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Question 14.9: Measuring g. A geologist uses a simple pendulum that has a l......

Measuring g. A geologist uses a simple pendulum that has a length of 37.10 \mathrm{~cm} and a frequency of 0.8190 \mathrm{~Hz} at a particular location on the Earth. What is the acceleration of gravity at this location?

APPROACH We can use the length \ell and frequency f of the pendulum in Eq. 14-12b, which contains our unknown, g.

f=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{g}{\ell}},\qquad [\theta \text{small}] (14-12b)

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We solve Eq.14-12b for g and obtain

g=(2 \pi f)^{2} \ell=\left(6.283 \times 0.8190 \mathrm{~s}^{-1}\right)^{2}(0.3710 \mathrm{~m})=9.824 \mathrm{~m} / \mathrm{s}^{2}

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