why is it minus y instead of plus y?) As a result, the net force on the mass when the mass is at a position y will be
F^{net} = k(LL_{0}  y)  mg.
Let's check that the signs of this are correct.
If the mass is above the equilibrium point (y is positive), the spring is not stretched as much as it would be at the equilibrium point, so the spring will not be pulling up as hard as it is at equilibrium. The objects weight (the pull of gravity downward) doesn't change, so the resultant will be down  the spring is stretched less than at equilibrium so gravity wins.
If the mass is below the equilibrium point (y is negative), the spring is stretched more than it would be at the equilibrium point, so the spring is pulling up harder than it is at equilibrium. The objects weight (the pull of gravity downward) doesn't change, so the resultant will be up  the spring is stretched more than at equilibrium so the spring wins.
So this is qualitatively OK. If we lift the mass upward, our analysis tells us that the resultant forces will pull it down when we release it, and if we pull the mass downward, our analysis tells us that the resultant forces will pull it up when we release it, just as we know physically it must.
We can therefore write Newton's second law. We'll group the constant parts of the force together for convenience:
ma = F^{net} = k(LL_{0}  y)  mg = [k(LL_{0} )  mg]  ky.
But the part of the net force in square brackets cancels! That's just the equilibrium condition. (Remember that L is a specific length  the length of the spring at equilibrium  NOT a variable!) So our Newton's second law reduces to
ma = ky
or
a = (k/m)y
Writing k/m = ω_{0}^{2}, we get
the same equation as we had for the horizontal mass, but with y instead of x. So we of course will expect that we will get the same solution as we had before:
where we have written
 y(t) to remind us that y is being expressed as a function of t,
 A for the amplitude of the oscillation,
 ω_{0} for the angular frequency ( = 2π/period), and
 φ for the phase shift that tells us where the oscillation starts when t = 0.
When we make the connection with real data taking, we may have to choose our origin at the motion detector rather than at the equilibrium point. If this is a distance d below the equilibrium point, then the solution would simply add d onto the solution we have already found:
.
Joe Redish 3/14/12
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