Question #4c8b0

1 Answer
Mar 8, 2017

Answer:

The Period becomes #sqrt2# times the original one.

Explanation:

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.