Let's solve the spring-mass-damper-forced system step by step. Given: - Mass \( m = \frac{1}{6} \) lb - Stretches spring \( s = \frac{8}{3} \) ft - Released from 6 ft below equilibrium - Damping force = \( \frac{1}{3} \) × velocity - External force: \( F(t) = 20 \cos(3t) \) - \( g = 32 \) ft/s\(^2\) **Step 1: Find mass \( m \) in slugs and spring constant \( k \)** - Weight (lb) = mass (slugs) × \( g \) - So, \( m = \frac{1}{6} \) lb ⇒ mass in slugs \( = \frac{1}{6 \times 32} = \frac{1}{192} \) slugs - Spring stretches \( \frac{8}{3} \) ft under \( \frac{1}{6} \) lb: \[ k = \frac{\text{force}}{\text{displacement}} = \frac{1/6}{8/3} = \frac{1}{6} \cdot \frac{3}{8} = \frac{1}{16} \text{ lb/ft} \] **Explanation**: We converted the mass to slugs (since F = ma uses slugs), and found the spring constant using Hooke's law. --- **Step 2: Write the differential equation for motion** The general equation is: \[ m x'' + c x' + k x = F(t) \] where: - \( m = \frac{1}{192} \) (slugs) - Damping \( c = \frac{1}{3} \) (lb·s/ft) - \( k = \frac{1}{16} \) - \( F(t) = 20 \cos(3t) \) Plug in the values: \[ \frac{1}{192} x'' + \frac{1}{3} x' + \frac{1}{16} x = 20 \cos(3t) \] Multiply both sides by 192 to clear fractions: \[ x'' + 64 x' + 12 x = 3840 \cos(3t) \] **Explanation**: We set up the second-order ODE using Newton's second law and substituted all the values, clearing the fractions for easier solving. --- **Step 3: Initial conditions and final equation** Initial position: 6 ft below equilibrium ⇒ \( x(0) = -6 \) Initially at rest ⇒ \( x'(0) = 0 \) **Final answer**: \[ \boxed{ x'' + 64 x' + 12 x = 3840 \cos(3t), \quad x(0) = -6, \quad x'(0) = 0 } \] **Explanation**: We have constructed the full initial value problem (ODE + initial conditions) describing the motion of the mass-spring-damper system under external forcing. Solving this ODE gives the required motion \( x(t) \).