A particle undergoes onedimensional motion in a potential Ux

A particle undergoes one-dimensional motion in a potential U(x), shown on the next page. The horizontal line shows the total energy E of the system. At what lettered point is the particle\'s speed the smallest? Which points correspond to situations of equilibrium? Are these equilibria stable or unstable? Given a formula for U(x), what calculation could you do to prove your assertion?

Solution

For a particle moving in a potential U(x), the energy of the particle is stored in the form of kinetic energy of potential energy or combination of two.

a.) It needs to be understood that as the potential energy of the particle reduces, a greated part of its total energy gets transformed to kinetic energy hence leading to higher speeds.

That is, as the potential decreases, the kinetic energy increases, hence the speed will be maximum at minimum potential. Therefore the speed will be maximum at point D.

b.) For a system, it is said to be at equilbrium if the rate of change of the potential with respect to position is zero.

That would mean the points with first derivative of the function U(x) would be points of equilibrium. Hence, in the given graph, the points B, C and D are points of equilibrium.

Further, whether the equilibrium is stable or unstable is determined by the double derivative of U(x). If the double derivative is smaller than zero, it\'s an unstable equilibrium.

While if the double derivative is greater than zero, we get a stable equilibrium.

Looking at the graphs, we can say that for the points B and D, the double derivative would come out to be positive, hence they represent stable equilibrium while for point C, the double derivative would be negative, hence there is an unstable equilibrium there.