JEE Main 2019 · Physics · Simple Harmonic Motion · hard

A uniform thin rod of length L is pivoted at a point at a distance L/4 from its centre, so that it can oscillate in a vertical plane about a…

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A uniform thin rod of length

L

is pivoted at a point at a distance

L/4

from its centre, so that it can oscillate in a vertical plane about a horizontal axis through that point. The length of the equivalent simple pendulum for small oscillations is

A. $\dfrac{2L}{3}$
B. $\dfrac{7L}{12}$
C. $\dfrac{5L}{12}$
D. $\dfrac{L}{3}$

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Answer: (B) $\dfrac{7L}{12}$

For a physical pendulum,

T = 2\pi\sqrt{I/(mgd)}

where

d

is the distance of the pivot from the centre of mass. Here

d=L/4

. By the parallel axis theorem

I = \dfrac{mL^2}{12}+m\left(\dfrac L4\right)^2=\dfrac{7mL^2}{48}

. Comparing with

T=2\pi\sqrt{\ell/g}

gives the equivalent length

\ell = \dfrac{I}{md}=\dfrac{7L^2/48}{L/4}=\dfrac{7L}{12}

. The value

2L/3

would be obtained for a pivot at the end of the rod.

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