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Question

A thin uniform vertical rod of mass m and length l is hinged about the point O as shown in the figure. The combined stiffness of springs is equal to k. Considering the springs to be light, find the angular frequency of small oscillations.


A
ω=3km+3gl
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B
ω=3km+gl
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C
ω=km+3gl
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D
ω=3km+3g2l
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Solution

The correct option is D ω=3km+3g2l
Let the individual spring constants be k1 and k2 respectively. Let us rotate the bar by small angle θ.


From the FBD, restoring torque about point O
τ=[mgl2sinθ+k1x(lcosθ)+k2x(lcosθ)]
τ=[(k1+k2)lxcosθ+mgl2sinθ] .......1)
Since, the elongation in one spring equals the compression in the other spring, we can say that, Both springs are in parallel combinations i.e k1+k2=k
Since θ is small, sinθθ and cosθ1
From the diagram we can say that , x=lθ
From the above given data, (1) can be written as,
τ=[kl2+mgl2]θ
But we know that, τ=Iα
Using this we get,
Iα=[kl2+mgl2]θ
Moment of inertia of rod about an axis passing through the end point O is given by I=ml23
α=[kl2+mgl2]θml23
Comparing this with α=ω2θ, we get
ω=3km+3g2l
Thus, option (d) is the correct answer.

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