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NEET 2000 (AIPMT) · Simple Pendulum

Simple Pendulum question from NEET 2000 (AIPMT)

Q84NEET 2000 · AIPMTEasy

Simple Pendulum

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 relation

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Answer: (a)

$f = \frac{1}{2\pi}\sqrt{\frac{g}{l}}$ , independent of mass, so $l \propto \frac{1}{f^2}$ .
$f_A = 2f_B \Rightarrow \frac{l_A}{l_B} = \frac{1}{4}$
So the answer is (a) $l_A = \frac{l_B}{4}$ , does not depend on mass.

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