21.1 Nature of equilibrium  (Page 3/4)

Once past “B” in opposite direction i.e towards right, it again gains potential energy at the expense of kinetic energy. It has positive velocity as it is moving in the reference direction. From "F-x" plot, we see that force is acting in negative x-direction. The velocity and force are in opposite direction in this side of motion also. As such, pendulum bob is again decelerated. Ultimately bob comes to a stop and then reverses direction towards point “B”. In this reverse journey, it acquires kinetic energy at the expense of loosing potential energy.

For the given set up, maximum mechanical energy corresponds to position θ = 15°. What it means that pendulum never crosses the points “A” and “B”. These points, therefore, are the “turning points” for the given maximum mechanical energy of the set up (=3.4 J).

We, therefore conclude that the equilibrium of pendulum bob is a “stable” equilibrium at “B” within maximum angular displacement. If we give a small disturbance to the bob, its motion is bounded by the energy imparted during the disturbance. As the kinetic energy imparted is less than 3.4 J, it does not escape beyond the turning points specified for the set up. We should note that this situation with respect to pendulum bob is similar to the spherical ball placed inside a spherical shell.

From the discussion so far, we conclude that :

1: First derivative of potential energy function with respect to displacement is zero.

$$\frac{dU}{dx}=0$$

2: Potential energy of the body is minimum for stable equilibrium for a given potential energy function and maximum allowable mechanical energy.

$$U\left(x\right)=U_{\min }$$

For this, the second derivative of potential energy function is positive at equilibrium point,

$$\frac{d^{2}U}{d{x}^2}>0$$

3: The position of stable equilibrium is bounded by two turning points corresponding to maximum allowable mechanical energy.

4: Force acts to restore the original position of the body in stable equilibrium.

Example

Problem 1 : The potential energy of a particle in a conservative system is given by the potential energy function,

$$U\left(x\right)=\frac{1}{2}a{x}^2-bx$$

where “a” and “b” are two positive constants. Find the equilibrium position and determine the nature of equilibrium.

Solution : The system here is a conservative system. This means that only conservative forces are in operation. In order to determine the equilibrium position, we make use of two facts (i) negative of first differential of potential energy function gives the net force on the system and (ii) for equilibrium, net external force is zero.

$$\Rightarrow \frac{dU}{dx}=\frac{1}{2}X2ax-b$$

For equilibrium, external force is zero,

$$F=-\frac{dU}{dx}=-\left(\frac{1}{2}X2ax-b\right)=0$$

$$\Rightarrow x=\frac{b}{a}$$

Further, we need to find the second derivative of potential energy function and investigate the resulting value.

$$\Rightarrow \frac{d^{2}U}{d{x}^2}=a$$

It is given that “a” is a positive constant. It means that the particle possess minimum potential energy at $x=\frac{b}{a}$ . Hence, particle is having stable equilibrium at this position.

Unstable equilibrium

The nature of the potential energy plot for unstable equilibrium is inverse to that of stable equilibrium curve. A typical plot is shown here in the figure.