The average acceleration of a particle performing S.H.M. over two...

The average acceleration of a particle over a given time interval is defined as the total change in velocity divided by the total time taken.

$\vec{a}_{avg} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{\vec{v}_f - \vec{v}_i}{\Delta t}$

For a particle performing Simple Harmonic Motion (S.H.M.), the velocity is a periodic function of time. After any integer number of complete oscillations, the particle returns to its exact initial state, meaning its final velocity $\vec{v}_f$ is equal to its initial velocity $\vec{v}_i$.

Over two complete oscillations, the change in velocity is:

$\Delta \vec{v} = \vec{v}_f - \vec{v}_i = 0$

Therefore, the average acceleration over two complete oscillations is zero.

Alternatively, the acceleration in S.H.M. is given by $a = -A\omega^2 \sin(\omega t + \phi)$. The average value of a sine function over any complete number of cycles is zero.

Answer: zero