Question

In a certain field, the potential energy is U = ax^2 - bx^3, where a and b constants. The particle is in stable equilibrium at x equal to
(1) Zero
(2) a/3b
(3) 2a/3b
(4) 2a/b

Answer 1

Answer:

U=ax2-bx3

Explanation:

du/dx=2ax-3bx2

d2u/dx2=2a-3bx

2a-3bx=0

2a=3bx

x=2a/3b

Answer 2

Answer:

a/3b.

Explanation:

We know that for being in stable equilibrium the potential energy is given as the second derivative or d^2u/dx^2 < 0.

So, U = ax^2 - bx^3.

dU/dx = 2ax - 3bx^2.

d^2U/dx^2 = 2a -6bx.

Since, we know that the second derivative will be <0 so 2ax -3bx^2<0 or x will be on solving x = 2a/3b.

From the above value of x we see that the first derivative must be equal to zero and that value of x must be calculated.

Hence, to remain stable the body must be at x=2a/3b.