A particle and its position varies with time as $ x = A\sin \omega t $ . Its average speed during its motion from mean position to mid-point of mean and extreme position is:
(A) Zero
(B) $ \dfrac{{3A\omega }}{\pi } $
(C) $ \dfrac{{A\omega }}{{2\pi }} $
(D) $ \dfrac{{2A\omega }}{\pi } $

Hint : The mean position is the centre of oscillation, the point perfectly in the middle of the two extremes. The total distance covered in moving from the mean position to the mid-point between the midpoint and extreme position is half of the Amplitude.

Formula used: In this solution we will be using the following formula;
 $\Rightarrow v = \dfrac{x}{t} $ where $ v $ is the average speed, $ x $ is the distance covered, and $ t $ is the time taken to cover the distance.

Complete step by step answer
In the question a particle is oscillation harmonically, and its equation was given by
 $ x = A\sin \omega t $ $ x $ is the distance covered (in a quarter cycle) after a particular time $ t $ , and $ \omega $ is the angular speed of the oscillation.
Now, the mean position is the centre of oscillation of the particle, i.e. the middle point between the extreme.
Now, the particle moves from mean to the midpoint between mean and extreme. Hence, the distance covered is half of the amplitude. Thus
 $\Rightarrow x = \dfrac{A}{2} $
Now, we must calculate the time taken for the particle to traverse this distance. From the equation of the distance covered by the particle with respect to time we have
 $\Rightarrow x = A\sin \omega t $
 $ \Rightarrow \dfrac{A}{2} = A\sin \omega t $
By eliminating $ A $ , we have
 $\Rightarrow \dfrac{1}{2} = \sin \omega t $
By trigonometric inversion, we have that
 $\Rightarrow \omega t = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) $
 $ \Rightarrow \omega t = \dfrac{\pi }{6} $
Hence, dividing both sides by $ \omega $ we get
 $\Rightarrow t = \dfrac{\pi }{{6\omega }} $
Thus, calculating average speed which is given by
 $\Rightarrow v = \dfrac{x}{t} $ we have
 $\Rightarrow v = \left( {\dfrac{A}{2} \div \dfrac{\pi }{{6\omega }}} \right) $ . This is identical to
 $\Rightarrow v = \dfrac{A}{2} \times \dfrac{{6\omega }}{\pi } $
Hence, by computation, we have that
 $\Rightarrow v = \dfrac{{3A\omega }}{\pi } $
Hence, the correct answer is option B.

Note
For clarity, the distance travelled in a simple harmonic motion from the formula $ x = A\sin \omega t $ is limited to the amplitude, that is to say, the amplitude is the largest value the distance can take. The distance (neglecting the sign) repeats after every quarter cycle. Even the sign only gives information on which side of the mean position the particle is at a particular time.