Assume a body performing harmonic oscillation, with the oscillation period of T = 6 s; the initial phase is zero. Calculate how long (in seconds) after the beginning of the motion the body displacement from equilibrium position will be one half of the amplitude.

The motion of a body performing harmonic oscillation can be described by the equation:

x(t) = A * cos(wt + φ)

where:

• x(t) is the displacement from the equilibrium position at time t,
• A is the amplitude of the oscillation,
• w is the angular frequency, and
• φ is the initial phase.

Given that the initial phase φ is zero, the equation simplifies to:

x(t) = A * cos(wt)

We are asked to find the time t when the displacement x(t) is half of the amplitude A. So we set x(t) = A/2 and solve for t:

A/2 = A * cos(wt) 1/2 = cos(wt) wt = arccos(1/2)

The angular frequency w is related to the period T by the equation w = 2π/T. Substituting T = 6 s into this equation gives w = 2π/6 = π/3 rad/s. Substituting this into the equation for wt gives:

π/3 * t = arccos(1/2) t = (3/π) * arccos(1/2)

Using the fact that arccos(1/2) = π/3 rad, we find:

t = (3/π) * π/3 = 1 s

So, the body's displacement from the equilibrium position will be one half of the amplitude 1 second after the beginning of the motion.