1. **Problem statement:** A 15 kg mass projectile moving at an initial velocity of 20 m/s hits a shock tube system with a spring of stiffness $k=4000$ N/m and a damper with damping coefficient $c=5000$ Ns/m. We want to find: a) The maximum displacement of the projectile after impact. b) The time it takes to reach this maximum displacement. 2. **Formulas and concepts:** The system can be modeled as a damped harmonic oscillator with mass $m=15$ kg, spring constant $k=4000$ N/m, and damping coefficient $c=5000$ Ns/m. The equation of motion is: $$m\ddot{x} + c\dot{x} + kx = 0$$ Initial conditions: $$x(0) = 0, \quad \dot{x}(0) = v_0 = 20\ \text{m/s}$$ The damping ratio $\zeta$ and natural frequency $\omega_n$ are: $$\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{4000}{15}} = \sqrt{266.67} \approx 16.33\ \text{rad/s}$$ $$\zeta = \frac{c}{2m\omega_n} = \frac{5000}{2 \times 15 \times 16.33} = \frac{5000}{489.9} \approx 10.21$$ Since $\zeta > 1$, the system is overdamped. 3. **Maximum displacement:** For an overdamped system with initial velocity and zero initial displacement, the maximum displacement occurs at time $t_{max}$ where velocity is zero. The general solution is: $$x(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}$$ where $$r_{1,2} = -\zeta \omega_n \pm \omega_n \sqrt{\zeta^2 - 1}$$ Calculate $r_1$ and $r_2$: $$r_1 = -10.21 \times 16.33 + 16.33 \times \sqrt{10.21^2 - 1} \approx -166.7 + 166.2 = -0.5$$ $$r_2 = -10.21 \times 16.33 - 16.33 \times \sqrt{10.21^2 - 1} \approx -166.7 - 166.2 = -332.9$$ Using initial conditions: $$x(0) = C_1 + C_2 = 0$$ $$\dot{x}(0) = C_1 r_1 + C_2 r_2 = 20$$ From $x(0)$: $$C_2 = -C_1$$ Substitute into velocity: $$C_1 r_1 - C_1 r_2 = 20 \Rightarrow C_1 (r_1 - r_2) = 20$$ $$C_1 = \frac{20}{r_1 - r_2} = \frac{20}{-0.5 - (-332.9)} = \frac{20}{332.4} \approx 0.0602$$ $$C_2 = -0.0602$$ The displacement is: $$x(t) = 0.0602 e^{-0.5 t} - 0.0602 e^{-332.9 t}$$ 4. **Time to maximum displacement:** Velocity: $$\dot{x}(t) = 0.0602 (-0.5) e^{-0.5 t} - 0.0602 (-332.9) e^{-332.9 t} = -0.0301 e^{-0.5 t} + 20.04 e^{-332.9 t}$$ Set $\dot{x}(t) = 0$ for max displacement: $$-0.0301 e^{-0.5 t} + 20.04 e^{-332.9 t} = 0$$ $$20.04 e^{-332.9 t} = 0.0301 e^{-0.5 t}$$ $$\frac{20.04}{0.0301} = e^{(332.9 - 0.5) t}$$ $$666.1 = e^{332.4 t}$$ $$\ln(666.1) = 332.4 t$$ $$t = \frac{\ln(666.1)}{332.4} = \frac{6.5}{332.4} \approx 0.0196\ \text{seconds}$$ 5. **Maximum displacement value:** Substitute $t=0.0196$ s into $x(t)$: $$x(0.0196) = 0.0602 e^{-0.5 \times 0.0196} - 0.0602 e^{-332.9 \times 0.0196}$$ $$= 0.0602 e^{-0.0098} - 0.0602 e^{-6.53}$$ $$\approx 0.0602 \times 0.9902 - 0.0602 \times 0.00146 = 0.0596 - 0.000088 = 0.0595\ \text{m}$$ **Final answers:** - a) Maximum displacement $\approx 0.0595$ meters (5.95 cm). - b) Time to reach maximum displacement $\approx 0.0196$ seconds.