Dropping a Slinky

A short video of what it looks like to drop a slinky. It’s surprising and elegant.

So what’s the speed that waves propagate in a slinky? A slinky is a bit tricky, because as you pull on it, it stretches out so that the density goes down. Meanwhile, the tension goes up. Both these effects increase the speed of wave propagation, so waves travel much more quickly at the top of a hanging slinky than at the bottom.

Since the only material properties around are the linear density $\lambda$ and the tension $T$ , we must put these together to get a velocity, which we do by

$v = \sqrt{\frac{T}{\lambda}}$

As the slinky hangs, it should be in equilibrium, so the gradient of the tension at any point is equal to gravity times the density there. This yields the result that the tension is a square root of how far up you go from the bottom of the spring. As a rough estimate, though, the tension should on average be about half the spring’s weight, while the density on average is the spring’s weight divided by its length. Thus

$v = \sqrt{\frac{weight}{mass/length}} = \sqrt{g l}$

where $g$ is gravitational acceleration and $l$ is the slinky’s length. The characteristic time of such a slinky is

$t = \frac{l}{v} = \sqrt{\frac{l}{g}}$

For a one meter slinky we get a time of .3 seconds (and a speed of only a few meters per second), meaning it’s an effect we can see quite well even without high-speed photography!

The same basic mechanism is there in dropping anything else, but typical sound speeds are on the order of thousands of meters per second, so usually it’s much too fast.

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This entry was posted on July 30, 2012 at 2:10 pm and is filed under Uncategorized. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to “Dropping a Slinky”

Dropping a Slinky (calculation) « Arcsecond Says:
July 30, 2012 at 5:55 pm
[...] Arcsecond « Dropping a Slinky [...]

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