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A damped single-degree-of-freedom system has \(m=5 \mathrm{~kg}, \quad c=500 \mathrm{~N}-\mathrm{s} / \mathrm{m},\) and \(k=50,000 \mathrm{~N} / \mathrm{m} .\) Determine the undamped and damped natural frequencies of vibration and the damping ratio of the system.

Short Answer

Expert verified
The undamped natural frequency (ω_n) is 100 rad/s, the damping ratio (ζ) is \(\frac{5}{2\sqrt{5}}\), and the damped natural frequency (ω_d) is 50\(\sqrt{3}\) rad/s.

Step by step solution

01

Find the undamped natural frequency (ω_n)

The undamped natural frequency (ω_n) is given by the formula: ω_n = \(\sqrt{\frac{k}{m}}\) Where k is the stiffness and m is the mass. Using the given values, we can calculate the undamped natural frequency: ω_n = \(\sqrt{\frac{50,000}{5}}\)
02

Calculate the undamped natural frequency (ω_n) value

Now, let's find the value of the undamped natural frequency: ω_n = \(\sqrt{\frac{50,000}{5}}\) = \(\sqrt{10,000}\) = 100 rad/s
03

Find the damping ratio (ζ)

The damping ratio (ζ) is given by the formula: ζ = \(\frac{c}{2\sqrt{mk}}\) Where c is the damping coefficient, m is the mass, and k is the stiffness. Using the given values, we can calculate the damping ratio: ζ = \(\frac{500}{2\sqrt{5 \times 50,000}}\)
04

Calculate the damping ratio (ζ) value

Now, let's find the value of the damping ratio: ζ = \(\frac{500}{2\sqrt{5 \times 50,000}}\) = \(\frac{500}{2 \times 100\sqrt{5}}\) = \(\frac{500}{200\sqrt{5}}\) = \(\frac{5}{2\sqrt{5}}\)
05

Find the damped natural frequency (ω_d)

The damped natural frequency (ω_d) is given by the formula: ω_d = ω_n\(\sqrt{1 - ζ^2}\) Where ω_n is the undamped natural frequency and ζ is the damping ratio. Using the calculated values, we can determine the damped natural frequency: ω_d = 100\(\sqrt{1 - (\frac{5}{2\sqrt{5}})^2}\)
06

Calculate the damped natural frequency (ω_d) value

Now, let's find the value of the damped natural frequency: ω_d = 100\(\sqrt{1 - (\frac{5}{2\sqrt{5}})^2}\) = 100\(\sqrt{1 - \frac{25}{20}}\) = 100\(\sqrt{\frac{15}{20}}\) = 100\(\sqrt{\frac{3}{4}}\)= 50\(\sqrt{3}\) rad/s Now we have all the required values: - Undamped natural frequency (ω_n): 100 rad/s - Damping ratio (ζ): \(\frac{5}{2\sqrt{5}}\) - Damped natural frequency (ω_d): 50\(\sqrt{3}\) rad/s

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Key Concepts

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

Undamped Natural Frequency
The undamped natural frequency, often symbolized as \( \omega_n \), is a critical characteristic of a mechanical system. Think of it as the heartbeat of the system—how fast it would oscillate if there was nothing to slow it down, like friction or air resistance. It's a theoretical scenario because all real-world systems experience some form of damping.

In the context of our single-degree-of-freedom system, calculating the undamped natural frequency involves two main components: the mass (m) and the stiffness (k). The mass represents how much inertia the system has—think about pushing a shopping cart versus a car, which one takes more effort? Stiffness, on the other hand, is like a spring's resistance to being squashed or stretched. The stiffer the spring, the higher the natural frequency.

Using the formula \( \omega_n = \sqrt{\frac{k}{m}} \), we see that a system's natural frequency increases with stiffness but decreases as mass increases. It's this delicate balance between the push of stiffness and the pull of mass that determines how quickly a system will vibrate when undamped.
Damping Ratio
The damping ratio, represented by \( \zeta \), acts like the system’s 'brakes.' It's a dimensionless measure that describes how oscillations in a system die out over time. A higher damping ratio means the system returns to its rest position more quickly after being disturbed—imagine a car suspension system that smooths out the bumps in the road efficiently. On the flipside, a low damping ratio means the system will oscillate more before settling down.

Mathematically, the damping ratio is found using the formula \( \zeta = \frac{c}{2\sqrt{mk}} \), where c is the damping coefficient, m is the mass, and k is the stiffness. The damping coefficient c is a property that describes the resistance that the system offers against motion—in other words, it's what actually 'damps' the vibrations.

By calculating the damping ratio, which in our case is \( \frac{5}{2\sqrt{5}} \), engineers can determine whether a system is underdamped, critically damped, or overdamped. This information is crucial for designing machinery, vehicles, and buildings that respond to loads in an expected, controlled manner.
Single-Degree-of-Freedom System
A single-degree-of-freedom (SDOF) system is the simplest type of dynamic system, usually consisting of a mass, spring, and damper. One degree of freedom means the system can move in only one way—up and down, forward and backward, or side to side.

Analysing the dynamic response of such systems is foundational in mechanical engineering and physics because it provides a straightforward context for understanding vibration behaviour. It sets the stage for more complex analyses involving multiple degrees of freedom, where objects can move in several independent directions simultaneously.

An SDOF system is used as a base for teaching fundamental concepts like natural frequency and damping because it's both simple enough to be approachable and complex enough to demonstrate real-world phenomena. In engineering design, these concepts are the cornerstone of creating safe and efficient structures and machines, ensuring they can withstand the forces they encounter without excessive vibration or failure.

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Most popular questions from this chapter

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.

Derive an expression for the time at which the response of a critically damped system will attain its maximum value. Also find the expression for the maximum response.

A spring-mass system has a natural frequency of \(10 \mathrm{~Hz}\). When the spring constant is reduced by \(800 \mathrm{~N} / \mathrm{m}\), the frequency is altered by \(45 \%\). Find the mass and spring constant of the original system.

A simple pendulum is found to vibrate at a frequency of \(0.5 \mathrm{~Hz}\) in a vacuum and \(0.45 \mathrm{~Hz}\) in a viscous fluid medium. Find the damping constant, assuming the mass of the bob of the pendulum as \(1 \mathrm{~kg}\).

A single-degree-of-freedom system consists of a mass, a spring, and a damper in which both dry friction and viscous damping act simultaneously. The free- vibration amplitude is found to decrease by \(1 \%\) per cycle when the amplitude is \(20 \mathrm{~mm}\) and by \(2 \%\) per cycle when the amplitude is \(10 \mathrm{~mm}\). Find the value of \((\mu N / k)\) for the dry-friction component of the damping.

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