Is gravity a restorative force in oscillations?

[Elfrid Payton]

Problem: http://i.imgur.com/o8Q1RGZ.png

Solution: http://i.imgur.com/IYiXITJ.png

To get d2y/dt2+ρgLby/M=0, you have to sum the forces and set them equal to Ma. But what happens to Mg when you sum the forces?

 

[rude man]

Gravity is always acting and is always constant, so it acts only to set the initial depth, then all other forces are independent of it. In other words, gravity is not a restorative force, and a restorative force is needed to initiate oscillations.

Perhaps an analogous example explains better: consider a spring, constant k, one end attached so as to be immobile, the other attached to a mass m, the whole thing laying on a horizontal frictionless plane. You pull the spring a distance from its relaxed position and it will oscillate back & forth with frequency sqrt(k/m). Gravity not involved.
Now suspend the spring vertically from the immobile end. The spring will stretch due to gravity pulling on the mass to its equilibrium position. Then you pull the spring down a bit further and again it will oscillate with the same frequency sqrt(k/m). The spring-mass system is a lot easier to analyze. You can include gravity or not in your diff. eq.; you get the same result.

 

[Elfrid Payton]

Gravity is always acting and is always constant, so it acts only to set the initial depth, then all other forces are independent of it. In other words, gravity is not a restorative force, and a restorative force is needed to initiate oscillations.

Perhaps an analogous example explains better: consider a spring, constant k, one end attached so as to be immobile, the other attached to a mass m, the whole thing laying on a horizontal frictionless plane. You pull the spring a distance from its relaxed position and it will oscillate back & forth with frequency sqrt(k/m). Gravity not involved.
Now suspend the spring vertically from the immobile end. The spring will stretch due to gravity pulling on the mass to its equilibrium position. Then you pull the spring down a bit further and again it will oscillate with the same frequency sqrt(k/m). The spring-mass system is a lot easier to analyze. You can include gravity or not in your diff. eq.; you get the same result.

Ah, so only restorative forces are included in differential equations for oscillations, and gravity is never a restorative force?

 

[rude man]

Ah, so only restorative forces are included in differential equations for oscillations, and gravity is never a restorative force?

Yes, until someone invents variable gravity!
Actually, here's an example of variable and restorative gravity: suppose you drill a hole thru the Earth passing thru its center, and then drop a rock in at one end. As the rock falls into the hole and thru the center, once it's past the center gravity acts to restore the rock towards the center. One can show fairly easily that the restorative force is proportional to the distance of the rock from the Earth center, and always towards the center, making this a simple harmonic motion. (This assumes uniform-density Earth which of course is not really true, but it makes a dandyexercise!)

But in your (and most) cases, gravity is constant so there's no restorative force coming from it.

 

[HalfLostAndSearching]

Gravity is indeed a restorative force in oscillations. In the equation d2y/dt2+ρgLby/M=0, the term ρgLby/M represents the force of gravity, where ρ is the density of the object, g is the acceleration due to gravity, L is the length of the object, b is the width of the object, and M is the mass of the object. When summing the forces, the force of gravity (Mg) is included in the equation, as it is a force acting on the object. This force is then balanced by the restoring force (ρgLby/M) in order to maintain equilibrium and create oscillations. Therefore, gravity is an essential component of oscillatory motion and can be considered a restorative force in this context.