A block Q of mass 3.0 kg lies on a smooth horizontal floor…

Step 1. Period (a). $\omega^2 = \dfrac{k}{m + M} = \dfrac{200\ \text{N/m}}{4.0\ \text{kg}} = 50\ \text{s}^{-2}$, so $\omega = 7.071\ \text{rad/s}$ and $T = 2\pi/\omega = 0.889\ \text{s}$.

Step 2. Largest amplitude (b). The spring pulls $P$; $Q$ moves only because $P$ drags it by friction. So the friction on $Q$ is $f = Ma$. Its largest value is $\mu m g = 0.30 \times 1.0\ \text{kg} \times 10\ \text{m/s}^2 = 3.0\ \text{N}$, so $Q$ can have at most $a = 3.0\ \text{N} / 3.0\ \text{kg} = 1.0\ \text{m/s}^2$. The pair's largest acceleration is $\omega^2 A$, so

Step 3. Greatest speed (c). $v_{\max} = \omega A_{\max} = 7.071\ \text{rad/s} \times 0.020\ \text{m} = 0.141\ \text{m/s}$.

Step 4. Friction at a point (d). The friction depends on where the pair is, not on the amplitude: $f = M\omega^2 x = 3.0\ \text{kg} \times 50\ \text{s}^{-2} \times 1.0\ \text{cm} = 1.50\ \text{N}$. It is below $3.0\ \text{N}$, as it must be, since $1.5\ \text{cm}$ is less than $A_{\max}$.

Step 5. Spring on the other block (e). Now friction drives $P$, and the limit is $m a \le \mu m g$, so $a \le \mu g = 3.0\ \text{m/s}^2$ and $A_{\max} = \mu g/\omega^2 = 6.0\ \text{cm}$.