Question

What is the error in the estimation of g if the length and time period of an oscillating pendulum have errors of 1% and 3%?

Answer 1

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The formula for time period of pendulum is
$t = \frac{1}{2\pi} \sqrt{ \frac{l}{g} } \\ \sqrt{ \frac{l}{g} } = 2\pi \: t \\ \frac{l}{g} = 4 {\pi}^{2} {t}^{2} \\ g = \frac{l}{4 {\pi}^{2} {t}^{2} }$
The error % in length and time are 1% and 3% respectively.
When two quantities are multiplied or divided, their errors are added.
Also, when a quantity is raised to a power, the error is multiplied by that power.
Total error = 1 + 2×3 = 7%
There will be an error of 7% in the value of g.

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