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Chapter 17: Problem 46

A damped harmonic oscillator involves a block (m = \(1.91 \mathrm{~kg}\) ), a spring \((k=12.6 \mathrm{~N} / \mathrm{m})\), and a damping force \(F=\) \(-b v_{x} .\) Initially, it oscillates with an amplitude of \(26.2 \mathrm{~cm}\); because of the damping, the amplitude falls to three-fourths of this initial value after four complete cycles. (a) What is the value of \(b ?(b)\) How much energy has been "lost" during these four cycles?

Short Answer

Expert verified
The damping factor \(b\) is approximately \(0.474 Ns/m\) and the energy lost during the four cycles of oscillation is \(0.189 J\)

Step by step solution

01

Calculate natural frequency

Firstly, one needs to get the natural frequency \(\omega_{0}\) of the system using the equation \(\omega_{0} = sqrt{k / m}\). Here, \(k\) is the spring constant, \(12.6 N/m\), and \(m\) is the mass, \(1.91 kg\).
02

Get the angular logarithmic decrement

Next, we can compute the logarithmic decrement which represents the rate of decay of motion in underdamped systems. It can be calculated by \(Δ = ln(A_n / A_{n+1})\), where \(A_n\) and \(A_{n+1}\) are successive amplitudes. Here, \(A_n\) is the initial amplitude which is \(0.262m\), and the amplitude after one cycle \(A_{n+1}\) is \(0.75 * A_n = 0.75 * 0.262 = 0.1965m\). After four cycles, \(4 Δ = ln(A_0 / A_{4}) = ln(0.262 / 0.1965) = 0.2924\). Solve for \(Δ\) to find \(Δ = 0.2924 / 4 = 0.0731\).
03

Calculate the damping factor \(b\)

The damping constant \(b\) can be found using the formula \(\b = 2 m ζ ω_{0}\). However, solving for the damping coefficient \(b\) requires us to find \(ζ\). \(ζ\) can be obtained from the equation \(ζ = 1 / (2Q)\), where \(Q\) is the quality factor of the oscillator which can be computed by \(Q = π / Δ = π / 0.0731 = 43.082\). Therefore, \(ζ = 1 / (2(43.082)) = 0.0116\). The damping factor \(b\) can then be calculated as \(b = 2 m ζ ω_{0} = 2 * 1.91 kg * 0.0116 * sqrt(12.6 N/m / 1.91 kg) = 0.474 Ns/m\).
04

Calculate lost energy

The initial energy of system is \(1/2 * k * A_{0}^2 = 1/2 * 12.6 N/m * (0.262 m)^2 = 0.432 J\). After four cycles the energy is \(1/2 * k * A_{4}^2 = 1/2 * 12.6 N/m * (0.1965 m)^2 = 0.243 J\). Therefore, the energy loss in the four cycles is the difference between the initial and final energies, \(0.432 J - 0.243 J = 0.189 J\)

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Most popular questions from this chapter

A particle of mass \(m\) moves in a fixed plane along the trajectory \(\overrightarrow{\mathbf{r}}=\hat{\mathbf{i}} A \cos \omega t+\hat{\mathbf{j}} A \cos 3 \omega t .\) (a) Sketch the trajectory of the particle. (b) Find the force acting on the particle. Also find (c) its potential energy and \((d\) ) its total energy as functions of time. \((e)\) Is the motion periodic? If so, find the period.

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