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Let's go through the problem step-by-step to solve for both parts (a) and (b).

### Problem Breakdown

Given:
- A mass of **15 lb** stretches a spring **4 in**.
- The mass is displaced **2 in** downward from the equilibrium position and released.
- Initial upward velocity of **6 ft/s**.
- Acceleration due to gravity \( g = 32 \, \text{ft/s}^2 \).
- No damping is applied.

The goal:
1. **(a)** Find the position function \( u(t) \) in feet.
2. **(b)** Determine the period \( T \), amplitude \( R \), and phase \( \delta \) of the motion.

### Solution

#### Step 1: Converting Units and Finding Constants
1. **Weight** \( W = 15 \, \text{lb} \).
   - Converting to mass \( m \): 
     \[
     m = \frac{W}{g} = \frac{15}{32} \, \text{slugs}
     \]

2. **Spring Constant** \( k \):
   - The spring stretches **4 inches** (or **1/3 feet**) under the mass.
   - Using Hooke's Law \( W = k \Delta \), where \( \Delta = 1/3 \) feet,
     \[
     k = \frac{W}{\Delta} = \frac{15}{1/3} = 45 \, \text{lb/ft}
     \]

#### Step 2: Set Up the Differential Equation for \( u(t) \)
The motion of the mass can be described by the differential equation for simple harmonic motion:
\[
m u'' + k u = 0
\]
Substituting \( m = \frac{15}{32} \) and \( k = 45 \):
\[
\frac{15}{32} u'' + 45 u = 0
\]
Simplifying:
\[
u'' + \frac{45}{\frac{15}{32}} u = 0 \quad \Rightarrow \quad u'' + 96 u = 0
\]
This gives us the characteristic equation:
\[
r^2 + 96 = 0
\]
Solving for \( r \):
\[
r = \pm i \sqrt{96} = \pm 4 \sqrt{6} \, i
\]
Thus, the general solution for \( u(t) \) is:
\[
u(t) = A \cos(4\sqrt{6} \, t) + B \sin(4\sqrt{6} \, t)
\]

#### Step 3: Applying Initial Conditions
1. **Initial Displacement**: The mass is displaced **2 in** (or **1/6 ft**) downward from the equilibrium, so:
   \[
   u(0) = -\frac{1}{6}
   \]
   This implies \( A = -\frac{1}{6} \).

2. **Initial Velocity**: Given an initial velocity of **6 ft/s** upward, we have:
   \[
   u'(t) = -A \cdot 4\sqrt{6} \sin(4\sqrt{6} \, t) + B \cdot 4\sqrt{6} \cos(4\sqrt{6} \, t)
   \]
   At \( t = 0 \):
   \[
   u'(0) = 4\sqrt{6} B = 6
   \]
   Solving for \( B \):
   \[
   B = \frac{6}{4\sqrt{6}} = \frac{\sqrt{6}}{4}
   \]

So the position function is:
\[
u(t) = -\frac{1}{6} \cos(4\sqrt{6} \, t) + \frac{\sqrt{6}}{4} \sin(4\sqrt{6} \, t)
\]

### Part (b): Finding \( T \), \( R \),

A mass weighing 15 lb stretches a spring 4 in. The mass is pulled down an additional 2 in and then set in motion with an initial upward velocity of 6 ft/s. No damping is applied.

Determine:
a. The position u(t) of the mass at any time t. Use 32 ft/s^2 as the acceleration due to gravity.
b. The period, amplitude, and phase of the motion, where -π < δ < π.

Learn how to calculate the position function, period, amplitude, and phase of a mass-spring system set in motion. This problem involves simple harmonic motion, differential equations, and initial value conditions, suitable for college-level physics or differential equations students.

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