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Calculate the period of a simple pendulu...

Calculate the period of a simple pendulum of length 0.98 m at a place where acceleration due to gravity is `9.8ms^(-2)`.

Text Solution

Verified by Experts

Given, `L=0.98m, g=9.8ms^(-2)`
We know that,
`T=2pi sqrt(((L)/(g)))" i.e. "2xx3.142xx sqrt((0.98)/(9.8))=6.284xx0.3162`
`T=1.987s.`
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