1. Why both L of left hand and right hand figures are different?

2. Why mg is put on the right end, not middle of the bar, in the right hand figure?

1) is sloppy notation, probably, from wanting the formulas to look alike.

 

2) It may be that this is not supposed to be a free-body diagram. But I would expect the spring constant to show up in the answer, so I don't really trust anything about the problem.

It's a very poorly stated problem but I suspect that the spring constant might cancel out of the second system. L is related to the spring constant.

It's a very poorly stated problem but I suspect that the spring constant might cancel out of the second system. L is related to the spring constant.

It shouldn't be; the bar is horizontal when unperturbed. L (as labeled) is from an external push, so it's arbitrary.

If it's displaced down by L as a result of mg then it's a different kettle of fish.

However, it is ridiculously ill stated, so it's difficult to tell.

This is a known relation in accelerometers.

 

In the second figure, the deflection L does result from the weight, and then the resonant frequency can be written this way as a function of the deflection instead of the spring stiffness.

 

At accelerometers, it's used as a sensitivity figure (in terms of displacement) related to the resonance frequency.

This is a known relation in accelerometers.

 

In the second figure, the deflection L does result from the weight, and then the resonant frequency can be written this way as a function of the deflection instead of the spring stiffness.

 

At accelerometers, it's used as a sensitivity figure (in terms of displacement) related to the resonance frequency.

The problem explicitly states that it is horizontal at equilibrium. The deflection of L is therefore not the result of mg; the deflection from mg is unknown. The formulation/presentation of the problem is quite poor.

Maybe they considered it to be at eqm before they put a weight on it and pushed it down by L.

It's not possible to be certain but at least that interpretation would mean that the formula might be right.

In any event, it's spectacularly ill described.

Maybe they considered it to be at eqm before they put a weight on it and pushed it down by L.

It's not possible to be certain but at least that interpretation would mean that the formula might be right.

In any event, it's spectacularly ill described.

That's a possibility. Likely, even, since it explains why the mg is at the end of the rod — that's where the invisible mass is placed.