When the length of simple pendulum is dubuled ; find the ratio of new frequency to old frequency?

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Answer 1 · 2018-08-05T06:51:11

The answer is #=1/sqrt2#

The period #T# of a simple pendulum is

#T=2pisqrt(l/g)#

Where #l# is the length of the pendulum

And #g# is the acceleration due to gravity

The frequency is

#f=1/T=1/(2pi)sqrt(g/l)#

Squaring both sides

#f^2=1/(4pi^2)*g/l#

Suppose l_1=2l

Then,

#f_1^2=1/(4pi^2)*g/(2l)#

Then,

#f_1^2/(f^2)=(1/(4pi^2)*g/(2l))/(1/(4pi^2)*g/l)=1/2#

#f_1/f=1/sqrt2#