The distance covered by a particle undergoing SHM in one time...

In a simple harmonic motion (SHM) the particle oscillates about its mean position on a straight line. The particle moves from its mean position $(O)$ to an extreme position $(\mathrm{P})$ and then return to its mean position covering same distance of A. Then by the conservative force, it is moved in opposite direction to a point $Q$ by distance A and then back to mean position covering a distance of A. This comprises of one time period as shown below
Hence, in one time period it covers a distance of $\mathrm{x}=\mathrm{OP}+\mathrm{PO}+\mathrm{OQ}+$\mathrm{QO}=\mathrm{A}$+\mathrm{A}+\mathrm{A}+\mathrm{A}=$4 $\mathrm{~A}$