A particle of mass m moves in a force field such that its potential energy in force field is defined by the equation `U = +A (x-a)^(2) (x-b)^(2)`. Where A, a and b are `+ve` constants then body may oscillate simple harmonically about point.

Options

A. a only
B. a and b both
C. only `(a+b)/(2)`
D. b only

Explanation

To solve the problem, we need to analyze the potential energy function given by:

\[ U(x) = A (x - a)^2 (x - b)^2 \]

where \( A \), \( a \), and \( b \) are positive constants. We want to find the points about which the particle may oscillate simple harmonically.

### Step 1: Understand the Potential Energy Function
The potential energy function \( U(x) \) is a product of two squared terms, which means it will be zero at certain points. The potential energy is zero when either of the squared terms is zero.

### Step 2: Set the Potential Energy to Zero
To find the points where the potential energy is zero, we set the equation to zero:

\[ U(x) = A (x - a)^2 (x - b)^2 = 0 \]

Since \( A \) is a positive constant, we can ignore it for the purpose of finding the roots. Thus, we need to solve:

\[ (x - a)^2 (x - b)^2 = 0 \]

### Step 3: Solve for \( x \)
The product of two squares is zero if either square is zero. Therefore, we have:

1. \( (x - a)^2 = 0 \) → \( x - a = 0 \) → \( x = a \)
2. \( (x - b)^2 = 0 \) → \( x - b = 0 \) → \( x = b \)

### Step 4: Identify the Points of Oscillation
From the above calculations, we find that the potential energy \( U(x) \) is zero at two points:

- \( x = a \)
- \( x = b \)

These points are the equilibrium positions where the particle can oscillate.

### Conclusion
The particle may oscillate simple harmonically about the points \( x = a \) and \( x = b \).