# Hanging around with Hooke's LawJuly 11, 2019 8:44 AM   Subscribe

Hopefully this is a trivial physics question, but I can't figure it out from the descriptions of Hooke's Law I've found. How does the apparent weight of a hanging spring change as the spring oscillates?

Let's say you have an extension spring hanging from a wooden joist in your ceiling. The upper end of the spring is fixed to the joist. You attach a 1 KG ball to the bottom of the spring.

When the spring is at rest, the weight of the combined spring+ball is 1 KG plus the weight of the spring. Fine.

Then you pull the ball down 1 meter, hold it in that position. The apparent weight (from the perspective of the joist) has increased, according to the rules of Hooke's law. Is that correct? How much has it increased? What form of Hooke's law goes here?

Then, you let go of the ball. The spring pulls the ball up; it goes past its initial position; it reaches a peak where it seems like the ball wouldn't weigh anything at all? Then it falls down again. This oscillating movement continues for some time.

How does the weight of the combined spring+ball vary across those oscillations? We can discount the weight of the spring and the deterioration of the oscillation over time if those simplifications help.

BTW, this is not for a homework assignment or anything like that. It's for a personal project.
posted by Winnie the Proust to Science & Nature (5 answers total) 2 users marked this as a favorite

When you pull the weight down, you're just adding weight,as far as the spring is concerned. You can work out the total downward force on the joist from the extension of the spring, just as you would if there were no 1Kg weight.

When the spring is at rest with the ball hanging from it, the force on the joist can again be worked out from the extension of the spring. This time that extension is due to the weight alone, not a combination of a weight and an extra pulling force, but that's irrelevant.

When the weight oscillates and moves up past its at-rest position, the force on the joist can once again be calculated from the extension of the spring. If you get it just right and the ball moves up to a point where the extension is zero, the spring is no longer under tension, so the force on the joist is zero.

The one anomalous situation is where the ball moves up, hits the spring, and either compresses the spring (unlikely) or just knocks the spring to the side. In that scenario there could be a small, random upward force on the joist, or a bigger force if the ball hits the ceiling at speed. You probably don't want that.
posted by pipeski at 9:01 AM on July 11 [3 favorites]

Yes, there are two parts to the weight. The gravitational force, which is the mass times the gravitational acceleration, and the kinematic force, which is the mass times the kinematic acceleration. Both get added together, and stretch the spring an amount proportional to their sum divided by the spring stiffness.
posted by Maxwell's demon at 9:07 AM on July 11

I think what you're really interested in is the force on the joist. Weight is a kind of force, but not the force you're interested in here, I think. You can use Hooke's law to tell you what the forces are on the spring which will, in turn, tell you about the forces of the spring on the joist.

So, say you have a spring with a spring constant of 10 N/m. If you have a 1kg bowling ball attached to the spring at rest, the string will stretch 1m: a 1 kg bowling ball has a weight of ~10 N, so will exert 10 N of force down on the spring, which will cause the spring to stretch 1m. That spring, in turn, will exert 10 N of force down on the joist (ignoring the weight of the spring, for now). So, when the spring is stretched 2m, the spring will exert 20N of force down on the joist. When the spring is stretched -1m (i.e. 1 m ABOVE its equilibrium point), it will exert -10N of force on the joist (i.e. 10N up). You can just add the weight of the spring (in N) to the above numbers if you want to consider the effect the spring's weight has on the force on the joist.
posted by Betelgeuse at 9:09 AM on July 11 [2 favorites]

When the spring is stretched -1m (i.e. 1 m ABOVE its equilibrium point), it will exert -10N of force on the joist (i.e. 10N up).

To be clear, this would be correct for a compression spring that had been shortened by 1m from its no-load condition.

It would not be correct for a bowling ball hanging from a tension spring; the equilibrium condition for that total system would involve the spring being extended by 1m, therefore exerting 10N of force upwards on the bowling ball at the bottom end and 10N downward on the joist at the top.

For that system, applying enough external upward force to the bowling ball to raise it by 1m would return the spring to zero extension, thereby making it exert no upward force on the bowling ball at all, and no corresponding downward force on the joist.

Then, you let go of the ball. The spring pulls the ball up; it goes past its initial position; it reaches a peak where it seems like the ball wouldn't weigh anything at all? Then it falls down again. This oscillating movement continues for some time.

How does the weight of the combined spring+ball vary across those oscillations? We can discount the weight of the spring and the deterioration of the oscillation over time if those simplifications help.

Neglecting spring weight, deviations from Hooke's law, air resistance and conversion of flexion to heat in the spring, what you end up with here is a system where if you plot the tension in the spring (and therefore the upward acceleration of the ball) vs time, you get a sinusoidal curve. Plotting the upward velocity of the ball vs time yields another sinusoid that lags the tension curve by 90°, and plotting its height yields another that lags the velocity curve by another 90°. The applied force and acceleration curves for the ball are therefore 180° out of phase with the position curve, and this is pretty much the canonical example of a simple harmonic oscillator.

Given that the acceleration of the ball at any instant is proportional to the net force on it, acceleration being 180° out of phase with position is consistent with Hooke's Law for the relationship between force and extension in the spring i.e. the force exerted by the spring is proportional to its extension, and in the opposite direction to that extension (pull the end of the spring down and it will exert an upward force in response).

Any upward force exerted on the ball by the bottom end of the spring is matched by a downward force exerted by its top end on the joist, and it's this downward force that you're calling "apparent weight" in your question. And yes, when the ball is as high as it ever gets, that downward force will indeed be reduced by a certain amount. In fact it will be reduced by the same amount that it was increased by when you initially pulled the ball down below its resting position to get your oscillator started. And because all these curves are sinusoids, the maximum height that the ball reaches will likewise be the same distance above its initial resting position as you initially pulled it below that position.

Now, in order for Hooke's Law to apply for the entirety of the ball's motion, the spring must always be exerting some amount of upward force on the ball. If the ball is hanging from a tension spring and you initially pull it down with a force that's more than its own weight, then at some point before the top of the ball's rise, the spring will stop exerting any upward force on it at all and the assumptions supporting simple harmonic motion will break.
posted by flabdablet at 10:36 AM on July 11 [3 favorites]

Thanks for these explanations. Makes sense.
posted by Winnie the Proust at 9:33 AM on July 17

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