Two simple pendulums of length 1, m and 4, m | vedclass

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(A) Let $T_{1}$ and $T_{2}$ be the time periods of the two pendulums.$T_{1} = 2\pi \sqrt{\frac{1}{g}}$ and $T_{2} = 2\pi \sqrt{\frac{4}{g}}$.Since $\e

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$2$

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(A) Let $T_{1}$ and $T_{2}$ be the time periods of the two pendulums.$T_{1} = 2\pi \sqrt{\frac{1}{g}}$ and $T_{2} = 2\pi \sqrt{\frac{4}{g}}$.Since $\ell_{1} Let the shorter pendulum complete $n_{1}$ oscillations and the longer pendulum complete $n_{2}$ oscillations when they are in phase again.For the pendulums to be in phase,the time taken must be equal: $n_{1} T_{1} = n_{2} T_{2}$.Substituting the values: $n_{1} 	imes 2\pi \sqrt{\frac{1}{g}} = n_{2} 	imes 2\pi \sqrt{\frac{4}{g}}$.$n_{1} = 2n_{2}$.For the first time they are in phase after $t=0$,the difference in the number of oscillations must be the smallest integer,i.e.,$n_{1} - n_{2} = 1$.Substituting $n_{1} = 2n_{2}$ into the equation: $2n_{2} - n_{2} = 1 \Rightarrow n_{2} = 1$.Then $n_{1} = 2(1) = 2$.Thus,the shorter pendulum completes $2$ oscillations.

Two simple pendulums of length $1\, m$ and $4\, m$ respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed a number of oscillations equal to

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