Positions $x_2,x_3$, and $x_4$ in Figure 10.20, where the potential energy has a local minimum or maximum, are special positions. Consider a system with the total energy $E_2$ shown in FIGURE 10.21. The particle can be at rest at $x_2$, with $K=0$, but it cannot move away from $x_2$. In other words, a system with energy $E_2$ is in equilibrium at $x_2$. If you disturb the particle, giving it a small kinetic energy and a total energy just slightly larger than $E_2$, the particle will undergo a very small oscillation centered on $x_2$, like a marble in the bottom of a bowl. An equilibrium for which small disturbances cause small oscillations is called a point of stable equilibrium. You should recognize that any minimum in the PE curve is a point of stable equilibrium. Position $x_4$ is also a point of stable equilibrium, in this case for a particle with $E=0$.

FIGURE 10.21 Points of stable and unstable equilibrium.

Figure 10.21 also shows a system with energy $E_3$ that is tangent to the PE curve at $x_3$. If a particle is placed exactly at $x_3$, it will stay there at rest $(K=0)$. But if you disturb the particle at $x_3$, giving it an energy only slightly more than $E_3$, it will speed up as it moves away from $x_3$. This is like trying to balance a marble on top of a hill. The slightest displacement will cause the marble to roll down the hill. A point of equilibrium for which a small disturbance causes the particle to move away is called a point of unstable equilibrium. Any maximum in the PE curve, such as $x_3$, is a point of unstable equilibrium.

We can summarize these lessons as follows:

EXAMPLE 10.8 Balancing a mass on a spring

A spring of length $L_0$ and spring constant $k$ is standing on one end. A block of mass $m$ is placed on the spring, compressing it. What is the length of the compressed spring?

MODEL Assume an ideal spring obeying Hooke’s law. The block $\text{+}$ earth $\text{+}$ spring system has both gravitational potential energy $U_{\text{G}}$ and elastic potential energy $U_{\text{Sp}}$. The block sitting on top of the spring is at a point of stable equilibrium (small disturbances cause the block to oscillate slightly around the equilibrium position), so we can solve this problem by looking at the energy diagram.

VISUALIZE FIGURE 10.22a is a pictorial representation. We’ve used a coordinate system with the origin at ground level, so the displacement of the spring is $y-L_0$.

FIGURE 10.22 The block $\text{+}$ earth $\text{+}$ spring system has both gravitational and elastic potential energy.

FIGURE 10.22 Full Alternative Text

SOLVE FIGURE 10.22b shows the two potential energies separately and also shows the total potential energy:

$$\begin{array}{ccc}\hfill U_{\text{tot}}& \hfill =\hfill & U_{\text{G}}+U_{\text{Sp}}\hfill \\ \hfill & \hfill =\hfill & mgy+{\displaystyle \frac{1}{2}}k{(y-L_0)}^2\hfill \end{array}$$

The equilibrium position (the minimum of $U_{\text{tot}}$) has shifted from $L_0$ to a smaller value of $y$, closer to the ground. We can find the equilibrium by locating the position of the minimum in the PE curve. You know from calculus that the minimum of a function is at the point where the derivative (or slope) is zero. The derivative of $U_{\text{tot}}$ is

$${\displaystyle \frac{d{U}_{\text{tot}}}{dy}}=mg+k(y-L_0)$$

The derivative is zero at the point $y_{\text{eq}}$, so we can easily find

$$\begin{array}{l}mg+k(y_{\text{eq}}-L_0)=0\\ y_{\text{eq}}=L_0-{\displaystyle \frac{mg}{k}}\end{array}$$

The block compresses the spring by the length $mg\text{/}k$ from its original length $L_0$, giving it a new equilibrium length $L_0-mg\text{/}k$.