Let $G$ be a graph and let $\delta$ be a distribution of pebbles on $G$. A pebbling move on the graph $G$ consists of removing two pebbles from one vertex and then placing one pebble at an adjacent vertex. Given a positive integer $d$, if we can move pebbles to any target vertex $v$ in $G$ only from the vertices in the set $N_d[v]=\{u\in V(G):d(u,v)\le d\}$ by pebbling moves, where $d(u,v)$ is the distance between $u$ and $v$, then such a graph pebbling played on $G$ is said to be distance $d$-restricted. For each target vertex $v\in V(G)$, we use $m(\delta ,d,v)$ to denote the maximum number of pebbles that can be moved to $v$ only from the vertices in the set $N_d[v]$. If $m(\delta ,d,v)\ge t$ for each $v\in V(G)$, then we say that $\delta$ is $(d,t)$-solvable. The optimal $(d,t)$-pebbling number of $G$, denoted by ${\pi }_{(d,t)}^{\ast }(G)$, is the minimum number of pebbles needed so that there is a $(d,t)$-solvable distribution of $G$. In this article, we study distance $2$-restricted pebbling in cycles and show that for any $n$-cycle $C_n$ with $n\ge 6$, ${\pi }_{(2,t)}^{\ast }(C_n)={\pi }_{(2,t-10)}^{\ast }(C_n)+4n$ for $t\ge 13$. It follows that if $n\ge 6$, then ${\pi }_{(2,10k+r)}^{\ast }(C_n)={\pi }_{(2,r)}^{\ast }(C_n)+4kn$ for $k\ge 1$ and $3\le r\le 12$. Consequently, for $n\ge 6$, the problem of determining the exact value of ${\pi }_{(2,t)}^{\ast }(C_n)$ for all $t\ge 1$ can be reduced to the problem of determining the exact value of ${\pi }_{(2,r)}^{\ast }(C_n)$ for $r\in [1,12]$. We also consider $C_n$ with $3\le n\le 5$. When $n=3$, we have ${\pi }_{(2,t)}^{\ast }(C_3)={\pi }_{(1,t)}^{\ast }(C_3)$, since the diameter of $C_3$ is one. The exact value of ${\pi }_{(1,t)}^{\ast }(C_3)$ is known. When $n=4,5$, we determine the exact value of ${\pi }_{(2,t)}^{\ast }(C_n)$ for $t\ge 1$.