First slide

Simple hormonic motion

Question

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

Difficult

Solution

$r_{c}$ = Distance of C.O.M. of (rod + particle) system from $A = \frac{m \times \frac{l}{2} + m \times l}{m + m} = \frac{3 l}{4}$
$\begin{array}{l} I_{A} = \frac{1}{3} m l^{2} + m l^{2} = \frac{4}{3} m l^{2} \\ ∴ T = 2 π \sqrt{\frac{I_{A}}{m . g r_{c}}} = 2 π \sqrt{\frac{\frac{4}{3} m l^{2}}{m . g (3 l / 4)}} = 2 π \sqrt{\frac{16 l}{9 g}} = \frac{8 π}{3} \sqrt{\frac{l}{g}} \\ ∴ f = \frac{1}{T} = \frac{3}{8 π} \sqrt{\frac{g}{l}} \end{array}$

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