A very light rod of length `l` pivoted at `O` is connected with two springs of stiffness `k_1` & `k_2` at a distance of a & l from the pivot respectively. A block of mass ma attached with the spring `k_(2)` is kept on a smooth horizontal surface. Find the angular frequency of small oscillation of the block `m`.

Explanation

Let the block be pulled towards right through a distance x, then `x = x_(B) + x_(CB)"….."(i)`
where, `x_(CH) = ` displacement of C (the block) relative to B
Thus, `x_(CB) = (F)/(k_(2)) "…."(ii)` and `x_(B) = ((F')/(k_(1)))(l)/(a)`
Torque acting on the rod about point O.
`tau_(0) = F'a - Fl rArr I_(0) (d^(2)theta)/(dt^(2)) = F'a - Fl`
Since the rod is very light its moment of inertia
`I_(0)` about O is approximentely equal to zero `rArr F' = F(l/a)"....."(iv)`
Using (iii) and (iv) `rArr x_(B) = (F)/(k_(1)) (l/a)^(2)"............."`
Using (i), (ii) & (v) `rArr x = (F)/(k_(1)) (l/a)^(2) + (F)/(k_(2))`
As force F is opposite to displacement x, then
`rArr F=-(k_(1)k_(2))/(k_(2)(l/a)^(2)+k_(1))xrArr momega^(2)x=(k_(1)k_(2))/(k_(2)(l/a)^(2)+k_(1))xrArr omega = sqrt((k_(1)k_(2)a^(2))/(m(k_(1)a^(2)+k_(2)l^(2))))`