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In the case of light damping \(\left(c

Short Answer

Expert verified
The logarithmic decrement is \(\ln \frac{x_n}{x_{n+1}}=\frac{2 \pi\left(c / c_{c}\right)}{\sqrt{1-\left(c / c_{c}\right)^{2}}}\)."

Step by step solution

01

Identify Definitions and Variables

Given the context of lightly damped oscillations, we define relevant variables: let \( c\) be the damping coefficient, and \( c_c\) be the critical damping coefficient. According to the problem, we assume the successive maximum displacements \( x_n \) and \( x_{n+1} \) to be consecutive peaks in the damped oscillation.
02

Write Equation for Damping Oscillation

The displacement of a damped oscillator can be expressed as \[ x(t) = A e^{-\frac{c}{2m}t} \cos(\omega_d t + \phi), \] where \( A \) is the amplitude, and \( \omega_d = \sqrt{\frac{k}{m} - \left(\frac{c}{2m}\right)^2} \). This simplifies in our problem to represent the maximum or peak displacements by only the exponential factor affecting them, assuming phase \( \phi \) results in a peak.
03

Define the Ratio of Successive Maxima

The ratio of successive maximum displacements can be determined purely by the exponentially decaying factor since peaks occur at integer multiples of the natural period. Hence, \[ \frac{x_n}{x_{n+1}} = e^{\frac{c}{2m} T_d}, \] where \( T_d \) is the period of the damped oscillations given by \( T_d = \frac{2\pi}{\omega_d} \).
04

Calculate the Logarithm of the Ratio

Substituting \( \omega_d = \sqrt{\frac{k}{m} - \left(\frac{c}{2m}\right)^2} \) into the expression for \( T_d \) gives \[ \frac{x_n}{x_{n+1}} = e^{\frac{c}{2m} \cdot \frac{2\pi}{\sqrt{\frac{k}{m} - \left(\frac{c}{2m}\right)^2}}}. \] Taking the natural logarithm of both sides results in the logarithmic decrement expression.
05

Express in Terms of Known Quantities

Finally, determine the logarithmic decrement: \[ \ln \frac{x_n}{x_{n+1}} = \frac{2\pi\left(\frac{c}{2m}\right)}{\sqrt{\frac{k}{m} - \left(\frac{c}{2m}\right)^2}} = \frac{2\pi \left(\frac{c}{c_c}\right)}{\sqrt{1 - \left(\frac{c}{c_c}\right)^2}}, \] where \( \frac{c}{c_c} = \frac{c/2m}{\sqrt{k/m}} \) for a critically damped system with \( c_c = 2m\sqrt{k/m} \).

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Key Concepts

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

Logarithmic Decrement
Logarithmic decrement is a metric used to quantify the rate at which oscillations of a damped system decrease over time. When dealing with lightly damped systems (i.e., when the damping coefficient, \( c \), is less than the critical damping coefficient, \( c_c \)), the amplitude of oscillation reduces gradually.
Logarithmic decrement, \( \ln \frac{x_n}{x_{n+1}} \), represents the natural logarithm of the ratio of successive maximum displacements. It's essential because it allows us to assess the energy loss in each cycle of oscillation.
Understanding this ratio becomes particularly important when you are analyzing systems of oscillations subjected to light damping, as it helps predict the behavior of the system over time. Logarithmic decrement is expressed mathematically as:\[ \ln \frac{x_n}{x_{n+1}} = \frac{2 \pi \left( c / c_c \right)}{\sqrt{1 - \left( c / c_c \right)^2}} \]This equation involves the damping ratio, \( c / c_c \), and indicates how energy dissipates with each oscillation cycle.
Maximum Displacement Ratio
In oscillatory systems, especially those experiencing damping, the maximum displacement ratio \( \frac{x_n}{x_{n+1}} \) is an important aspect to consider. This ratio defines how the peak heights of consecutive oscillations compare. It reflects the inevitable energy loss in such systems.
Even in lightly damped systems, despite relatively small damping forces, each oscillation cycle experiences a reduction in peak amplitude. When maximum displacements decay successively, they do so in a predictable pattern.
The ratio of successive maximum displacements is given by:\[ \frac{x_n}{x_{n+1}} = e^{\frac{c}{2m} T_d} \]where \( T_d \) is the period of damped oscillations. This formula gives insight into the energy lost due to damping forces during each cycle, effectively illustrating a key characteristic of damped harmonic motion.
Light Damping
Light damping refers to a scenario in oscillatory systems where the damping forces are small compared to the restoring forces. It is characterized by a damping coefficient \( c \) that is less than the critical damping coefficient \( c_c \). This ensures that the system's oscillations gradually diminish without ceasing all at once.
In systems with light damping, the oscillations persist within a narrow range over multiple cycles. This type of damping is typical in many real-world mechanical and structural systems where some degree of damping is unavoidable but desired to be minimal for purposes of sustainability and efficiency.
The mathematical expression for lightly damped systems reveals that their displacement over time is governed by an exponentially decaying amplitude:\[ x(t) = A e^{-\frac{c}{2m}t} \cos(\omega_d t + \phi) \]Using this expression, we can extract crucial information about how gently energy is drained from the system, allowing for sustained yet gradually diminishing oscillations.

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Most popular questions from this chapter

Show that for a small value of the damping factor \(c / c_{c}\), the maximum amplitude of a forced vibration occurs when \(\omega_{f} \approx \omega_{n}\) and that the corresponding value of the magnification factor is \(\frac{1}{2}\left(c / c_{c}\right) .\)

A particle moves in simple harmonic motion. Knowing that the maximum velocity is \(200 \mathrm{mm} / \mathrm{s}\) and the maximum acceleration is \(4 \mathrm{m} / \mathrm{s}^{2},\) determine the amplitude and frequency of the motion.

From mechanics of materials, it is known that for a simply supported beam of uniform cross section, a static load \(P\) applied at the center will cause a deflection of \(\delta_{A}=P L^{3} / 48 E I,\) where \(L\) is the length of the beam, \(E\) is the modulus of elasticity, and \(I\) is the moment of inertia of the cross-sectional area of the beam. Knowing that \(L=15 \mathrm{ft}, E=30 \times 10^{6}\) psi, and \(I=2 \times 10^{-3} \mathrm{ft}^{4}\), determine \((a)\) the equivalent spring constant of the beam, \((b)\) the frequency of vibration of a \(1500-\mathrm{lb}\) block and to the center of the beam. Neglect the mass of the beam and assume that the load remains in contact with the beam.

A 2 -kg block is supported by a spring with a constant of \(k=128 \mathrm{N} / \mathrm{m}\) and a dashpot with a coefficient of viscous damping of \(c=0.6 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}\). The block is in equilibrium when it is struck from below by a hammer that imparts to the block an upward velocity of \(0.4 \mathrm{m} / \mathrm{m} / \mathrm{s}\). Determine \((a)\) the logarithmic decrement, (b) the maximum upward displacement of the block from equilibrium after two cycles.

The 8 -kg uniform bar \(A B\) is hinged at \(C\) and is attached at \(A\) to a spring of constant \(k=500 \mathrm{N} / \mathrm{m}\). If end \(A\) is given a small displacement and released, determine \((a)\) the frequency of small oscillations, (b) the smallest value of the spring constant \(k\) for which oscillations will occur.

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