Final answer:
A particle in simple harmonic motion takes a quarter of its time period T to reach a displacement of 1/√2 times its amplitude. This corresponds to the position where the cosine function is at π/4 or 3π/4, which represents a quarter-cycle. Hence, the correct answer is A. T/4.
Step-by-step explanation:
The question concerns the time it takes for a particle undergoing simple harmonic motion (SHM) to travel from its mean position to a displacement of 1/√2 times its amplitude. Given that SHM can be represented by a cosine function, the position x(t) of the particle at time t is given by A cos(2πt/T), where A is the amplitude and T is the time period.
Since we want the displacement to be 1/√2 of the amplitude, we set x(t) = A/√2. This corresponds to the cosine of π/4 or 3π/4, as cos(π/4) = cos(3π/4) = 1/√2. This state is reached at a quarter of the period, as the complete period represents a cycle of 360 degrees or 2π radians. Since a full cycle of cosine from peak to peak takes a time T, and π/4 or 3π/4 corresponds to a quarter of this cycle, the time taken to reach A/√2 is T/4.
Therefore, the correct option for the time taken by the particle to travel from the mean position to 1/√2 times its amplitude is A. T/4.