The average acceleration in one time period in a simple harmonic motion is
A. $A{{\omega }^{2}}$
B. $\dfrac{A{{\omega }^{2}}}{2}$
C. $\dfrac{A{{\omega }^{2}}}{\sqrt{2}}$
D. Zero

Hint: The displacement in one time period in a simple harmonic motion is zero as the body returns to its mean position after every oscillation in one time period.

Complete step by step answer:
Acceleration of a particle executing SHM is defined as the rate of change of its velocity with respect to time at any particular instant. The formula for an accelerating particle is given by: $a=-A{{\omega }^{2}}$
So, A particle is said to be executing SHM if its acceleration at any instant is directed towards the mean position and is proportional to its displacement from the mean position. The negative sign indicated that acceleration and displacement are in opposite directions.
We know that the net displacement of the particle under SHM is zero as the body returns to its mean position after one time period.
The acceleration of the particle at two extreme positions during the SHM in one time period is:
\[a=-A{{\omega }^{2}}\]
\[a=A{{\omega }^{2}}\]
Also, the acceleration is proportional to displacement, the accelerations on both the extreme positions are directed in the opposite direction.
The average of these acceleration is:
\[a=0\]
So, the correct answer is option D. zero

Note: Another way of calculating the correct answer can be: Since we already know that the average displacement is zero therefore the average velocity and consequently the average acceleration would also be zero.