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Answer 1 · 2017-03-08T05:38:39

The Period becomes $sqrt2$ times the original one.

Let, $T, l and g$ be the Period, Length and Gravitational

Acceleration of a Simple Pendulum.

$:. T=2pisqrt(l/g).......................(1).$

Suppose that, the period is $T'$ when, the length $l'=2l.$

Then, by $(1), T'=2pisqrt{(l')/g}=2pisqrt{(2l)/g}..............(2).$

$:. (T')/T=[2pisqrt{(2l)/g}]/{2pisqrt(l/g)}$ .

$rArr (T')/T=sqrt{(2l)/l}=sqrt2, i.e.,$

$T'=(sqrt2)T,$ meaning that the Period becomes $sqrt2$ times the

original one.