Do we ignore weight of the rod in vertical spring-block system

free-body-diagramharmonic-oscillatornewtonian-gravitynewtonian-mechanicsspring

Problem: The first spring is placed at the
middle of the rod and the second one at the end as shown in the figure. If the end of the rod is slightly pulled up and released,
determine the angular frequency of small oscillations.

Figure

My professor said "We ignore the weight since the initial elongation of the spring will be cancelled by the weight of the rod", and hence her differential equations of motion don't contain any gravity term while mine do .

Could someone tell or hopefully explain which one of us is correct?

Best Answer

The static forces in the structure are indeterminate. The unstretched lengths of the springs and the position of the fixed points is not given, so you don't know the three reaction forces on the rod, or the bending moments inside the rod.

However, all those forces cancel out in dynamic equations of motion, in exactly the same way as they do in the statics equation.

Maybe your confusion is because you included some of the static forces (i.e. the weight) but ignored the others, and therefore you think the motion depends on the weight.

Note, of course the motion depends on the mass of the rod (and presumably $I$ in the figure is the moment of inertia of the rod about the pivot at the left hand end), but mass and weight are two different things!