The second hand of a wall clock is 12 cm long…

Step 1. SHM. The shadow is the projection of uniform circular motion, so it moves in SHM with $A = 12\ \text{cm}$ and $T = 60\ \text{s}$. Then $\omega = \dfrac{2\pi}{60\ \text{s}} = 0.105\ \text{rad/s}$.

Step 2. After 10 s. It starts at the top, so $x = A\cos\omega t$. The phase is $0.105\ \text{rad/s} \times 10\ \text{s} = \pi/3$. So $x = 12\ \text{cm} \times \cos 60^\circ = 6\ \text{cm}$.

Step 3. Centre to 6 cm. From the centre, $x = A\sin\omega t$. We want $t$ when $\sin\omega t = \dfrac{6\ \text{cm}}{12\ \text{cm}} = 0.5$. So $\omega t = \pi/6$, which is a twelfth of a turn: $t = \dfrac{60\ \text{s}}{12} = 5\ \text{s}$.