A simple pendulum of length l has a brass bob | vedclass

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(C) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$.From this formula,we can see that the time period $T$ i

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$4l$

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(C) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$.From this formula,we can see that the time period $T$ is directly proportional to the square root of the length $l$,i.e.,$T \propto \sqrt{l}$.The time period of a simple pendulum is independent of the mass,size,or density of the bob.Let the initial length be $l_1 = l$ and the initial time period be $T_1 = T$.Let the new length be $l_2$ and the new time period be $T_2 = 2T$.Using the proportionality $T \propto \sqrt{l}$,we have $\frac{T_2}{T_1} = \sqrt{\frac{l_2}{l_1}}$.Substituting the given values: $\frac{2T}{T} = \sqrt{\frac{l_2}{l}}$.$2 = \sqrt{\frac{l_2}{l}}$.Squaring both sides,we get $4 = \frac{l_2}{l}$.Therefore,the new length $l_2 = 4l$.

$A$ simple pendulum of length $l$ has a brass bob attached at its lower end. Its period is $T$. $A$ steel bob of the same size,having density $x$ times that of brass,replaces the brass bob. Its length is then changed such that the period becomes $2T$. What is the new length?

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