Let's solve the problem in 3 steps, **explaining each step**, and finally provide the equation of motion \( x(t) \) for the mass-spring-damper system with external force. --- ## **Step 1: Find the parameters (mass, damping, spring constant)** **Given:** - Mass \( m = \frac{1}{6} \) lb - Weight stretches spring \( s = \frac{8}{3} \) ft - Damping force = \( \frac{1}{3} \) of velocity - External force: \( F(t) = 20 \cos(3t) \) - Gravity \( g = 32 \) ft/s² **Convert mass from pounds to slugs:** \[ \text{mass in slugs} = \frac{\text{weight in pounds}}{g} = \frac{1/6}{32} = \frac{1}{192} \text{ slugs} \] **Spring Constant \( k \):** \[ \text{Weight} = k \cdot \text{stretch} \implies \frac{1}{6} = k \cdot \frac{8}{3} \implies k = \frac{1}{6} \cdot \frac{3}{8} = \frac{1}{16} \] **Damping coefficient \( c \):** \[ c = \frac{1}{3} \] **Summary of parameters:** - \( m = \frac{1}{192} \) - \( c = \frac{1}{3} \) - \( k = \frac{1}{16} \) **Explanation:** We converted all units, found the spring constant using Hooke's Law, and identified the damping coefficient. --- ## **Step 2: Write the Differential Equation** The standard forced damped oscillator equation is: \[ m x'' + c x' + k x = F(t) \] Plugging in values: \[ \frac{1}{192} x'' + \frac{1}{3} x' + \frac{1}{16} x = 20\cos(3t) \] Multiply both sides by 192 to clear denominators: \[ x'' + 64 x' + 12 x = 3840\cos(3t) \] **Explanation:** We substituted the parameters into the standard form and cleared denominators for a simpler equation. --- ## **Step 3: Initial Conditions and Find \( x(t) \)** - **Initial position:** released from 6 feet below equilibrium: \( x(0) = -6 \) - **Initial velocity:** at rest, \( x'(0) = 0 \) **Final equation with initial conditions:** \[ x'' + 64 x' + 12 x = 3840\cos(3t), \quad x(0) = -6, \quad x'(0) = 0 \] **Explanation:** We set the initial conditions as given and wrote the full initial value problem for \( x(t) \). --- ## **Final Answer** ### **The equation of motion is:** \[ \boxed{ x'' + 64 x' + 12 x = 3840\cos(3t), \quad x(0) = -6, \quad x'(0) = 0 } \] This is the ODE for the given mass-spring-damper system with external forcing. To find the explicit form of \( x(t) \), you would solve this ODE, typically using the method of undetermined coefficients or Laplace transforms. If you'd like the explicit solution, let me know!

Let's clarify and correct the solution process based on the updated information that the mass is 16 lb. We'll proceed step-by-step, explaining each part and then providing the final equation of motion. --- ### **Step 1: Determine system parameters (mass, spring constant, damping coefficient)** **Given:** - Mass \( m = 16 \) lb - Spring stretches \( s = \frac{8}{3} \) ft - Damping force is \( \frac{1}{3} \) of instantaneous velocity - External force \( F(t) = 20 \cos(3t) \) - Gravity \( g = 32\, \text{ft/s}^2 \) **Convert mass to slugs:** Mass in slugs: \[ m = \frac{\text{weight}}{g} = \frac{16}{32} = 0.5 \text{ slugs} \] **Calculate spring constant \( k \):** Hooke's Law relates weight and spring stretch: \[ \text{Weight} = k \times \text{stretch} \] \[ 16 = k \times \frac{8}{3} \] \[ k = \frac{16 \times 3}{8} = 6 \] **Damping coefficient \( c \):** Given damping force \( = \frac{1}{3} \times v \), so: \[ c = \frac{1}{3} \] **Summary:** | Parameter | Value | |------------|---------| | \( m \) | 0.5 slugs | | \( k \) | 6 lb/ft | | \( c \) | \( \frac{1}{3} \) lb·s/ft | --- ### **Step 2: Write the differential equation** The standard form: \[ m x'' + c x' + k x = F(t) \] Plugging in the values: \[ 0.5 x'' + \frac{1}{3} x' + 6 x = 20 \cos(3t) \] Multiply through by 6 to clear fractions: \[ 6 \times 0.5 x'' + 6 \times \frac{1}{3} x' + 6 \times 6 x = 6 \times 20 \cos(3t) \] \[ 3 x'' + 2 x' + 36 x = 120 \cos(3t) \] **Final ODE:** \[ \boxed{ 3 x'' + 2 x' + 36 x = 120 \cos(3t) } \] --- ### **Step 3: Initial conditions and form of solution** - **Initial position:** \( x(0) = -6 \) ft (since 6 ft below equilibrium) - **Initial velocity:** \( x'(0) = 0 \) Thus, the initial value problem is: \[ \boxed{ \begin{cases} 3 x'' + 2 x' + 36 x = 120 \cos(3t) \\ x(0) = -6 \\ x'(0) = 0 \end{cases} } \] --- ### **Final answer:** **The governing differential equation with initial conditions is:** \[ \boxed{ \quad 3 x'' + 2 x' + 36 x = 120 \cos(3t), \quad x(0) = -6, \quad x'(0) = 0 } \] This provides the complete setup for solving the problem. To find the explicit solution \( x(t) \), you'd proceed with solving this nonhomogeneous ODE, which typically involves finding the homogeneous solution plus a particular solution.