**Solution:**

Given, time taken by the particle in passing from A to B = 3 s
Also, the particle returns to B after 3 seconds.

Let's assume the amplitude of SHM is A, and the time period is T.

**Finding the position of the particle at A and B:**

At position A, the particle has maximum displacement (i.e., A) and zero velocity.

At position B, the particle has zero displacement and maximum velocity.

**Using the equations of SHM:**

Displacement equation of SHM is given by:

x = A sin(ωt + φ)

where, ω is the angular frequency and φ is the initial phase angle.

Velocity equation of SHM is given by:

v = Aω cos(ωt + φ)

where, ω is the angular frequency and φ is the initial phase angle.

**Finding the angular frequency of SHM:**

At point A, the velocity of the particle is zero.

Therefore, v = Aω cos(φ) = 0

Hence, cos(φ) = 0 or φ = π/2

At point B, the displacement of the particle is zero.

Therefore, x = A sin(ωt + π/2) = 0

Hence, sin(ωt + π/2) = 0 or ωt + π/2 = nπ, where n is an integer.

Since the particle takes 3 seconds to reach from A to B, we have:

3ω + π/2 = mπ, where m is an integer.

Solving for ω, we get:

ω = (2mπ - π/2)/3

**Finding the time period of SHM:**

The time period of SHM is given by:

T = 2π/ω

Substituting the value of ω, we get:

T = 2π/(2mπ - π/2)/3

T = 6π/(4mπ - π)

T = 6/4m - 1/4

T = (24 - 1)/4m

T = 23/4m

Since m can take any integer value, the time period can take any value of 23/4m seconds.

Therefore, the time period of SHM is 23/4 seconds.

Hiii Vasanthan ... Is the answer 4 or 18sec???