A body is made of two identical rods $A B$ and $C D$ rigidly joined together to form a T shape. Two springs are attached to this body and the body is free to rotate in a vertical plane about a fixed horizontal axis passing through $C$ as shown in the figure. Then, what is the angular frequency (in rad/s) of small oscillation of this system? Mass of each rod is $m = 10 k g,$ their lengths are $L = 50 c m$ and spring constant for each spring is $K = 100 N / m . (g = 10 m / s^{2})$

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Detailed Solution

Let the body be displaced through an angle $θ$ about its mean position.
Question Image
Net torque on the body at this position about C.
$τ = - (2 K a θ a \cos θ + m g a \sin θ)$
$= - (2 K a^{2} + m g a) θ$
(where $θ$ is small, $\sin θ = θ a n d \cos θ = 1$ )
Angular acceleration
$α = \frac{τ}{I} = - \frac{(2 K a^{2} + m g a) θ}{\frac{m \times {(2 a)}^{2}}{12} + \frac{m \times {(2 a)}^{2}}{3}}$
or $α = - \frac{(2 K a^{2} + m g a) θ}{\frac{5 m a^{2}}{3}}$
$\Rightarrow ω = \sqrt{\frac{(2 K a^{2} + m g a)}{5 m a^{2} / 3}} = \sqrt{(\frac{6 K}{5 m} + \frac{3 g}{5 a})} = \sqrt{\frac{6}{5} (\frac{K}{m} + \frac{g}{L})} = \sqrt{\frac{6}{5} (\frac{100}{10} + \frac{10}{0.5})} = \sqrt{\frac{6}{5} (30)} = 6 r a d / s$