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Chapter 2: Problem 37

A single degree of freedom system is represented as a \(2 \mathrm{~kg}\) mass attached to a spring possessing a stiffness of \(4 \mathrm{~N} / \mathrm{m}\). If the coefficients of static and kinetic friction between the block and the surface it moves on are respectively \(\mu_{s}=\) \(0.12\) and \(\mu_{k}=0.10\), determine the drop in amplitude between successive periods during free vibration. What is the frequency of the oscillations?

Short Answer

Expert verified
Answer: The drop in amplitude between successive periods is \(A_n (1 - e^{-0.512})\), and the frequency of oscillations is 0.225 Hz.

Step by step solution

01

Calculate the damping constant

To find the damping constant, we need to use the kinetic friction coefficient, mass of the block, and gravitational acceleration (9.81 m/s²). The damping constant (c) can be calculated using the formula: \(c = \mu_k * m * g\) Where: \(m = 2 \mathrm{~kg}\) (mass of the block) \(\mu_k = 0.10\) (kinetic friction coefficient) \(g = 9.81 \mathrm{~m/s^2}\) (gravitational acceleration) \(c = 0.10 * 2 * 9.81 = 1.962 \mathrm{~N \cdot s / m}\)
02

Calculate the natural frequency

To find the natural frequency (wn) of the system, we will use the formula: \(w_n = \sqrt{\frac{k}{m}}\) Where: \(k = 4 \mathrm{~N / m}\) (spring stiffness) \(m = 2 \mathrm{~kg}\) (mass of the block) \(w_n = \sqrt{\frac{4}{2}} = \sqrt{2} \mathrm{~rad/s}\)
03

Calculate the damping ratio

To find the damping ratio (ζ), we will use the formula: \(\zeta = \frac{c}{2 \sqrt{mk}}\) Where: \(c = 1.962 \mathrm{~N \cdot s / m}\) (damping constant) \(m = 2 \mathrm{~kg}\) (mass of the block) \(k = 4 \mathrm{~N / m}\) (spring stiffness) \(\zeta = \frac{1.962}{2 \sqrt{2 \cdot 4}} = 0.245\)
04

Calculate the logarithmic decrement

To find the logarithmic decrement (δ) in amplitude, we will use the formula: \(\delta = 2 \pi \frac{\zeta}{\sqrt{1-\zeta^2}}\) Where: \(\zeta = 0.245\) (damping ratio) \(\delta = 2 \pi \frac{0.245}{\sqrt{1 - 0.245^2}} = 0.512\)
05

Calculate the drop in amplitude between successive periods

To find the drop in amplitude between successive periods, we will use the formula: \(A_{n+1} = A_n e^{-\delta}\) Where: \(A_n\) is the amplitude at period n \(A_{n+1}\) is the amplitude at period n+1 \(\delta = 0.512\) (logarithmic decrement) \(A_{n+1} = A_n e^{-0.512}\) The drop in amplitude between successive periods is equal to \(A_n - A_{n+1} = A_n (1 - e^{-0.512})\).
06

Calculate the frequency of oscillations

To find the frequency of oscillations, we will use the formula: \(f = \frac{w_n}{2 \pi}\) Where: \(w_n = \sqrt{2} \mathrm{~rad/s}\) (natural frequency) \(f = \frac{\sqrt{2}}{2 \pi} = 0.225 \mathrm{~Hz}\) The frequency of oscillations is 0.225 Hz. In conclusion, the drop in amplitude between successive periods is \(A_n (1 - e^{-0.512})\), and the frequency of oscillations is 0.225 Hz.

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