Given information:
A particle of mass 1 kg is undergoing small oscillation about the equilibrium point in the potential
V(x) = (1/2)x^12 - (1/x^6) for x > 0 m.

To find:
The time period (T) of the oscillation.

Explanation:
1. Equilibrium point:
The equilibrium point is where the potential energy is at a minimum. To find the equilibrium point, we differentiate the potential energy function with respect to x and set it equal to zero.

dV(x)/dx = 0
6x^11 + 6/x^7 = 0
Multiplying both sides by x^7, we get:
6x^18 + 6 = 0
x^18 = -1 (which has no real solution)

Since there is no real solution for x, the equilibrium point does not exist.

2. Harmonic approximation:
Since the potential energy function does not have a simple harmonic form, we need to use the harmonic approximation to find the time period of oscillation.

For small oscillations, the potential energy function can be approximated as a quadratic function around the equilibrium point.

V(x) ≈ V0 + (1/2)kx^2

Where V0 is the potential energy at the equilibrium point and k is the spring constant.

3. Taylor expansion:
To find the coefficients V0 and k, we can expand the potential energy function using a Taylor series expansion.

V(x) = V0 + (1/2)kx^2 + higher-order terms

We can neglect the higher-order terms as they are negligible for small oscillations.

4. Finding V0:
To find V0, we substitute x = 0 into the potential energy function.

V(0) = V0 + (1/2)k(0)^2
V0 = V(0)

Since the potential energy function is undefined at x = 0, we cannot determine the value of V0.

5. Finding k:
To find k, we differentiate the potential energy function with respect to x and evaluate it at the equilibrium point.

dV(x)/dx = kx
k = dV(x)/dx (at the equilibrium point)

But as we found earlier, the equilibrium point does not exist, so we cannot determine the value of k.

6. Time period of oscillation:
Since we cannot determine the values of V0 and k, we cannot directly find the time period (T) of the oscillation using the simple harmonic motion formula T = 2π√(m/k).

Therefore, it seems that the given solution of T = π/3 is incorrect.