Two simple pendulums of length 0.5, m and 2.0 | vedclass

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(C) Let $T_S$ be the time period of the shorter pendulum and $T_L$ be the time period of the longer pendulum.Given lengths are $l_S = 0.5\, m$ and $l_

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$2$

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(C) Let $T_S$ be the time period of the shorter pendulum and $T_L$ be the time period of the longer pendulum.Given lengths are $l_S = 0.5\, m$ and $l_L = 2.0\, m$.The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.So,$T_S = 2\pi \sqrt{\frac{0.5}{g}}$ and $T_L = 2\pi \sqrt{\frac{2.0}{g}}$.Taking the ratio,$\frac{T_L}{T_S} = \sqrt{\frac{2.0}{0.5}} = \sqrt{4} = 2$.Thus,$T_L = 2T_S$.Let $N_S$ be the number of oscillations of the shorter pendulum and $N_L$ be the number of oscillations of the longer pendulum when they are in the same phase again.At this time $t$,$t = N_S T_S = N_L T_L$.Substituting $T_L = 2T_S$,we get $N_S T_S = N_L (2T_S)$.Therefore,$N_S = 2N_L$.For the first time they are in phase again,we take the smallest integer values,$N_L = 1$,which gives $N_S = 2$.

Two simple pendulums of length $0.5\, m$ and $2.0\, m$ respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed .... oscillations.

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