A trolley of mass 1 kg is joined to a wall by a spring of 5 N/m, and a damper gives a force of 2 N…

Step 1. The steady part. With $y_p = a\cos 2t + b\sin 2t$ the coefficient of $\cos 2t$ gives $a + 4b = 10$ and that of $\sin 2t$ gives $b - 4a = 0$. So $b = 4a$, $a = \dfrac{10}{17}$ and $b = \dfrac{40}{17}$.

Step 2. The amplitude. $\sqrt{a^2 + b^2} = \dfrac{\sqrt{1700}}{17} = 2.4254\ \text{m}$.

Step 3. The full solution. $x = e^{-t}(A\cos 2t + B\sin 2t) + \dfrac{10}{17}\cos 2t + \dfrac{40}{17}\sin 2t$.

Step 4. The constants. From $x(0) = 0$: $A + \dfrac{10}{17} = 0$, so $A = -\dfrac{10}{17}$. From $x'(0) = 0$: $-A + 2B + 2b = 0$, so $B = -\dfrac{45}{17}$.

Step 5. At $t = \dfrac{\pi}{2}$. $\cos\pi = -1$ and $\sin\pi = 0$. So $x = e^{-\pi/2}\left(\dfrac{10}{17}\right) - \dfrac{10}{17} = \dfrac{10}{17}(0.20788 - 1) = -0.4660\ \text{m}$.