Let's solve the spring-mass-damper system **step by step**, as requested. --- ### **1. Write the Differential Equation** The general equation for a mass-spring-damper system with an external force is: \[ m \ddot{x} + c \dot{x} + kx = f(t) \] where: - \( m \): mass (in slugs) - \( c \): damping coefficient - \( k \): spring constant - \( f(t) \): external force --- ### **2. Identify Each Parameter** #### **Mass (\( m \))** Given **weight** \( W = 16 \) lb. \[ m = \frac{W}{g} = \frac{16}{32} = 0.5\ \text{slugs} \] --- #### **Spring Constant (\( k \))** The mass stretches the spring \( \Delta x = \frac{8}{3} \) ft under its own weight. Hooke's Law: \( W = k \Delta x \implies k = \frac{W}{\Delta x} \) \[ k = \frac{16}{8/3} = 16 \cdot \frac{3}{8} = 6\ \text{lb/ft} \] --- #### **Damping Coefficient (\( c \))** Damping force is **numerically equal to \( \frac{1}{2} \) the instantaneous velocity**: \( c = \frac{1}{2} \) (in lb·s/ft). --- #### **External Force (\( f(t) \))** \[ f(t) = 10 \cos(3t) \] --- ### **3. Write the ODE with numbers plugged in** \[ 0.5 \ddot{x} + 0.5 \dot{x} + 6x = 10\cos(3t) \] Divide both sides by 0.5: \[ \ddot{x} + \dot{x} + 12x = 20\cos(3t) \] --- ### **4. Initial Conditions** The mass is **released from rest** at a point **3 ft below equilibrium**. Let \( x = 0 \) be equilibrium position (down is positive): \[ x(0) = 3 \] \[ \dot{x}(0) = 0 \] --- ### **5. Solve the Homogeneous Equation** \[ \ddot{x} + \dot{x} + 12x = 0 \] **Characteristic equation:** \[ r^2 + r + 12 = 0 \] \[ r = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 12}}{2} = \frac{-1 \pm \sqrt{1 - 48}}{2} = \frac{-1 \pm \sqrt{-47}}{2} \] \[ r = \frac{-1}{2} \pm \frac{\sqrt{47}}{2}i \] Homogeneous solution: \[ x_h(t) = e^{-t/2} \left[ A \cos\left(\frac{\sqrt{47}}{2}t\right) + B \sin\left(\frac{\sqrt{47}}{2}t\right) \right] \] --- ### **6. Find a Particular Solution (\( x_p \))** Try \( x_p = C \cos(3t) + D \sin(3t) \): Derivatives: - \( \dot{x}_p = -3C \sin(3t) + 3D \cos(3t) \) - \( \ddot{x}_p = -9C \cos(3t) - 9D \sin(3t) \) Plug into the ODE: \[ \ddot{x}_p + \dot{x}_p + 12x_p = 20 \cos(3t) \] \[ (-9C \cos 3t - 9D \sin 3t) + (-3C \sin 3t + 3D \cos 3t) + 12(C \cos 3t + D \sin 3t) = 20 \cos 3t \] Group terms: - **\(\cos(3t)\) coefficients:** - \(-9C + 3D + 12C = (3C + 3D)\) - **\(\sin(3t)\) coefficients:** - \(-9D - 3C + 12D = (3D - 3C)\) So: \[ (3C + 3D)\cos(3t) + (3D - 3C)\sin(3t) = 20\cos(3t) + 0\sin(3t) \] Set equal to the right-hand side: \[ 3C + 3D = 20 \\ 3D - 3C = 0 \] From the second equation: \[ 3D - 3C = 0 \implies D = C \] From the first equation: \[ 3C + 3D = 20 \implies 3C + 3C = 20 \implies 6C = 20 \implies C = \frac{10}{3} \] Then \( D = C = \frac{10}{3} \). So the particular solution is: \[ x_p(t) = \frac{10}{3} \cos(3t) + \frac{10}{3} \sin(3t) \] Or factor out: \[ x_p(t) = \frac{10}{3} [\cos(3t) + \sin(3t)] \] --- ### **7. General Solution** \[ x(t) = x_h(t) + x_p(t) \] \[ x(t) = e^{-t/2} \left[ A \cos\left(\frac{\sqrt{47}}{2}t\right) + B \sin\left(\frac{\sqrt{47}}{2}t\right) \right] + \frac{10}{3} \left[ \cos(3t) + \sin(3t) \right] \] --- ### **8. Apply Initial Conditions** #### **At \( t = 0 \):** \[ x(0) = e^{0} [A \cdot 1 + B \cdot 0] + \frac{10}{3} [1 + 0] = A + \frac{10}{3} \] Given \( x(0) = 3 \): \[ A + \frac{10}{3} = 3 \implies A = 3 - \frac{10}{3} = \frac{9 - 