Find the ratio of time periods of two pendulums whose lengths are in the ratio $1 : 4$ .

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Solution

Given:
The ratio of length, of the pendulum = $1 : 4$
Let l₁be the length of the first pendulum,
l₂be the length of the second pendulum,
It is given,
$\frac{l_{1}}{l_{2}} = \frac{1}{4} S o, 4 l_{1} = l_{2}$
Calculation:
The time period of the first pendulum,
$T_{1} = \frac{1}{2 π} \sqrt{\frac{l_{1}}{g}}$
The time period of the second pendulum,
$T_{2} = \frac{1}{π} \sqrt{\frac{l_{2}}{g}}$
The ratio of the time period is given by,
$\frac{T_{1}}{T_{2}} = \frac{1}{2 π} \sqrt{\frac{l_{1}}{g}} \frac{1}{2 π} \sqrt{\frac{4 l_{1}}{g}}$
$H e r e 2 π, \frac{l_{1}}{g}$ gets cancel each other
So,
$\frac{T_{1}}{T_{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
The ratio of the time period of the two pendulums is 1:2.

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