Two simple pendulums of lengths 1.44 , m and | vedclass

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.Since $T \propto \sqrt{l}$,the frequency $n = \frac{1}{T}$ is propo

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$6$ oscillations of smaller pendulum

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.Since $T \propto \sqrt{l}$,the frequency $n = \frac{1}{T}$ is proportional to $\frac{1}{\sqrt{l}}$.Let $l_1 = 1.44 \, m$ and $l_2 = 1 \, m$.Let $n_1$ and $n_2$ be the number of oscillations of the pendulums with lengths $l_1$ and $l_2$ respectively.For them to swing together again,the time taken must be equal: $t = n_1 T_1 = n_2 T_2$.Thus,$\frac{n_2}{n_1} = \frac{T_1}{T_2} = \sqrt{\frac{l_1}{l_2}} = \sqrt{\frac{1.44}{1}} = \frac{1.2}{1} = \frac{6}{5}$.This implies $5 n_2 = 6 n_1$.For the smallest integer values,$n_1 = 5$ (oscillations of the longer pendulum) and $n_2 = 6$ (oscillations of the shorter pendulum).Therefore,they will swing together again after $6$ oscillations of the smaller pendulum.

Two simple pendulums of lengths $1.44 \, m$ and $1 \, m$ start swinging together. After how many vibrations will they again start swinging together?

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