10}{3} = -\frac{1}{3} \] --- #### **For \( \dot{x}(0) \):** Compute derivative: First, recall: \[ x(t) = e^{-t/2} \left[ A \cos\left(\omega t\right) + B \sin\left(\omega t\right) \right] + \frac{10}{3} \cos(3t) + \frac{10}{3} \sin(3t) \] where \( \omega = \frac{\sqrt{47}}{2} \). Derivative: \[ \dot{x}(t) = \frac{d}{dt} \bigg\{ e^{-t/2} [A \cos(\omega t) + B \sin(\omega t)] \bigg\} + \frac{10}{3} [-3 \sin(3t) + 3 \cos(3t)] \] Let's calculate the derivative of the first term using the product rule: \[ \frac{d}{dt}[e^{-t/2} F(t)] = e^{-t/2} F'(t) - \frac{1}{2} e^{-t/2} F(t) \] Let \( F(t) = A \cos(\omega t) + B \sin(\omega t) \): \[ F'(t) = -A \omega \sin(\omega t) + B \omega \cos(\omega t) \] So, at \( t = 0 \): - \( e^{-t/2} \to 1 \) - \( \sin(0) = 0 \) - \( \cos(0) = 1 \) \[ F(0) = A \\ F'(0) = B \omega \] So, \[ \left. \frac{d}{dt} e^{-t/2} F(t) \right|_{t=0} = F'(0) - \frac{1}{2} F(0) = B \omega - \frac{1}{2} A \] Now, the derivatives of the particular solution at \( t = 0 \): \[ \frac{d}{dt} \left[ \frac{10}{3} \cos(3t) \right] = -10 \sin(3t) \] \[ \frac{d}{dt} \left[ \

Let's continue from where we left off to find \( \dot{x}(0) \): ### **8. Apply Initial Conditions (continued)** Recall the derivative: \[ \dot{x}(t) = e^{-t/2} [B \omega - \tfrac{1}{2} A] + \frac{10}{3} \left[ -3 \sin(3t) + 3 \cos(3t) \right] \] At \( t=0 \): \[ \sin(0) = 0,\quad \cos(0) = 1 \] So, \[ \dot{x}(0) = (1)\left[ B \omega - \frac{1}{2} A \right] + \frac{10}{3} \times 3 \times 1 = B \omega - \frac{1}{2} A + 10 \] Given initial velocity \( \dot{x}(0) = 0 \): \[ 0 = B \omega - \frac{1}{2} A + 10 \] Recall \( A = -\frac{1}{3} \), so: \[ 0 = B \omega - \frac{1}{2} \left(-\frac{1}{3}\right) + 10 \] \[ 0 = B \omega + \frac{1}{6} + 10 \] \[ B \omega = - \frac{1}{6} - 10 = - \frac{1}{6} - \frac{60}{6} = - \frac{61}{6} \] Recall: \[ \omega = \frac{\sqrt{47}}{2} \] Thus, \[ B = - \frac{61/6}{\sqrt{47}/2} = - \frac{61/6}{\sqrt{47}/2} = - \frac{61/6} \times \frac{2}{\sqrt{47}} = - \frac{61 \times 2}{6 \sqrt{47}} = - \frac{122}{6 \sqrt{47}} = - \frac{61}{3 \sqrt{47}} \] --- ### **9. Final Equation of Motion** Putting everything together: \[ \boxed{ x(t) = e^{-t/2} \left[ -\frac{1}{3} \cos\left( \frac{\sqrt{47}}{2} t \right) + \left( - \frac{61}{3 \sqrt{47}} \right) \sin\left( \frac{\sqrt{47}}{2} t \right) \right] + \frac{10}{3} \left[ \cos(3t) + \sin(3t) \right] } \] Alternatively, simplifying notation: \[ \boxed{ \begin{aligned} x(t) &= e^{-t/2} \left[ -\frac{1}{3} \cos\left( \frac{\sqrt{47}}{2} t \right) - \frac{61}{3 \sqrt{47}} \sin\left( \frac{\sqrt{47}}{2} t \right) \right] \\ &\quad + \frac{10}{3} \left[ \cos(3t) + \sin(3t) \right] \end{aligned} } \] This is the **equation of motion \( x(t) \)** for the damped, driven harmonic oscillator with initial displacement and velocity as specified. --- ### **Final notes:** - The first term represents the **homogeneous (transient) response** due to damping and initial conditions. - The second term is the **particular (steady-state) response** driven by the external force. - The damping causes the transient to decay exponentially over time. **Answer:** \[ \boxed{ x(t) = e^{-t/2} \left[ -\frac{1}{3} \cos\left( \frac{\sqrt{47}}{2} t \right) - \frac{61}{3 \sqrt{47}} \sin\left( \frac{\sqrt{47}}{2} t \right) \right] + \frac{10}{3} \left[ \cos(3t) + \sin(3t) \right] } \]