# Question #72b8f

Given that the amplitude of SHM is $a = 10 c m$ and its time period is $T = 4 s$. So the particle executing SHM completes one oscillation in $4 s$ and 2 oscillations in $8 s$
In one complete oscillation it covers 4 times the the length of its amplitude i.e. $4 \times 10 c m = 40 c m$
So in $8 s$ the particle will cover $2 \times 40 c m = 80 c m$ distance