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, L0 L zero . If you attach the mass to the spring in this position, the mass will stay at rest and not move - if left unperturbed. In fact, the only forces acting on the mass in this configuration (its weight, Fw, and the normal force from the table, Fn) are in the vertical direction and they cancel each other out.

This image is divided in two showing the difference between a spring with and without an object at the end of it. In the upper part of the image a spring alone is showed with an arrow that labels the length of the spring as L zero. Here, the total force is 0. The lower part of the image consists of a spring with a square-shaped body at the end of it. in this diagram, the object is affected by two forces. Fn, which acts pointing upwards from the center of the body and Fw, which acts pointing downwards. in this system, the total force is equal to Fn plus Fw, which it all equals to zero.

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 (FTOT=Fw+Fn=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, L0 L zero . If you attach a mass to the spring in this configuration (Figure 2 (b)), the total force acting on the mass is given by its weight pointing downwards. The mass will start moving downward stretching the spring. In this situation, the forces acting on the mass are its weight (Fw, which is constant during the oscillation) and the elastic restoring force from the spring, defined by Hooke's Law, Fel F el . The latter is not constant but changes linearly with displacement. The point at which the elastic force is equal to the weight of the mass is the equilibrium position of the mass (Figure 2 (c)). At this point, in fact, the total force is zero,

FTOT = Fw + Fel = mg - kx = 0 F tot is equal to F w plus F el, minus K per x, which all equals 0

Figure 2: Vertical Spring

This point depends on the magnitude of the elastic force (Fel=-kx) and the weight of the mass (Fw=mg), 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, xeq), 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.