/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 162 A damped system has the followin... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 162

A damped system has the following parameters: \(m=2 \mathrm{~kg}, c=3 \mathrm{~N}-\mathrm{s} / \mathrm{m},\) and \(k=40 \mathrm{~N} / \mathrm{m} .\) Determine the natural frequency, damping ratio, and the type of response of the system in free vibration. Find the amount of damping to be added or subtracted to make the system critically damped.

Short Answer

Expert verified
The natural frequency of the system is \(2\sqrt{5}\, \text{rad/s}\) and the damping ratio is \(\frac{3\sqrt{5}}{20}\) which indicates an underdamped system. To make the system critically damped, the damping coefficient needs to be increased by \((4\sqrt{5}-3)\, \text{N路s/m}\).

Step by step solution

01

Determine the natural frequency of the undamped system

To calculate the natural frequency, we will use the formula: \[蠅_n = \sqrt{\frac{k}{m}}\] where: - \(蠅_n\) is the natural frequency - \(k\) is the stiffness - \(m\) is the mass Given: - \(m = 2 \,\text{kg}\) - \(k = 40 \,\text{N/m}\) Using these values in the formula, we get: \[ 蠅_n = \sqrt{\frac{40\,\text{N/m}}{2\,\text{kg}}} \]
02

Calculate the natural frequency

Calculating our value for the natural frequency: \[蠅_n = \sqrt{20}\] \[蠅_n = \sqrt{4 \times 5}\] \[蠅_n = 2 \sqrt{5}\, \text{rad/s}\]
03

Determine the damping ratio

To calculate the damping ratio, we will use the formula: \[ 尉 = \frac{c}{2 \sqrt{m k}} \] where: - \(尉\) is the damping ratio - \(c\) is the damping coefficient - \(m\) is the mass - \(k\) is the stiffness Given: - \(c = 3\, \text{N路s/m}\) - \(m = 2\, \text{kg}\) - \(k = 40\, \text{N/m}\) Using these values in the formula, we get: \[ 尉 = \frac{3\, \text{N路s/m}}{2 \sqrt{(2\, \text{kg})(40\, \text{N/m})}} \]
04

Calculate the damping ratio

Calculating our value for the damping ratio: \[ 尉 = \frac{3}{2 (2 \sqrt{5})} \] \[ 尉 = \frac{3}{4 \sqrt{5}} \] \[ 尉 = \frac{3 \sqrt{5}}{20} \]
05

Determine the type of response

Based on the damping ratio calculated, we can determine the type of response: - If \(尉 < 1\), the system is underdamped - If \(尉 = 1\), the system is critically damped - If \(尉 > 1\), the system is overdamped Since our calculated damping ratio \(尉 = \frac{3 \sqrt{5}}{20} < 1\) (approximately 0.335), the system is underdamped.
06

Find the amount of damping to be added or subtracted for critical damping

For the system to be critically damped, the damping ratio (\(尉\)) should equal 1. We will use the formula for the damping ratio to find the critical damping coefficient: \[ 尉 = \frac{c_{\text{critical}}}{2 \sqrt{m k}} \] Rearranging the formula for \(c_{\text{critical}}\): \[ c_{\text{critical}} = 2 \sqrt{m k} \] Given: - \(m = 2\, \text{kg}\) - \(k = 40\, \text{N/m}\) Using these values in the formula, we get: \[ c_{\text{critical}} = 2 \sqrt{(2\, \text{kg})(40\, \text{N/m})} \] \[c_{\text{critical}} = 4 \sqrt{5}\, \text{N路s/m}\] Now we can determine the amount of damping to be added or subtracted: \[螖c = c_{\text{critical}} - c\] \[螖c = 4 \sqrt{5}\, \text{N路s/m} - 3 \,\text{N路s/m}\]
07

Calculate the amount of damping to be added or subtracted for critical damping

Calculating the value of 螖c: \[螖c = (\, 4 \sqrt{5}-3) \text{N路s/m}\] Hence, the damping coefficient needs to be increased by \((4 \sqrt{5} - 3)\, \text{N路s/m}\) to make the system critically damped.

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Key Concepts

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

Natural Frequency
Understanding the natural frequency of a mechanical system is essential for predicting how the system will respond to external forces. Natural frequency, typically denoted by \(\omega_n\), is the rate at which a system oscillates when it is not subjected to external forces other than those required to hold the system in place.

