If P.E. of a system is given by relation `U= (A)/(x^(2))-(B)/(x)`, where 'A'and'B' are positive constant, then the mean positive of S.H.M.is

Options

A. `x=(A)/(B)`
B. `x=(B)/(A)`
C. `x= (B)/(2A)`
D. `x= (2A)/(B)`

Answer

To find the mean position of Simple Harmonic Motion (S.H.M) given the potential energy \( U = \frac{A}{x^2} - \frac{B}{x} \), we will follow these steps:

### Step 1: Differentiate the Potential Energy
To find the force acting on the system, we need to differentiate the potential energy with respect to \( x \):
\[ F = -\frac{dU}{dx} \]
Calculating the derivative:
\[ U = \frac{A}{x^2} - \frac{B}{x} \]
Differentiating \( U \):
\[ \frac{dU}{dx} = -\frac{2A}{x^3} + \frac{B}{x^2} \]
Thus, the force \( F \) becomes:
\[ F = -\left(-\frac{2A}{x^3} + \frac{B}{x^2}\right) = \frac{2A}{x^3} - \frac{B}{x^2} \]

### Step 2: Set the Force Equal to Zero
At the mean position of S.H.M, the net force acting on the system is zero:
\[ F = 0 \]
This gives us the equation:
\[ \frac{2A}{x^3} - \frac{B}{x^2} = 0 \]

### Step 3: Solve for \( x \)
Rearranging the equation:
\[ \frac{2A}{x^3} = \frac{B}{x^2} \]
Multiplying both sides by \( x^3 \):
\[ 2A = Bx \]
Now, solving for \( x \):
\[ x = \frac{2A}{B} \]

### Conclusion
Thus, the mean position of S.H.M is:
\[ x = \frac{2A}{B} \]