two particle of mass $m$ each are fixed to a mass-less rod of length $2 ℓ$ . The rod is smoothly hinged at one end to a ceiling . It performs oscillation of small angle in vertical plane . The length of the equivalent simple pendulum is :

\frac{3 ℓ}{2}

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\frac{10 ℓ}{3}

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\frac{5 ℓ}{3}

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none f these

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Solution

The correct option is C $\frac{5 ℓ}{3}$
$T = 2 π \sqrt{\frac{I}{M g r}}$
$I = M l^{2} + 4 M l^{2} = 5 m l^{2}$
$r = \frac{M l + 2 M l}{2 M} = \frac{3 ℓ}{2}$
$T = 2 π \sqrt{\frac{5 m L^{2}}{2 M \cdot g \cdot \frac{3 l}{2}}} = 2 π \sqrt{\frac{5 l}{3 g}}$
$Now on comparing with the Time period of Simple penduluan$
$\begin{matrix} T = 2 π \sqrt{\frac{l}{g}} we get L = \frac{5 l}{3} \end{matrix}$

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two particle of mass m each are fixed to a mass-less rod of length 2ℓ. The rod is smoothly hinged at one end to a ceiling . It performs oscillation of small angle in vertical plane . The length of the equivalent simple pendulum is :

The correct option is C 5ℓ3T=2π√IMgr I=Ml2+4Ml2=5ml2 r=Ml+2Ml2M=3ℓ2 T=2π√5mL22M⋅g⋅3l2=2π√5l3g Now on comparing with the Time period of Simple penduluan T=2π√lg we get L=5l3

two particle of mass $m$ each are fixed to a mass-less rod of length $2 ℓ$ . The rod is smoothly hinged at one end to a ceiling . It performs oscillation of small angle in vertical plane . The length of the equivalent simple pendulum is :