The time period of the two pendulums is 1.44 s and 0.36 s respectively. Calculate the ratio of their lengths.

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Solution

Step 1:Given data
Let the two pendulums be A and B
Time period of pendulum A is $T_{1} = 1.44 s e c$
Time period of pendulum B is $T_{2} = 0.36 s e c$
Step 2: Relationship between time period and length
We know that time period is directly proportional to the square root of the length
$T \propto \sqrt{L} w e r e, T = t i m e p e r i o d o f a p e n d u l u m L = e f f e c t i v e l e n g t h$
Hence it can be written as,
$T^{2} = L w h e r e T = t i m e p e r i o d o f a p e n d u l u m L = e f f e c t i v e l e n g t h$
Step 3:Calculating corresponding ratio of their length
$\sqrt{\frac{L_{1}}{L_{2}}} = \frac{T_{1}}{T_{2}} \sqrt{\frac{L_{1}}{L_{2}}} = \frac{1.44 s e c}{0.36 s e c} \sqrt{\frac{L_{1}}{L_{2}}} = \frac{144}{36} \frac{L_{1}}{L_{2}} = \frac{4}{1} T a k i n g s q u a r e o n b o t h s i d e s, {(\sqrt{\frac{L_{1}}{L_{2})}})}^{2} = {(\frac{4}{1})}^{2} \frac{L_{1}}{L_{2}} = \frac{16}{1} = 16 : 1$
Hence, the ratio of their length is 16 :1.

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