Is that SHM ?

Let $x$ be the distance of the rod's center of mass from the left wheel, $M$ be the mass of the rod, and $N_{1}$ and $N_{2}$ be the magnitude of normal force of the rod on the left wheel and right wheel, respectively. Then, we know that

$N_{1} +N_{2} =Mg$ and $N_{2} x=N_{1} (l-x)$ and $M\ddot { x } =\mu ({ N }_{ 2 }-{ N }_{ 1 })$

From the first two equations we get $N_{1}=Mg \frac{x}{l}$ and $N_{2}=Mg(1-\frac{x}{l})$ . Hence,

$\ddot { x } =-\frac { 2\mu g }{ l } x+\mu g$ , implying that

$\omega =\sqrt { \frac { 2\mu g }{ l } } \approx \pi$

Since $\omega =\frac { 2\pi }{ T }$ , thus $T\approx \boxed { 2 }$ .