A rod of length 'l' and mass m, pivoted at one end, is held by a spring at its mid point and a spring at far end, both pulling in opposite directions. The springs have spring constant K, and equilibrium their pull is perpendicular to the rod. Find the frequency of small oscillations about the equilibrium position

A rod of length 'l' and mass m, pivoted at one end, is held by a sprin

1.	A rod of length 'l' and mass m, pivoted at one end, is held by a spring at its mid point and a spring at far end, both pulling in opposite directions. The springs have spring constant K, and equilibrium their pull is perpendicular to the rod. Find the frequency of small oscillations about the equilibrium position
Answer» Solution :Let `theta` be the SMALL ANGULAR displacement of the rod Let `x_1 and x_2` be the changes in the lengths of the springs . Coresponding restoring forces will ACT as shown. Here `x_2 = L theta and x_1 =(L theta)/(2)` Net restoning torque on the rod will be `tau =(K x_2)L + (Kx_1) L/2 =-(KL^2 + K (L^2)/(4))theta` So, `(ML^2)/(3) ALPHA = K ((5L^2)/(4)) theta` `alpha = - ((15L^2)/(4mL^2)K)theta , alpha =- ((15K)/(4m)) theta` frequency `=1/(2PI) sqrt((alpha)/(theta)) = 1/(2pi) sqrt((15K)/(4m))`