# Equilibrium Position

The equilibrium position of a mechanical system is the position where the total force acting on it is zero. In this page, you'll learn what affects the equilibrium position of a mass oscillating on a spring, and how to determine it. Before treating the vertical spring case, we'll start with the simpler case of a mass attached to a horizontal spring.

### Horizontal Spring

Think of a mass attached to a horizontal spring placed on a table (under ideal conditions, so we can neglect friction). The position where the spring is not stretched or compressed is the equilibrium position (x=0 in Figure 1) and it corresponds to the natural length of the spring, _{0} **F _{w}**, and the normal force from the table,

**F**) are in the vertical direction and they cancel each other out.

_{n}**Figure 1**: Horizontal Spring

If you attach a different mass to the spring in that position, the vertical forces change but they still cancel out, so the total force will still be zero (**F _{TOT}**=

**F**+

_{w}**F**=

_{n}**0**). This means that the equilibrium position of a horizontal spring is independent of the mass attached to it. The same reasoning applies if you, for example, change the spring for a stiffer one but with the same natural length: the equilibrium position will be unaffected.

### Vertical Spring

Let’s now consider a mass attached to a vertical ideal spring, like the one of this simulation. When the spring is unloaded (Figure 2 (a)), its equilibrium configuration is again the one where it is not stretched or compressed, so its natural length, _{0} **F _{w}**, which is constant during the oscillation) and the elastic restoring force from the spring, defined by Hooke's Law,

**F**

_{el }**F _{TOT}** =

**F**+

_{w}**F**= m

_{el}**g**- k

**x**=

**0**

**Figure 2**: Vertical Spring

This point depends on the magnitude of the elastic force (**F _{el}**=-k

**x**) and the weight of the mass (

**F**=m

_{w}**g**), so it will be affected by changes in the stiffness of the spring (k) and different masses (m). Its distance from the spring equilibrium position x=0 (here called the equilibrium displacement or equilibrium extension, x

_{eq}), is given by:

Where m is the mass, **g** is the gravity and k is the spring constant. The heavier the mass, the larger is the equilibrium displacement. The stiffer the spring, the smaller the equilibrium displacement.