For a particle undergoing simple harmonic motion, the displacement...

The displacement of a particle in simple harmonic motion is given by $x(t) = A \sin(\omega t + \phi)$.

After one complete time period $T = \dfrac{2\pi}{\omega}$, the position of the particle is:

$x(t+T) = A \sin(\omega(t+T) + \phi)$

$x(t+T) = A \sin(\omega t + 2\pi + \phi) = A \sin(\omega t + \phi) = x(t)$

Since the initial and final positions of the particle are the same after one time period, the net displacement is zero.

The total distance traveled over one time period is $4A$, but the displacement is zero.