A particle executing SHM has a period of 6 s. The time taken by the particle to move form mean position tohalf of the amplitude, starting from the mean position is

The motion of a particle in Simple Harmonic Motion (SHM) can be described by the equation x = A sin(wt + φ), where A is the amplitude, w is the angular frequency, t is the time, and φ is the phase constant.

Given that the period T of the SHM is 6 seconds, we can find the angular frequency w using the formula w = 2π/T. Substituting T = 6 s, we get w = 2π/6 = π/3 rad/s.

We are asked to find the time taken by the particle to move from the mean position to half of the amplitude. This means we are looking for the time t when x = A/2.

Substituting x = A/2 into the equation of SHM, we get A/2 = A sin(wt + φ). Since the particle starts from the mean position, the phase constant φ = 0. So the equation simplifies to 1/2 = sin(wt).

We can now solve for t: sin(wt) = 1/2. The solutions to this equation are wt = π/6 + 2nπ and wt = 5π/6 + 2nπ, where n is an integer.

Since we are looking for the first time the particle reaches half the amplitude, we take the smallest positive solution, which is wt = π/6. Solving for t gives t = π/6w = π/6 * 3/π = 1/2 s.

So, the time taken by the particle to move from the mean position to half of the amplitude, starting from the mean position, is 1/2 second.