For a mechanical system with a mass \(m\) and stiffness \(k\), natural frequency can be found using the formula \(\omega_n = \sqrt{\frac{k}{m}}\). In the context of our exercise, the system's natural frequency was calculated using the given mass \(2\) kg and stiffness \(40\) N/m, leading to a result of \(2\sqrt{5}\) rad/s. This frequency is a key characteristic that determines how quickly the system would oscillate without damping forces.
Damping Ratio
The damping ratio \(尉\), often symbolized by the Greek letter zeta, is a dimensionless measure that describes the extent to which oscillations in a system diminish over time due to non-conservative forces, such as friction or air resistance. It is calculated using the formula \(尉 = \frac{c}{2\sqrt{mk}}\), where \(c\) is the damping coefficient.

In the given exercise, we determined the system's damping ratio using a mass of \(2\) kg, a stiffness of \(40\) N/m, and a damping coefficient of \(3\) N路s/m. The resulting damping ratio of approximately \(0.335\) tells us the system is underdamped. An underdamped system will oscillate with progressively smaller amplitude over time, indicating that some energy is lost in each oscillation cycle but not enough to prevent the oscillations altogether.
Critically Damped System
A critically damped system is one that returns to equilibrium as quickly as possible without oscillating. For a system to be critically damped, its damping ratio \(尉\) needs to be exactly 1. If the damping ratio is less than 1, the system is underdamped, leading to oscillations, and if it is greater than 1, the system is overdamped, where it returns to equilibrium more slowly and without oscillations.

To achieve critical damping, the damping coefficient \(c\) may need to be adjusted. Using the rearranged formula \(c_{\text{critical}} = 2\sqrt{mk}\), we find the critical damping coefficient. In our exercise, we found that by increasing the existing damping coefficient by \(4\sqrt{5} - 3\) N路s/m, the underdamped system can be adjusted to be critically damped, effectively optimizing the system's return to rest.

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Most popular questions from this chapter

The characteristic roots of a single-degree-of-freedom system are given below. Find all the applicable features of the system among the characteristic equation, time constant, undamped natural frequency, damped frequency, and damping ratio. a. \(s_{1,2}=-4 \pm 5 i\) b. \(s_{1,2}=4 \pm 5 i\) c. \(s_{1,2}=-4,-5\) d. \(s_{1,2}=-4,-4\)

A panel made of fiber-reinforced composite material is observed to behave as a single-degreeof-freedom system of mass \(1 \mathrm{~kg}\) and stiffness \(2 \mathrm{~N} / \mathrm{m}\). The ratio of successive amplitudes is found to be \(1.1 .\) Determine the value of the hysteresis-damping constant \(\beta,\) the equivalent viscous-damping constant \(c_{\mathrm{eq}},\) and the energy loss per cycle for an amplitude of \(10 \mathrm{~mm} .\)

Determine the displacement, velocity, and acceleration of the mass of a spring-mass system with \(k=500 \mathrm{~N} / \mathrm{m}, m=2 \mathrm{~kg}, x_{0}=0.1 \mathrm{~m},\) and \(\dot{x}_{0}=5 \mathrm{~m} / \mathrm{s}\)

A mass of \(20 \mathrm{~kg}\) is suspended from a spring of stiffness \(10,000 \mathrm{~N} / \mathrm{m} .\) The vertical motion of the mass is subject to Coulomb friction of magnitude \(50 \mathrm{~N}\). If the spring is initially displaced downward by \(5 \mathrm{~cm}\) from its static equilibrium position, determine (a) the number of half cycles elapsed before the mass comes to rest, (b) the time elapsed before the mass comes to rest, and (c) the final extension of the spring.

The rotor of a dial indicator is connected to a torsional spring and a torsional viscous damper to form a single-degree-of-freedom torsional system. The scale is graduated in equal divisions, and the equilibrium position of the rotor corresponds to zero on the scale. When a torque of \(2 \times 10^{-3} \mathrm{~N}-\mathrm{m}\) is applied, the angular displacement of the rotor is found to be \(50^{\circ}\) with the pointer showing 80 divisions on the scale. When the rotor is released from this position, the pointer swings first to -20 divisions in one second and then to 5 divisions in another second. Find (a) the mass moment of inertia of the rotor, (b) the undamped natural time period of the rotor, (c) the torsional damping constant, and (d) the torsional spring stiffness.
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