# Horizontal and vertical oscillations of a loaded spring

#### logearav

Joined Aug 19, 2011
243
Revered Members,
I have attached two images which explain horizontal and vertical oscillations of a loaded spring. In horizontal oscillations the restoring force is taken as F = -kx.
But in vertical oscillations the restoring force is taken as F = kdl.
Restoring force is opposite to the direction of displacement so negative sign is included. But why in vertical oscillations the restoring force is
F = kdl. Why not F = -kdl?

#### BillO

Joined Nov 24, 2008
990
If the system in the vertical case is at equilibrium, we have two forces acting on the body in opposite directions such that there is not net force on the body.

$$F_{e}\ =\ kdl-mg\ =\ 0 \ therefore \ kdl=mg$$

However, kdl is the static restoring force on the system and is not really of interest. For that reason, the values being used in this example are the magnitudes of the forces.

The value of interest is the resulting net force on the body caused by the displacement, y, from the equilibrium position. That net force is opposite to the direction of displacement and as explained in the attachment;

$$F\ =\ -ky$$

If this was a statics problem, the other forces directions might be of more interest.

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#### logearav

Joined Aug 19, 2011
243
If the system in the vertical case is at rest, we have two forces acting on the body in opposite directions such that there is not net force on the body.

$$F_{r}\ =\ kdl-mg\ =\ 0 \ therefore \ kdl=mg$$

However, kdl is the static restoring force on the system and is not really of interest. For that reason, the values being used in this example are the magnitudes of the forces.

The value of interest is the resulting net force on the body caused by the displacement, y, from the equilibrium position. That net force is opposite to the direction of displacement and as explained in the attachment;

$$F\ =\ -ky$$

If this was a statics problem, the other forces directions might be of more interest.
Why F = kdx - mg was not applicable in horizontal oscillations of the spring?

#### BillO

Joined Nov 24, 2008
990
There is no static force on the spring in the horizontal example you gave. There could be a case where there is such a force, but your particular example was not such a case.

#### logearav

Joined Aug 19, 2011
243
In the horizontal case too, there is a mass suspended to the spring, same as vertical case. So why there is no static force on horizontal case?

#### BillO

Joined Nov 24, 2008
990
It's not suspended, it is attached. Do not read things into the system that are not explicitly given. There is no rest, or static, force implicit or implied in the horizontal system.

#### logearav

Joined Aug 19, 2011
243
BillO, now i got it almost 90%. Thanks a ton for your beautiful explanation. But some doubt persists. I can't understand your line " There is no rest, or static, force implicit or implied in the horizontal system"
Both horizontal and vertical cases are springs, one end of which is attached to a mass. Both are at rest when undisturbed. They oscillate only when the attached mass is pulled. Then how your statement of no rest is applicable only to horizontal case alone.

#### studiot

Joined Nov 9, 2007
4,998

I will try again here.

What do you think the -ve sign means in the equation?

#### logearav

Joined Aug 19, 2011
243
Studiot,
- ve means opposite direction. Am I right?

#### studiot

Joined Nov 9, 2007
4,998
Yes that's the idea.

The restoring force acts in a direction opposite to the displacement.
That is why it is called a restoring force.

In the case of your horizontal example the displacement is in the positive (x) direction so is positive.
Therefore the restoring force is in the negative x direction so is negative.

In the case of your vertical example the positive y direction is upwards so the displacement is in the negative y direction and the restoring force is in the positive y direction.

simple as that.

#### BillO

Joined Nov 24, 2008
990
@Studiot,

WADR, the displacement from equilibrium in both cases is in a positive direction, and in both cases the resulting restoring force is negative.

i.e. in the horizontal case F=-kx and in the vertical case F=-ky

(What's with the -ve? Negative?)

@logearav

In the horizontal case there is no tension on the spring in the static (or equilibrium) state because there is no intrinsic force acting on the mass in the dimension of free motion.

In the vertical case, there is the force of gravity acting on the mass in the dimension of free motion.

However, for small displacements from equilibrium, the vertical and horizontal cases can be treated exactly the same.

I apologize if I am using terms you are not familiar with, but I know no other way of explaining this without getting into long examples.

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#### logearav

Joined Aug 19, 2011
243
If the system in the vertical case is at equilibrium, we have two forces acting on the body in opposite directions such that there is not net force on the body.

$$F_{e}\ =\ kdl-mg\ =\ 0$$
$$BillO, You are good at explaining the concepts in a fascinating way with a lot of patience.I came back to your first post, because i now could understand this post after your last post. Thanks for that. Now i am getting the concept, though a little bit slowly. 1)Why not mg - k dl = 0? However, kdl is the static restoring force on the system and is not really of interest. For that reason, the values being used in this example are the magnitudes of the forces. 2) So, Static restoring force will come into play, only when force of gravity acting on the system. In the horizontal case, there is no force of gravity, so we dont take into account static restoring force. Did i get it right?$$

#### BillO

Joined Nov 24, 2008
990
BillO,
You are good at explaining the concepts in a fascinating way with a lot of patience.I came back to your first post, because i now could understand this post after your last post. Thanks for that. Now i am getting the concept, though a little bit slowly.
1)Why not mg - k dl = 0?
As I said, the static, or equilibrium, case is not essentail to the understanding of the dynamic problem. So either one is fine. The authors had to choose one, but both are essentially the same, Whether you say kdl=mg (kdl-mg=0), or mg=kdl (mg-kdl=0), is not relevant. All they are trying to say is that the forces cancel each other out.

2) So, Static restoring force will come into play, only when force of gravity acting on the system. In the horizontal case, there is no force of gravity, so we dont take into account static restoring force. Did i get it right?
Yes, I think you possibly do understand, but in either case, the equilibrium forces need not be taken into account to solve the dynamic problem. You only need to know that before the dynamic perturbation (or displacement), that everything else was in equilibrium.

• logearav