Solution:

Given, two particles performing SHM with the same angular frequency.

One particle is at x = A and the other is at x = 0.

We need to find the maximum distance between the particles and the time period after which it will happen for the first time.

Maximum distance between the particles:

- The two particles will be at maximum distance from each other when one is at the extreme position and the other is at the mean position.
- Let's assume that the particle at x = A is at its extreme position at time t = 0.
- The equation of motion for this particle can be written as x1 = A sin(ωt)
- The particle at x = 0 is at its mean position at time t = 0.
- The equation of motion for this particle can be written as x2 = 0 + B sin(ωt + φ)
- Since the particles are moving away from each other, the phase difference between them will be π.
- Therefore, φ = π/2
- The equation of motion for the second particle can be written as x2 = B cos(ωt)
- At the time t = T/4, the particle at x = A will be at its mean position and the particle at x = 0 will be at its extreme position.
- Therefore, the maximum distance between the particles can be calculated as follows:

Dmax = x1(T/4) - x2(T/4)
= A sin(π/2) - B cos(π/2)
= A

- Therefore, the maximum distance between the particles is equal to A.

Time period after which the particles will be at maximum distance:

- The time period of the SHM is given by T = 2π/ω.
- Therefore, the time period after which the particles will be at maximum distance can be calculated as follows:

T/4 = (π/2)/ω
= (π/2)/√(k/m)

- Where k is the spring constant and m is the mass of the particle.
- Therefore, the time period after which the particles will be at maximum distance is given by (π/2)/√(k/m).

Conclusion:

- The maximum distance between the particles is equal to A.
- The time period after which the particles will be at maximum distance is given by (π/2)/√(k/m).