If the angular frequency of small oscillations of a thin uniform vertical rod of mass `m` and length `l` hinged at the point `O` (Fig.) is `sqrt((n)/(l))`, then find `n`. The force constant for each spring is `K//2` and take `K = (2mg)/(l)`. The springs are of negiligible mass. `(g = 10 m//s^(2))`

Answer

75

Explanation

For small Angular displacement
Restoring torque about point `O`.
`(mg l/2 sintheta) + (Klsintheta)l = (ml^(2))/(3)alpha`, (for small `theta sintheta ~~ theta`)
`alpha = ((mgl/2 + Kl^(2))theta)/((ml^(2))/(3)) rArr alpha = [(3g)/(2l) + (3k)/(m)]theta = -omega^(2)theta`

So angular speed `omega = sqrt((3g)/(2l) + (3k)/(m)]theta = -omega^(2)theta`
`omega = sqrt((3g)/(2l)(1 + (2kl)/(mg))) = sqrt((75)/(L)) :. N = 75`