**Problem Statement:**A mass weighing 56 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 207 lbs when the mass has a velocity of 6 ft/sec.Suppose the object is displaced an additional 3 inches and released.Find an equation for the object's displacement, $ u(t) $, in feet after $ t $ seconds.\[ u(t) = \quad \text{\_\_\_\_\_\_\_\_\_\_\_\_} \]**Explanation:**This problem involves an object connected to a spring within a medium that provides viscous resistance. This forms a classic physics problem often modeled by differential equations.- **Given Data:** - Weight of the mass: 56 lbs - Stretch caused by mass: 3 inches - Viscous resistance: 207 lbs at 6 ft/sec- **Assumptions:** - The spring follows Hooke's Law. - The medium provides linear viscous resistance.**Steps to Solve:**1. **Convert Units:** - Convert the 3 inches stretch to feet (3 inches = 0.25 feet).2. **Determine Parameters:** - Compute the spring constant $ k $ using Hooke’s Law: $ F = kx $ $ 56 = k \times 0.25 $ $ k = 224 $ lbs/ft - Compute the damping coefficient $ c $ using the given viscous resistance: $ c \times 6 = 207 $ $ c = 34.5 $ lbs-sec/ft3. **Determine the Displacement Equation:** - The differential equation modeling the system is: $ m \frac{d^2 u}{dt^2} + c \frac{du}{dt} + k u = 0 $ Here, $ m = \frac{56}{32} = 1.75 $ slugs (using $ g = 32 $ ft/sec$^2$) - Plug in the values: $ 1.75 \frac{d^2 u}{dt^2} + 34.5 \frac{du}{dt} + 224 u = 0 $ - Solve the differential equation to find $ u(t) $.**Note

A mass weighting 56 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 207 lbs when the mass has a velocity of 6 ft/sec. Suppose the object is displaced an additional 3 inches and released. Find an equation for the object's displacement, u(t), in feet after t seconds.

A mass weighting 56 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 207 lbs when the mass has a velocity of 6 ft/sec. Suppose the object is displaced an additional 3 inches and released. Find an equation for the object's displacement, u(t), in feet after t seconds.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

**Problem Statement:**

A mass weighing 56 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 207 lbs when the mass has a velocity of 6 ft/sec.

Suppose the object is displaced an additional 3 inches and released.

Find an equation for the object's displacement, $ u(t) $, in feet after $ t $ seconds.

\[ u(t) = \quad \text{\_\_\_\_\_\_\_\_\_\_\_\_} \]

**Explanation:**

This problem involves an object connected to a spring within a medium that provides viscous resistance. This forms a classic physics problem often modeled by differential equations.

- **Given Data:**
- Weight of the mass: 56 lbs
- Stretch caused by mass: 3 inches
- Viscous resistance: 207 lbs at 6 ft/sec

- **Assumptions:**
- The spring follows Hooke's Law.
- The medium provides linear viscous resistance.

**Steps to Solve:**

1. **Convert Units:**
- Convert the 3 inches stretch to feet (3 inches = 0.25 feet).

2. **Determine Parameters:**
- Compute the spring constant $ k $ using Hooke’s Law: $ F = kx $
$ 56 = k \times 0.25 $
$ k = 224 $ lbs/ft

- Compute the damping coefficient $ c $ using the given viscous resistance:
$ c \times 6 = 207 $
$ c = 34.5 $ lbs-sec/ft

3. **Determine the Displacement Equation:**

- The differential equation modeling the system is:
$ m \frac{d^2 u}{dt^2} + c \frac{du}{dt} + k u = 0 $
Here, $ m = \frac{56}{32} = 1.75 $ slugs (using $ g = 32 $ ft/sec$^2$)

- Plug in the values:
$ 1.75 \frac{d^2 u}{dt^2} + 34.5 \frac{du}{dt} + 224 u = 0 $

- Solve the differential equation to find $ u(t) $.

**Note

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