To solve this problem, we need to understand the concept of average velocity in simple harmonic motion (SHM).$\\$In SHM, a particle oscillates back and forth about an equilibrium position.$\\$The displacement of the particle as a function of time is given by:$\\ x(t) = A \cos(\omega t + \phi) \\$where:$\\$• $A$ is the amplitude of the motion,$\\$• $\omega$ is the angular frequency, and$\\$• $\phi$ is the phase constant.$\\$The velocity of the particle is the derivative of the displacement with respect to time:$\\ v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi) \\$The average velocity over one complete vibration (one full cycle) is given by:$\\ v_{avg} = \frac{Total \ displacement}{Total \ time} \\$In one complete cycle of SHM, the particle returns to its starting point,$\\$so the total displacement is zero.$\\$Therefore, the average velocity over one complete vibration is:$\\ v_{avg} = \frac{0}{T} = 0 \\$where $T$ is the time period of the motion.$\\$Thus, the average velocity of a particle executing SHM in one complete vibration is zero.$\\$This corresponds to Option 4.