At equilibrium, let elongation and compression of spring 1 and 2 be
x1 and
x2 respectively.
Then,
k1x1bcosθ+k2x2lcosθ=mgl2cosθ…(i)
[balancing torques about
′O′]
Let us rotate the rod by an angle
θ. Then, spring 1 gets elongated further by
bθ and spring 2 is compressed further by
lθ.
Writing torque about point
′O′ again,
τ0=−[k1bcosθ(x1+bθ)+k2lcosθ(x2+lθ)]+mgl2cosθ
[taking clockwise positive]
Using (i), restoring torque about point
′O′ becomes
τ0=−(k1bθ×bcosθ+k2lθ×lcosθ)
From this, we can say that
τ=−(k1b2cosθ+k2l2cosθ)θ
But we know that,
τ=Iα
Moment of inertia of rod about point
′O′ is
I=ml23
When
θ is very small,
cosθ≈1
From this, we can say that,
ml23α=−(k1b2+k2l2)θ
⇒α=−3(k1b2+k2l2)ml2θ
Comparing this with
α=−ω2θ, we get
ω=√3(k1b2+k2l2)ml2
From the data given in the question,
ω=
⎷3×[16×(12√2)2+1×(1)2]164×12 rad/s
⇒ω=24 rad/s