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(Underdamped) A body weighing \(100 \mathrm{lb}\) (mass \(m=\) \(3.125\) slugs in fps units) is oscillating attached to a spring and a dashpot. Its first two maximum displacements of \(6.73\) in. and \(1.46\) in. are observed to occur at times \(0.34\) \(s\) and \(1.17 \mathrm{~s}\), respectively. Compute the damping constant (in pound-seconds per foot) and spring constant (in pounds per foot).

Short Answer

Expert verified
Damping constant \( c \approx 31.43 \) lb-s/ft and spring constant \( k \approx 145.96 \) lb/ft.

Step by step solution

01

Identify Given Values

Recognize the known values: Weight = 100 lb, Mass \( m = 3.125 \) slugs, First maximum displacement \( x_1 = 6.73 \) in = 0.561 ft, Time of first maximum \( t_1 = 0.34 \) s, Second maximum displacement \( x_2 = 1.46 \) in = 0.1217 ft, Time of second maximum \( t_2 = 1.17 \) s.
02

Formulate Damped Oscillation Equations

The displacement for underdamped motion is given by the equation: \( x(t) = A e^{-\beta t} \cos(\omega_d t + \phi) \), where \( \beta = \frac{c}{2m} \) is the damping factor and \( \omega_d \) is the damped angular frequency.
03

Determine Damping Ratio

The amplitude ratio between two peaks is given by \( \frac{x_2}{x_1} = e^{-\beta (t_2 - t_1)} \). Substitute \( \beta = \frac{c}{2m} \) and solve for \( c \): \( 0.1217/0.561 = e^{-\frac{c}{2 \times 3.125}(1.17 - 0.34)} \).
04

Solve for the Damping Constant

Taking the natural logarithm of both sides, \( \ln(0.1217/0.561) = -\frac{c}{2 \times 3.125} \times (1.17 - 0.34) \). Solve for \( c \) to find \( c \approx 31.43 \) lb-s/ft.
05

Use Period to Compute Natural Frequency

The time between successive maximum displacements corresponds to the damped period \( T_d = t_2 - t_1 = 1.17 - 0.34 \approx 0.83 \) s. The damped frequency is \( \omega_d = \frac{2\pi}{T_d} \approx 7.57 \) rad/s.
06

Solve for Natural Frequency and Spring Constant

The relationship between natural frequency and damped frequency is \( \omega_d^2 = \omega_0^2 - \beta^2 \). Knowing \( \omega_0 = \sqrt{\frac{k}{m}} \), solve for \( k \) using \( \omega_d \) and \( \beta \). \( k = m(\omega_d^2 + (\frac{c}{2m})^2) \approx 145.96 \) lb/ft.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Damping Constant
In an oscillating system, especially when attached to a spring and dampened by a dashpot, the damping constant is a crucial parameter. It measures how much the damping force opposes the motion and is typically expressed in pound-seconds per foot (lb-s/ft). In a simple damped oscillating system, like the one described, the damping constant helps determine how quickly the oscillations will die out.

The damping constant, represented by the symbol \( c \), appears in the equation \( \beta = \frac{c}{2m} \), where \( \beta \) is the damping factor and \( m \) is the mass. The primary function of \( c \) is to account for non-conservative forces, causing energy dissipation in the motion. A higher damping constant means quicker energy loss, leading to faster reduction in the amplitude of oscillation.

To determine \( c \), we used the amplitude decay observed in a sequence of maximum displacements over time. By solving \( \ln(0.1217/0.561) = -\frac{c}{2 \times 3.125} \times (1.17 - 0.34) \), we derived \( c \approx 31.43 \) lb-s/ft. This value shows how effectively the system resists oscillations, balancing its store of kinetic and potential energy.
Spring Constant
The spring constant, denoted by the symbol \( k \), characterizes the stiffness of a spring in a mechanical system. Its units are pounds per foot (lb/ft), and it describes how much force is needed to stretch or compress the spring by a certain distance.

In the context of the problem, the spring constant is derived indirectly through the relationship with the natural frequency of the system. The equation \( \omega_0 = \sqrt{\frac{k}{m}} \) relates the natural frequency \( \omega_0 \) to the spring constant. By understanding that the natural frequency \( \omega_0 \) is connected to the mass and how stiff the system is, \( k \) can be calculated more effectively, with energy conservation principles playing a crucial role.

Using the formula \( k = m(\omega_d^2 + (\frac{c}{2m})^2) \) where \( \omega_d \) is the damped frequency, we determined that \( k \approx 145.96 \) lb/ft. This value reflects the inherent resistance of the spring to being deformed by external forces, directly affecting oscillatory behavior.
Damped Oscillation Equations
Damped oscillation equations play a pivotal role in understanding how oscillatory systems behave when external forces, such as friction or air resistance, act upon them. The fundamental expression for the displacement of a body in underdamped motion is \( x(t) = A e^{-\beta t} \cos(\omega_d t + \phi) \).

Here, \( A \) is the initial amplitude, \( \beta = \frac{c}{2m} \) is the damping factor, \( \omega_d \) is the damped angular frequency, and \( \phi \) is the phase angle. This equation encapsulates the essence of damped motion, combining exponential decay with a cosine wave to describe how the system's motion decreases over time.

Several important relationships arise in such analyses:
  • Greater damping (higher \( c \)) results in faster decay of oscillations.
  • \( \omega_d = \frac{2\pi}{T_d} \), where \( T_d \) is the damped period.
  • Natural frequency \( \omega_0 \) impacts how quickly the system would oscillate without damping.
Understanding these equations and relationships allows students to draw connections between theoretical models and real-world behavior, demystifying how systems transition from rest to motion and back, in the presence of damping forces.

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