InterviewSolution
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InterviewSolution
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A bob of mass `M` is hung using a string of length `l`. A mass `m` moving with a velocity `u` pierces thorugh the bob and emerges out with velocity `(u)/(3)` horizontally. The frequency of small oscillations of the bib, considering `A` as amplitude is,A. `(1)/(2pi)sqrt((3mu)/(2MA))`B. `(1)/(2pi)sqrt((2m)/(3MA))`C. `(1)/(2pi)((2mu)/(3MA))`D. `(1)/(2pi)((3mu)/(2MA))` |
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Answer» Correct Answer - C Using momentum conservation we get, `mu =Mv +m.(u)/(3) :. V=(2)/(3)(mu)/(M)` `K.E.` at mean position of the bob `=(1)/(2)Mv^(2) = (1)/(2)M(4)/(9)(m^(2)u^(2))/(M^(2)) =(2m^(2)u^(2))/(9M)` Comparing with `K.E. = (1)/(2) M omega^(2) A^(2)` at mean position, We get` omega^(2) = (4m^(2)u^(2))/(9M^(2)A^(2))` or `omega = (2mu)/(3MA)` `= 2piv rArr v =(1)/(2) ((2mu)/(3MA))` `:.` Frequency (C ) is correct. |
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