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T=2π[itex]\sqrt{}l/g[/itex]

- Thread starter Bibigul
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T=2π[itex]\sqrt{}l/g[/itex]

- #2

Doc Al

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In terms of the period? Just solve for g in terms of T and L.

- #3

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No, only in terms of L.

- #4

Simon Bridge

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Why should the length L depend on the strength of gravity? This is usually something you just cut to whatever length you like - 1m on Jupiter is the same as 1m on the moon... unless, I suppose, you are doing the measuring from a different gravity from the one the meter is in? (This forum I don't want to make too many assumptions!)

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