Correct Answer - C
If the rod is rotated through an angle `theta`, extension in one spring=compression in the other spring,
i.e., `x=l theta//2`
therefore, force acting on each of
the ends of the rod,
`F=kx=k(ltheta//2)`
Restoring torque on the rod,
`tau=-Fl=-k(ltheta//2)l=-kl^(2)(theta//2)`
As `tau=Ialpha=(Ml^(2)//12)alpha`,
`-kl^(2)(theta//2)=(Ml^(2)//12)alpha`
or `alpha=-(6k)/(M)theta`, i.e., `alpha proptheta`
thus, the motion of the rod is simple harmonic with
`omega^(2)=(6k)/(M)` or `omega=sqrt((6k)/(M))` or `upsilon=(1)/(2pi)sqrt((6k)/(M))`