Given information:
- Maximum displacement (A) = 4 cm
- Acceleration (a) = 3 cm/s² at a distance of 1 cm from the mean position

To find:
Velocity (v) when the particle is at a distance of 2 cm from its mean position

Explanation:
1. Displacement (x):
The equation for simple harmonic motion is given by:
x = A * sin(ωt) ...(1)

where,
x = displacement of the particle from its mean position at time t
A = amplitude (maximum displacement)
ω = angular frequency

In the given problem, the maximum displacement (A) is given as 4 cm. Therefore, the equation becomes:
x = 4 * sin(ωt) ...(2)

2. Acceleration (a):
The acceleration of a particle in simple harmonic motion is given by:
a = -ω² * x ...(3)

where,
a = acceleration of the particle at displacement x
ω = angular frequency

We are given that at a distance of 1 cm from the mean position, the acceleration (a) is 3 cm/s². Plugging in the values in equation (3), we get:
3 = -ω² * 1
ω² = -3 ...(4)

3. Finding ω:
From equation (4), we can find ω by taking the square root of both sides:
ω = √(-3)
ω = √3i ...(5)

4. Finding velocity (v):
The velocity of the particle at displacement x is given by:
v = ω * A * cos(ωt) ...(6)

To find the velocity when the particle is at a distance of 2 cm from its mean position, we need to substitute x = 2 cm and solve for v. Plugging in the values in equation (6), we get:
v = √3i * 4 * cos(√3i * t)
v = 4√3i * cos(√3i * t) ...(7)

5. Finding time (t):
To find the time (t) when the particle is at a distance of 2 cm, we can use equation (2):
2 = 4 * sin(√3i * t)
sin(√3i * t) = 1/2
√3i * t = π/6
t = π/(6√3i) ...(8)

6. Substituting t in velocity equation:
Substituting the value of t from equation (8) into equation (7), we get:
v = 4√3i * cos(√3i * π/(6√3i))
v = 4√3i * cos(π/6)
v = 4√3i * √3/2
v = 6 cm/s ...(9)

Answer:
The velocity of the particle when it is at a distance of 2 cm from its mean position is 6 cm/s.