Two sphrical bob of masses $ M_A $ and $ M_B $ are hung vertically from two strings of length $ \ell_A $ and $ \ell_B $ respectively. They are executing SHM with frequency relation $ f_A = 2f_B $, Then: (2000)
$ \ell_A = \frac{\ell_B}{4} $
$ \ell_A = 2\ell_B $ & $ M_A = 2M_B $
$ \ell_A = \frac{\ell_B}{2} $ & $ M_A = \frac{M_B}{2} $
Solution:
Frequency of a simple pendulum is $ f = \frac{1}{2\pi}\sqrt{\frac{g}{\ell}} $, which is independent of mass. Given $ f_A = 2f_B $, we have $ \frac{1}{\sqrt{\ell_A}} = \frac{2}{\sqrt{\ell_B}} $. Squaring both sides yields $ \ell_A = \frac{\ell_B}{4} $.
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