Two masses M_{A} and M_{B} are hung from two | vedclass

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(B) The frequency of a simple pendulum is given by $f = \frac{1}{2 \pi} \sqrt{\frac{g}{l}}$.Given the frequency relation $f_{A}=2 f_{B}$.Substituting

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$l_{A}=\frac{l_{B}}{4},$ does not depend on mass

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(B) The frequency of a simple pendulum is given by $f = \frac{1}{2 \pi} \sqrt{\frac{g}{l}}$.Given the frequency relation $f_{A}=2 f_{B}$.Substituting the formula for frequency:$\frac{1}{2 \pi} \sqrt{\frac{g}{l_{A}}} = 2 	imes \frac{1}{2 \pi} \sqrt{\frac{g}{l_{B}}}$Squaring both sides:$\frac{g}{l_{A}} = 4 	imes \frac{g}{l_{B}}$$\frac{1}{l_{A}} = \frac{4}{l_{B}}$Therefore,$l_{A} = \frac{l_{B}}{4}$.Since the frequency of a simple pendulum depends only on the length $l$ and acceleration due to gravity $g$,it does not depend on the mass $M$ of the bob.

Two masses $M_{A}$ and $M_{B}$ are hung from two strings of length $l_{A}$ and $l_{B}$ respectively. They are executing $SHM$ with frequency relation $f_{A}=2 f_{B}$,then the relation is:

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