A particle executing simple harmonic motion is given by $y=8\sin \pi \left(10t-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$ where y  is in metre, and t is in seconds. Then the average speed of the particle during one  full oscillation is 

Average speed of a particle executing simple harmonic motion can be written as $\frac{\text{Total\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}Distance}}{\text{Total\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}time}}$   
In a simple harmonic motion, the total distance travelled by the particle is 4 times  the amplitude. The total time taken is one full time period of oscillation.
Therefore, $\text{Average\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}speed=}\frac{\text{Total\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}Distance}}{\text{Total\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}time}}=\frac{\text{4A}}{\text{T}}$ 
Comparing the given equation $y=8\sin \pi \left(10t-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$ with standard equation,  $y=A\sin \left(\omega t+\varphi \right)$ , we get  $A=8m$ and   $\omega =10\pi$.
Using the relation  $\omega =\frac{2\pi }{T}$, we get  $10\pi =\frac{2\pi }{T}$ 
i.e. $T=\frac{1}{5}s$.
Hence  $\text{Average\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}speed=}\frac{\text{4A}}{\text{T}}=\frac{4\times 8}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}=160m/s$.