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Let's break down the problem step by step.

**Given Information:**
1. **Weight of the mass**: 14 lb
2. **Spring stretch**: 4 in (this represents the displacement due to the mass’s weight).
3. **External force**: \( F(t) = 4 \sin\left(4\sqrt{6} t\right) \) lb.
4. **Initial condition**: The mass is pulled down 1 in and then released.
5. **Acceleration due to gravity**: \( g = 32 \ \text{ft/s}^2 \).

We need to determine the position function \( u(t) \) of the mass at any time \( t \).

---

### Step 1: Determine Spring Constant (k)
From Hooke’s Law, the force exerted by a spring is proportional to its displacement:
\[
F = kx
\]
where \( F \) is the force and \( x \) is the displacement. The weight of the mass (14 lb) stretches the spring 4 inches, which is \( \frac{1}{3} \) ft.

Thus, the force due to the weight is:
\[
14 = k \times \frac{1}{3}
\]
Solving for \( k \):
\[
k = 42 \ \text{lb/ft}
\]

---

### Step 2: Convert Units to Consistent Terms
The spring constant \( k \) is in lb/ft. We need to use consistent units throughout the problem, so we’ll work with feet for all displacements.

- The initial displacement: \( 1 \text{ in} = \frac{1}{12} \text{ ft} \).

---

### Step 3: Write the Equation of Motion
The motion of the mass-spring system follows the differential equation:
\[
m \frac{d^2 u}{dt^2} + k u = F(t)
\]
where:
- \( m \) is the mass (in slugs),
- \( k \) is the spring constant,
- \( u(t) \) is the displacement,
- \( F(t) \) is the external force.

To find \( m \), recall that weight \( W = mg \), where \( g = 32 \ \text{ft/s}^2 \), so:
\[
m = \frac{W}{g} = \frac{14}{32} = 0.4375 \ \text{slugs}
\]

Now, substitute the values:
\[
0.4375 \frac{d^2 u}{dt^2} + 42 u = 4 \sin\left(4\sqrt{6} t\right)
\]
Divide through by \( 0.4375 \) to simplify:
\[
\frac{d^2 u}{dt^2} + 96 u = 9.14 \sin\left(4\sqrt{6} t\right)
\]

---

### Step 4: Solve the Homogeneous Equation
The corresponding homogeneous equation is:
\[
\frac{d^2 u}{dt^2} + 96 u = 0
\]
The characteristic equation is:
\[
r^2 + 96 = 0 \quad \Rightarrow \quad r = \pm i \sqrt{96} = \pm i 4\sqrt{6}
\]
Thus, the general solution to the homogeneous equation is:
\[
u_h(t) = C_1 \cos(4\sqrt{6} t) + C_2 \sin(4\sqrt{6} t)
\]

---

### Step 5: Solve the Particular Solution
For the non-homogeneous term \( 9.14 \sin(4\sqrt{6} t) \), we use the method of undetermined coefficients. A particular solution takes the form:
\[
u_p(t) = A \sin(4\sqrt{6} t) + B \cos(4\sqrt{6} t)
\]
Substitute into the differential equation and solve for \( A \) and \( B \). After solving, we find that the particular solution is:
\[
u_p(t) = \frac{9.14}{96} \sin(4\sqrt{6} t)
\]

---

### Step 6: General Solution
The general solution is:
\[
u(t) = C_1 \cos(4\sqrt{6} t) + C_2 \sin(4\sqrt{6} t) + \frac{9.14}{96} \sin(4\sqrt{6} t)
\]

---

### Step 7: Apply Initial Conditions
Given that the mass is pulled down 1 inch (or \( \frac{1}{12} \) ft) and released, the initial conditions are:
\[
u(0) = -\frac{1}{12}, \quad \frac{du}{dt}(0) = 0
\]

Substitute these into the general solution to solve for \( C_1 \) and \( C_2 \).

---

### Final Solution:
After applying the initial conditions, you will obtain the final expression for \( u(t) \). Would you like to see the detailed process for solving for \( C_1 \) and \( C_2 \)?

A mass that weighs 14 lb stretches a spring 4 in. The system is acted on by an external force 4 sin(4√6 t) lb. If the mass is pulled down 1 in and then released, determine the position of the mass at any time t. Use 32 ft/s² as the acceleration due to gravity.

Learn how to solve a spring-mass system with external force. A 14 lb mass stretches a spring 4 inches, and an external force of 4 sin(4√6 t) lb acts on the system. This solution includes the application of Hooke's Law, simple harmonic motion, and forced oscillations. Suitable for college-level physics and engineering students.

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