Motion in one-dimensional potential

where is the potential energy of the particle at position .

Let the curve take the form shown in Figure 2.4. For instance, this curve might represent the gravitational potential energy of a cyclist freewheeling in a hilly region. Observe that we have set the potential energy at infinity to zero (which we are generally free to do, because potential energy is undefined to an arbitrary additive constant). This is a fairly common convention. What can we deduce about the motion of the particle in this potential?

Well, we know that the total energy, --which is the sum of the kinetic energy, , and the potential energy, --is a constant of the motion. [See Equation (2.22).] Hence, we can write

However, we also know that a kinetic energy can never be negative, because , and neither nor can be negative. Hence, the preceding expression tells us that the particle's motion is restricted to the region (or regions) in which the potential energy curve falls below the value . This idea is illustrated in Figure 2.4. Suppose that the total energy of the system is . It is clear, from the figure, that the particle is trapped inside one or other of the two dips in the potential; these dips are generally referred to as

The preceding discussion suggests that the motion of a particle moving in a potential
generally becomes less bounded as the total energy
of the system increases.
Conversely, we would expect the motion to become more bounded as
decreases.
In fact, if the energy becomes sufficiently small then it appears likely that the
system will settle down in some *equilibrium state* in which the particle remains stationary.
Let us try to identify any prospective equilibrium states in Figure 2.4.
If the particle remains stationary then it must be subject to zero force (otherwise,
it would accelerate). Hence, according to Equation (2.45), an equilibrium
state is characterized by

(2.47) |

In other words, an equilibrium state corresponds to either a maximum or a minimum of the potential energy curve, . It can be seen that the curve shown in Figure 2.4 has three associated equilibrium states located at , , and .

Let us now make a distinction between *stable* equilibrium points
and *unstable* equilibrium points. When the particle is slightly
displaced from a stable equilibrium point then the resultant force acting
on it
must always be such as to return it to this point.
In other words, if
is an equilibrium point then we require

(2.48) |

for stability; that is, if the particle is displaced to the right, so that , then the force must act to the left, so that , and vice versa. Likewise, if

(2.49) |

then the equilibrium point is unstable. It follows, from Equation (2.45), that stable equilibrium points are characterized by

(2.50) |

In other words, a stable equilibrium point corresponds to a minimum of the potential energy curve, . Likewise, an unstable equilibrium point corresponds to a maximum of the curve. Hence, we conclude that, in Figure 2.4, and are stable equilibrium points, whereas is an unstable equilibrium point. Of course, this makes perfect sense if we think of as a gravitational potential energy curve, so that is directly proportional to height. In this case, all we are saying is that it is easy to confine a low energy mass at the bottom of a smooth valley, but it is very difficult to balance the same mass on the top of a smooth hill (because any slight displacement of the mass will cause it to slide down the hill). Finally, if

(2.51) |

at any point (or in any region) then we have what is known as a

The equation of motion of a particle moving in one dimension under the action of a conservative force is, in principle, integrable. Because , the energy conservation equation, Equation (2.46), can be rearranged to give

(2.52) |

where the signs correspond to motion to the left and to the right, respectively. However, because , this expression can be integrated to give

where . For sufficiently simple potential functions, , Equation (2.53) can be solved to give as a function of . For instance, if , , and the plus sign is chosen, then

(2.54) |

which can be inverted to give

(2.55) |

where and . Note that the particle reverses direction each time it reaches one of the so-called