Frequency of a rod pivoted at one end with springs at center and end

[uriwolln]

A rod with a length of L and mass m, which had its axis at one end, has two springs attached to it, one at the center and one at the end. The whole system is horizontal so no use of gravitation here.
The question is, what is the frequency of the oscillation.


I know the momentum of inertia of rod around its axis at the end is 1/3(ML^2). And I think because, on the center of mass, a spring is attached it does not add it should not be considered when using Ix(alpha)=R X F. Which F in this case F= -KX. So i tried this thing out, and it just does not come right, what am i missing?

 

[SammyS]

A rod with a length of L and mass m, which had its axis at one end, has two springs attached to it, one at the center and one at the end. The whole system is horizontal so no use of gravitation here.

I take it that you mean that the whole system moves in a horizontal plane.

The question is, what is the frequency of the oscillation.

I know the momentum of inertia of rod around its axis at the end is 1/3(ML^2). And I think because, on the center of mass, a spring is attached it does not add it should not be considered when using Ix(alpha)=R X F.

The spring at the center DOES produce torque about the axis at the end of the rod, so this is not correct.

Which F in this case F= -KX. So i tried this thing out, and it just does not come right, what am i missing?

F, as you labeled it is the force the spring at the free-end of the rod exerts at the free-end of the rod. This assumes that this end of the rod is displaced a distance x from equilibrium.

If the free-end of the rod has moved a distance x, then the center of the rod has moved a distance x/2. Then the spring at center exerts a force -kx/2 which is F/2. This force is exerted at a distance, L/2 from the axis of rotation.

Thus the total torque about the axis is ‒k x L ‒ k (x/2) (L/2).

 

[uriwolln]

Thanks!
Helped me to work this out