A uniform rod of mass $M$ and length $L$ hangs from a frictionless pivot and is free to oscillate about its pivot. It is connected at its bottom by a horizontal light spring to a vertical wall. The spring constant of the spring is $k$ . When the rod is vertical, the spring is unstretched. When the free end of the rod is slightly displaced along the length of the spring and released. Find $N$ if its oscillation frequency is $\frac{1}{2 π} \sqrt{\frac{N (M g + 2 K L)}{2 M L}}$ .

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Detailed Solution

When the bar is slightly displaced (say) by an angle $θ, x = L θ$

$τ =$ restoring torque on the bar about $O = l a$
$= (M g) (\frac{L}{2} \sin θ) + K (L θ) L$
$= L θ (\frac{M g}{2} + K L) (∵ \cos θ \to 1 when θ \to 0)$
Angular acceleration, $α = - ω^{2} θ$ ,
Where $ω^{2} = \frac{L (\frac{M g}{2} + K L)}{l}$
$l =$ moment of inertia of bar about $O$
$= M (\frac{L^{2}}{12}) + M {(\frac{L}{2})}^{2}$
$= \frac{1}{3} M L^{2}$

frequency is given by $f = \frac{1}{2 π} \sqrt{\frac{N (M g + 2 K L)}{2 M L}}$

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