prove that a position of equilibrium is stable if the potential energy is minimum
Answers
Answered by
1
Oscillation is intimately related with stable equilibrium.
To illustrate it, let us consider a typical curve between the position (x) of the particle and its potential energy (U) for a one dimensional particle motion in a conservative field.

Tangents drawn at B, C, D and E are parallel to the x-axis. This means, at these points, slope (dU/dx) is zero.
Recalling F = − (dU/dx) , we can further say that at B , C , D and E , force acting on the particle is zero i.e. these are equilibrium positions.
For portions BC and DE, an increase in the value of x corresponds to an increase in the value of U.
The slope of the curve at any point in this portion is positive and hence , force( F = − dU/dx ) is negative.
It means, in BC and DE region, the force acting on the particle tends to pull it in a region of lower potential energy.
Similarly it can be shown that for the portions AB and CD (where slope is negative and hence force is positive) again the force pulls the particle in the region of lower potential energy.
Thus any slight displacement of the particle, either way from the position of minimum potential energy results into a force tending to bring the particle back to its original position.
This force is often referred to as restoring force and site of minimum potential energy, as recalled earlier, is the position of stable equilibrium.
To illustrate it, let us consider a typical curve between the position (x) of the particle and its potential energy (U) for a one dimensional particle motion in a conservative field.

Tangents drawn at B, C, D and E are parallel to the x-axis. This means, at these points, slope (dU/dx) is zero.
Recalling F = − (dU/dx) , we can further say that at B , C , D and E , force acting on the particle is zero i.e. these are equilibrium positions.
For portions BC and DE, an increase in the value of x corresponds to an increase in the value of U.
The slope of the curve at any point in this portion is positive and hence , force( F = − dU/dx ) is negative.
It means, in BC and DE region, the force acting on the particle tends to pull it in a region of lower potential energy.
Similarly it can be shown that for the portions AB and CD (where slope is negative and hence force is positive) again the force pulls the particle in the region of lower potential energy.
Thus any slight displacement of the particle, either way from the position of minimum potential energy results into a force tending to bring the particle back to its original position.
This force is often referred to as restoring force and site of minimum potential energy, as recalled earlier, is the position of stable equilibrium.
Similar questions