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A thin rod AB of mass m and length l is hinged at end A and suspended freely. A particle of mass m is attached at the lower end B. The lower end of the rod is slightly pulled to one side and released. Then frequency of oscillation of the rod is

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Detailed Solution

rc = Distance of C.O.M. of (rod + particle) system from A=m× l2 + m × lm+m=3l4IA=13 ml2+ml2=43ml2∴ T=2πIAm.g rc=2π43ml2m.g(3l/4) = 2π16l9g = 8π3lg∴ f=1T=38π gl

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