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Question

Two springs of spring constant mgl each are attached to the end of a uniform rod of a mass m as which is hinged as shown, As the rod is rotated slightly about the hinge, it undergoes SHM with angular frequency ω. Find ω(in rad/s).
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A
18
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B
9
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C
12
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D
6
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Solution

The correct option is D 6
When the rod is displaced by an angular displacement θ
Net torque about axis:
τ=(kxlcosθ)+(kxlcosθ)+mgl2sinθ
τ=(kxl)+(kxl)+mgl2θ
Since displacement is small:
τ=(kl2θ)+(kl2θ)+mgl2θ
Here
Keff=2kl2mgl2
Thus
ω=KeffI
ω=    2kl2mgl2ml23
Putting l=1.25 m and g=10ms2
We get ω=6rad/s

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