A rod of mass m and length l is connected by two spring of spring constants `k_(1) and k_(2)`, so that it is horizontal at equilibrium. What is the natural frequency of the system?

Options

A. `(1)/(2pi)sqrt(k_(1)b^(2)+k_(2)l^(2))/(ml^(2))`
B. `(1)/(2pi)sqrt(2k_(1)b^(2)+k_(2)l^(2))/(ml^(2))`
C. `(1)/(2pi)sqrt((k_(1)b^(2)+k_(2)l^(2))/(2ml^(2)))`
D. `(1)/(2pi)sqrt((3(k_(2)b^(2)+k_(2)l^(2)))/(ml^(2)))`

Answer

D

Explanation

Let rod is displaced by an angle `theta`, taking torque about edge of rod
`k_(b)bthetaxxbcostheta+k_(2)lthetaxxlcostheta=-(Id^(2)theta)/(dt^(2))`
Here, `I=(ml^(2))/(3) and theta` is small `costheta~~1`
`therefore(ml^(2)alpha)/(3)+(k_(1)b^(2)+k_(2)l^(2))theta=0`
`impliesalpha=(-3(k_(1)b^(2)+k_(2)l^(2))theta)/(ml^(2))`
`thereforeomega=sqrt((3(k_(1)b^(2)+k_(2)l^(2)))/(ml^(2)))`