The potential enery of a particle varies with posiion `x` according to the relation `U(x) = 2x^(4) - 27 x` the point `x = (3)/(2)` is point of

Options

A. (a)unstable equilibrium
B. (b)stable equilibrium
C. (c)neutral equilibrium
D. (d)none of these

Answer

To solve the problem, we need to analyze the potential energy function given by \( U(x) = 2x^4 - 27x \) and determine the nature of the point \( x = \frac{3}{2} \).

### Step 1: Find the force from potential energy
The force \( F \) acting on the particle can be derived from the potential energy function using the relation:
\[ F = -\frac{dU}{dx} \]
We will differentiate \( U(x) \) with respect to \( x \).

### Step 2: Differentiate the potential energy function
Differentiating \( U(x) \):
\[ \frac{dU}{dx} = \frac{d}{dx}(2x^4 - 27x) = 8x^3 - 27 \]
Thus, the force becomes:
\[ F = - (8x^3 - 27) = -8x^3 + 27 \]

### Step 3: Evaluate the force at \( x = \frac{3}{2} \)
Now, we substitute \( x = \frac{3}{2} \) into the force equation:
\[ F\left(\frac{3}{2}\right) = -8\left(\frac{3}{2}\right)^3 + 27 \]
Calculating \( \left(\frac{3}{2}\right)^3 \):
\[ \left(\frac{3}{2}\right)^3 = \frac{27}{8} \]
Now substituting this back into the force equation:
\[ F\left(\frac{3}{2}\right) = -8 \cdot \frac{27}{8} + 27 = -27 + 27 = 0 \]
Since \( F = 0 \), this indicates that \( x = \frac{3}{2} \) is a point of equilibrium.

### Step 4: Determine the nature of equilibrium
To determine whether this point is a point of stable or unstable equilibrium, we need to check the second derivative of the potential energy function:
\[ \frac{d^2U}{dx^2} = \frac{d}{dx}(8x^3) = 24x^2 \]
Now we evaluate the second derivative at \( x = \frac{3}{2} \):
\[ \frac{d^2U}{dx^2}\left(\frac{3}{2}\right) = 24\left(\frac{3}{2}\right)^2 = 24 \cdot \frac{9}{4} = 54 \]
Since \( \frac{d^2U}{dx^2} > 0 \), this indicates that the point is a point of stable equilibrium.

### Conclusion
Thus, the point \( x = \frac{3}{2} \) is a point of **stable equilibrium**.

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