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A rod of mass m and length l is connecte...


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?

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)))`

Text Solution

Verified by Experts

The correct Answer is:
D
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)))`
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