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Two pendulums of different lengths are in the mean position at a certain instant. The minimum time after which they will be again in phase is 5 / 4 T where T is time period of shorter pendulum. Find the ratio of lengths of the two pendulums:

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$1 : 25$

$1 : 2$

$1 : 16$

$1 : 4$

answer is D.

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Detailed Solution

$T_{1} = T; T_{2} = KT$

$ϕ_{1} - ϕ_{2} = \frac{2 π}{T_{1}} t - \frac{2 π}{T_{2}} t = 2 π 2 πt [\frac{1}{T} - \frac{1}{KT}] = 2 π$

$\frac{T}{t} [\frac{K - 1}{K}] = 1$

$t [\frac{K - 1}{K}] = T$

$t = \frac{5 T}{4}$

$\frac{5 T}{4} [\frac{K - 1}{K}] = T; 5 K - 5 = 4 K$

$K = 5 T T α \sqrt{L}$

$\frac{L_{1}}{L_{2}} = {(\frac{T_{1}}{T_{2}})}^{2} = {(\frac{T}{5 T})}^{2} = \frac{1}{25}$

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