Question

The period of a pendulum is given by the formula T=2pi(sqrt(L/g))
where L is the length of the pendulum in feet, g = 32 ft/s2
is the acceleration due to gravity, and T is the length of one period in
seconds.

(a) If the length of the pendulum is measured to be 3 feet long with a max error of 1/8th of an inch, what is the approximate max error in the period, T?
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(b) Find the relative error in the period, in terms of the relative error in the length.
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(c) Find the relative error in the period, for part a.
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Answer 1

\[\Delta L=0.125\]\[T(L)=2\pi \sqrt{\frac{L}{g}}=\frac{2\pi}{\sqrt{g}}.\sqrt{L}=k \sqrt{L}\], where \[k=\frac{2\pi}{\sqrt{g}}\], a constant
Then approximate small change in T with small change in L is given by
\[\Delta T=\frac{dT}{dL} . \Delta L\]