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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 \(k=5000 \mathrm{~N} / \mathrm{m}\). Determine the undamped and damped natural frequencies of vibration and the damping ratio of the system.

Short Answer

Expert verified
The undamped natural frequency (ωn) is 31.62 rad/s, the damping ratio (ζ) is 0.1, and the damped natural frequency (ωd) is 31.35 rad/s.

Step by step solution

01

Calculate the undamped natural frequency (ωn)

To determine the undamped natural frequency (ωn), we can use the following formula: ωn = \( \sqrt{\frac{k}{m}} \) Plugging in the given values of mass (m) and stiffness coefficient (k): ωn = \( \sqrt{\frac{5000}{5}} \) ωn = \( \sqrt{1000} \) ωn = 31.62 rad/s So, the undamped natural frequency of the system is 31.62 rad/s.
02

Calculate the damping ratio (ζ)

To find the damping ratio (ζ), we can use the following formula: ζ = \( \frac{c}{2 \sqrt{mk}} \) Plugging in the given values of mass (m), damping coefficient (c), and stiffness coefficient (k): ζ = \( \frac{500}{2 \sqrt{5 \times 5000}} \) ζ = \( \frac{500}{2 \sqrt{25000}} \) ζ = 0.1 So, the damping ratio of the system is 0.1.
03

Calculate the damped natural frequency (ωd)

To determine the damped natural frequency (ωd), we can use the following formula: ωd = ωn \( \sqrt{1 - ζ^2} \) Plugging in the values of undamped natural frequency (ωn) and damping ratio (ζ): ωd = 31.62 \( \sqrt{1 - (0.1)^2} \) ωd = 31.62 \( \sqrt{1 - 0.01} \) ωd = 31.62 \( \sqrt{0.99} \) ωd = 31.35 rad/s So, the damped natural frequency of the system is 31.35 rad/s. To summarize: - Undamped natural frequency (ωn): 31.62 rad/s - Damping ratio (ζ): 0.1 - Damped natural frequency (ωd): 31.35 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 denoted by \(\omega_n\), is a critical parameter in understanding the dynamic behavior of mechanical systems. It represents the rate at which a system would oscillate if there were no damping forces present, such as friction or air resistance. In simpler terms, imagine if you pushed a swing and it continued swinging back and forth without ever slowing down—that's what undamped motion looks like.

The calculation of undamped natural frequency is straightforward. It's determined by the system's stiffness, represented by the constant \(k\), and its mass, \(m\), using the formula \(\omega_n = \sqrt{\frac{k}{m}}\). The result is expressed in radians per second (rad/s), which tells you how many oscillations, in terms of angular displacement, the system can make in one second if undamped.

Knowing the undamped natural frequency is essential for engineers and designers as it helps predict how the system will respond to external forces and assists in avoiding resonant frequencies that could cause excessive vibrations and potential damage.
Damping Ratio
The damping ratio, represented by the Greek letter \(\zeta\), is a dimensionless measure that describes how oscillations in a system decay over time. It indicates the level of damping in the system relative to critical damping, which is the minimum amount of damping necessary to bring a system to rest without oscillating.

Think of the damping ratio as a way to gauge the effectiveness of your car's shock absorbers when driving over a bump—the higher the damping ratio, the quicker the car settles back to a steady state without bouncing.

The formula \(\zeta = \frac{c}{2 \sqrt{mk}}\) helps calculate the damping ratio using the damping coefficient \(c\), the mass \(m\), and the stiffness coefficient \(k\). A \(\zeta\) value less than 1 indicates underdamped motion where the system will oscillate with decreasing amplitude, a value greater than 1 is overdamped (no oscillations, just a return to equilibrium), while a value exactly equal to 1 represents critical damping. For example, a damping ratio of 0.1, as in the exercise, suggests a lightly damped system that would gradually come to rest.
Single-Degree-of-Freedom System
A single-degree-of-freedom (SDOF) system is the simplest mechanical system that can exhibit vibratory motion. It is characterized by its ability to move in only one direction or along a single axis. Practical examples of SDOF systems could be a mass on a spring or a pendulum. These types of systems can serve as fundamental building blocks for understanding more complex systems with multiple degrees of freedom.

In the context of vibration analysis, understanding SDOF systems is key for grasping the basics of how structures respond to dynamic loads. The terms and calculations presented in the exercise, such as the undamped natural frequency, damping ratio, and damped natural frequency, are all related to SDOF systems. By mastering these concepts, students and engineers can predict how single-mass systems behave under various conditions, aiding in design and safety evaluation.

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Most popular questions from this chapter

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.

Using the MATLAB Program2 \(m,\) plot the free-vibration response of a viscously damped system with \(m=4 \mathrm{~kg}, k=2500 \mathrm{~N} / \mathrm{m}, x_{0}=100 \mathrm{~mm}, \dot{x}_{0}=-10 \mathrm{~m} / \mathrm{s}, \Delta t=0.01 \mathrm{~s}, n=50\) for the following values of the damping constant: a. \(c=0\) b. \(c=100 \mathrm{~N}-\mathrm{s} / \mathrm{m}\) c. \(c=200 \mathrm{~N}-\mathrm{s} / \mathrm{m}\) d. \(c=400 \mathrm{~N}-\mathrm{s} / \mathrm{m}\)

The equation of motion of a rocket, of mass \(m,\) traveling vertically under a thrust \(F\) and air resistance or drag \(D\) is given by $$ m \dot{\nu}=F-D-m g $$ If \(m=1000 \mathrm{~kg}, F=50,000 \mathrm{~N}, D=2000 \mathrm{v}\), and \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\), find the time variation of the velocity of the rocket, \(\nu(t)=\frac{d x(t)}{d t},\) using the initial conditions \(x(0)=0\) and \(v(0)=0\), where \(x(t)\) is the distance traveled by the rocket in time \(t\).

Describe how the phase angle \(\phi_{0}\) in Eq. (2.23) is to be computed for different combinations of positive and negative values of the initial displacement \(\left(x_{0}\right)\) and the initial velocity \(\left(\dot{x}_{0}\right)\).

A mass of \(5 \mathrm{~kg}\) is attached to the top of a helical spring, and the system is made to vibrate by giving to the mass an initial deflection of \(25 \mathrm{~mm}\). The amplitude of the mass is found to reduce to \(10 \mathrm{~mm}\) after 100 cycles of vibration. Assuming a spring rate of \(200 \mathrm{~N} / \mathrm{m}\) for the helical spring, find the value of the hysteretic-damping coefficient \((h)\) of the spring.

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