Let's solve this spring-mass problem step by step:

Given Information:

1. Mass $m$:

   • Weight $W = 3 \ \text{lb}$.
   • Convert weight to mass using $m = \frac{W}{g}$, where $g = 32.2 \ \text{ft/s}^2$.
   • $m = \frac{3}{32.2} \ \text{slugs}$.

2. Spring constant $k$:

   • The spring is stretched 3 inches (or $\frac{1}{4} \ \text{ft}$) by a 3 lb weight.
   • Using Hooke's Law $W = kx$, we get $k = \frac{3}{\frac{1}{4}} = 12 \ \text{lb/ft}$.

3. Initial Conditions:

   • Initial displacement: $u(0) = -\frac{1}{3} \ \text{ft}$ (since the mass is pushed upward 4 inches from equilibrium).
   • Initial velocity: $u'(0) = -2 \ \text{ft/s}$ (downward velocity).

Step 1: Find the Equation of Motion

For a spring-mass system without damping or external forces, the equation of motion is:

$m u'' + k u = 0$

Substitute $m = \frac{3}{32.2} \ \text{slugs}$ and $k = 12 \ \text{lb/ft}$:

$\frac{3}{32.2} u'' + 12 u = 0 \implies u'' + \frac{12 \cdot 32.2}{3} u = 0 \implies u'' + 128.8 u = 0$

Step 2: Determine the Angular Frequency $\omega_0$

The equation $u'' + \omega_0^2 u = 0$ indicates that:

$\omega_0^2 = 128.8 \implies \omega_0 = \sqrt{128.8} \approx 11.35 \ \text{rad/s}$

Step 3: Find the General Solution

The general solution for the differential equation is:

$u(t) = A \cos(\omega_0 t) + B \sin(\omega_0 t)$

Step 4: Apply Initial Conditions

1. Initial Displacement $u(0) = -\frac{1}{3}$:

   $-\frac{1}{3} = A \cos(0) + B \sin(0) \implies A = -\frac{1}{3}$

2. Initial Velocity $u'(0) = -2$:

   $u'(t) = -A \omega_0 \sin(\omega_0 t) + B \omega_0 \cos(\omega_0 t)$

   At $t = 0$:

   $-2 = B \omega_0 \implies B = \frac{-2}{\omega_0} = \frac{-2}{11.35} \approx -0.176 \ \text{ft}$

   So the solution becomes:

   $u(t) = -\frac{1}{3} \cos(11.35 t) - 0.176 \sin(11.35 t)$

Step 5: Determine Amplitude $R$ and Phase $\delta$

The solution can be rewritten in the form:

$u(t) = R \cos(\omega_0 t - \delta)$

where $R$ is the amplitude and $\delta$ is the phase shift. To find $R$ and $\delta$:

$R = \sqrt{A^2 + B^2} = \sqrt{\left(-\frac{1}{3}\right)^2 + (-0.176)^2} \approx \sqrt{0.1111 + 0.031} \approx 0.384 \ \text{ft}$

$\tan(\delta) = \frac{B}{A} = \frac{-0.176}{-\frac{1}{3}} = \frac{0.176}{0.333} \approx 0.528$

$\delta \approx \arctan(0.528) \approx 0.489 \ \text{radians}$

Step 6: Determine the Period $T$

The period $T$ of the oscillation is given by:

$T = \frac{2\pi}{\omega_0} \approx \frac{2\pi}{11.35} \approx 0.554 \ \text{s}$

Final Solution:

• Position function: $u(t) = 0.384 \cos(11.35 t - 0.489) \ \text{ft}$
• Angular frequency $\omega_0$: $11.35 \ \text{rad/s}$
• Period $T$: $0.554 \ \text{s}$
• Amplitude $R$: $0.384 \ \text{ft}$
• Phase $\delta$: $0.489 \ \text{radians}$

Would you like to see a more detailed explanation of any step, or have any other questions?

Related Questions:

1. How would the solution change if there was damping in the system?
2. What happens to the period if the spring constant $k$ is doubled?
3. How does the amplitude of the motion relate to the initial conditions?
4. What would the motion look like if an external force was applied?
5. How can energy be analyzed in this spring-mass system?

Tip:

In undamped spring-mass systems, the period $T$ only depends on the mass $m$ and spring constant $k$, not on initial displacement or velocity.