Reading a Potential Energy Curve — Question 4

Question 4

Horizontal tangents with different stability

Assume one-dimensional classical motion with $K=E-U(x)$ and $F_x=-dU/dx$, with no dissipation. Compare $U_A(x)=x^4$ J with $U_B(x)=x^3$ J near x=0, with x in meters.

1. Find the force at x=0 for both potentials.

2. Determine the force directions just to either side of zero.

3. Classify the equilibrium at zero for each potential and explain why the second derivative alone is inconclusive.

Question 4 - Solution

Horizontal tangents with different stability

(a) Differentiate once.

Differentiate the potentials: $$F_A=-4x^3,\qquad F_B=-3x^2\Rightarrow\boxed{F_A(0)=F_B(0)=0}.$$

(b) Check the signs on both sides.

For A, the force restores from both sides: $$x<0\Rightarrow F_A>0,\qquad x>0\Rightarrow F_A<0.$$ For B, $F_B<0$ on either side of zero, so a negative displacement is driven farther left.

(c) Use the local shape, not only one derivative test.

The second derivatives satisfy $$U_A''(0)=0=U_B''(0),$$ but A has a strict flat minimum and B a stationary inflection. Hence $\boxed{A:\text{ stable};\quad B:\text{ unstable}}$. A zero second derivative requires inspection of higher-order terms or restoring directions.