The correct option is
A 12π
⎷k1a2+k2b2L2(m+M3)A is the point where the rod is hinged.
Let θ be the angular displacement from mean position.
The extension of spring connected at point B is =bθ (θ is so small that we take it as linear displacement)
So restoring force at point B=k2bθ
Moment of this force at point A=k2bθ.b=k2b2θ (b is the distance of force from A)
The contraction of spring connected at point c is:
=aθ
So, restoring force is =k1aθ
and moment of force about A is k1aθ.a=k1a2θ (a is the distance from point A of restoring force)
So, total restoring force =(k1a2θ+k2b2θ)θ
Now the total moment of inertia of the system about point A is:
I= Moment of inertia about point A +Mooment of inertial of mass m about A
=ML23+ML2
So, the angular acceleration
d2θdt2=−(k1a2+k2b2)θI=−(k1a2+k2b2)θL2(M3+M)
⇒d2θdt2+(k1a2+k2b2)L2(M3+M)θ=0
So angular frequency ω=
⎷(k1a2+k2b2)L2(M3+M)
and frequency, f=ω2π=12π
⎷(k1a2+k2b2)L2(M3+M)