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Determine the damping ratio \(\zeta\) for the system shown. The system parameters are \(m=4 \mathrm{kg}, k=\) \(500 \mathrm{N} / \mathrm{m},\) and \(c=100 \mathrm{N} \cdot \mathrm{s} / \mathrm{m} .\) Neglect the mass and friction of all pulleys, and assume that the cord remains taut throughout a motion cycle.

Short Answer

Expert verified
The damping ratio \( \zeta \) is approximately 1.118.

Step by step solution

01

Identify the given parameters

We are given the mass \( m = 4 \, \mathrm{kg} \), the spring stiffness \( k = 500 \, \mathrm{N/m} \), and the damping coefficient \( c = 100 \, \mathrm{N\cdot s/m} \).
02

Define the formula for the natural frequency

The natural frequency \( \omega_n \) can be calculated using the formula: \[ \omega_n = \sqrt{\frac{k}{m}} \].
03

Calculate the natural frequency

Substitute the values of \( k \) and \( m \) into the formula: \[ \omega_n = \sqrt{\frac{500}{4}} = \sqrt{125} = 11.18 \, \mathrm{rad/s} \].
04

Define the formula for the damping ratio

The damping ratio \( \zeta \) is given by the formula: \[ \zeta = \frac{c}{2m\omega_n} \].
05

Calculate the damping ratio

Substitute the values of \( c \), \( m \), and \( \omega_n \) into the formula: \[ \zeta = \frac{100}{2 \times 4 \times 11.18} = \frac{100}{89.44} = 1.118 \].

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Key Concepts

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

Natural Frequency
The natural frequency of a system is a concept that comes into play when we deal with oscillations, especially in mechanical systems. At its simplest, the natural frequency is the rate at which a system tends to oscillate in the absence of any driving or damping force. It is determined by the system's inherent properties, like mass and stiffness.

Calculating the natural frequency involves a crucial formula:
  • \( \omega_n = \sqrt{\frac{k}{m}} \)
This formula highlights the relationship between the spring stiffness \( k \) and the mass \( m \). The greater the stiffness, the higher the natural frequency. Conversely, a greater mass leads to a lower natural frequency.

In our exercise, with a mass \( m = 4 \, \text{kg} \) and spring stiffness \( k = 500 \, \text{N/m} \), we find the natural frequency by substituting these values, giving us \( \omega_n = 11.18 \, \text{rad/s} \). This value tells us how quickly the mass-spring system would oscillate naturally if it were moving without any damping.
Spring Stiffness
Spring stiffness, denoted by \( k \), is a measure of how much force is required to stretch or compress a spring by a unit length. It's a fundamental parameter in determining a system's ability to resist deformation. The stiffer the spring (higher \( k \)), the more force it takes to cause a displacement.

Understanding spring stiffness helps in analyzing mechanical systems' potential energy storage. A stiffer spring can store more energy when compressed or extended compared to a less stiff spring. In the exercise, with \( k = 500 \, \text{N/m} \), the spring's stiffness directly affects the system's natural frequency and response to forces.

Adding spring stiffness into our calculations, as we see from the natural frequency formula, it dictates how quickly or slowly a system oscillates. For systems involving springs, this value is crucial in designing stable and efficient mechanisms that don’t oscillate uncontrollably.
Damping Coefficient
The damping coefficient \( c \) plays a key role in describing how oscillations in a system die out over time due to resistive forces. This applies to real-world systems where there’s friction or any form of resistance, be it air resistance or material friction.

Mathematically, the damping effect is characterized by the damping ratio \( \zeta \), which is determined using:
  • \( \zeta = \frac{c}{2m\omega_n} \)
This equation shows the relationship between the damping coefficient, mass, and natural frequency. The damping coefficient's value tells us how quickly a system returns to rest. In our exercise, the damping coefficient was \( c = 100 \, \text{N}\cdot\text{s/m} \), which leads to a damping ratio \( \zeta = 1.118 \).

A damping ratio greater than 1 indicates an overdamped system, meaning the system returns to equilibrium without oscillating. It’s crucial for systems that need precise control without any oscillations, like in automobile suspensions or precision instruments.

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Most popular questions from this chapter

The equilibrium position of the mass \(m\) occurs where \(y=0\) and \(y_{B}=0 .\) When the attachment \(B\) is given a steady vertical motion \(y_{B}=b \sin \omega t,\) the mass \(m\) will acquire a steady vertical oscillation. Derive the differential equation of motion for \(m\) and specify the circular frequency \(\omega_{c}\) for which the oscillations of \(m\) tend to become excessively large. The stiffness of the spring is \(k\), and the mass and friction of the pulley are negligible.

A 90 -kg man stands at the end of a diving board and causes a vertical oscillation which is observed to have a period of 0.6 s. What is the static deflection \(\delta_{\mathrm{st}}\) at the end of the board? Neglect the mass of the board.

Determine the amplitude of vertical vibration of the car as it travels at a velocity \(v=40 \mathrm{km} / \mathrm{h}\) over the wavy road whose contour may be expressed as a sine or cosine function with a double amplitude \(2 b=50 \mathrm{mm} .\) The mass of the car is \(1800 \mathrm{kg}\) and the stiffness of each of the four car springs is \(35 \mathrm{kN} / \mathrm{m}\) Assume that all four wheels are in continuous con- tact with the road, and neglect damping. Note that the wheelbase of the car and the spatial period of the road are the same at \(L=3 \mathrm{m},\) so that it may be assumed that the car translates but does not rotate. At what critical speed \(v_{c}\) is the vertical vibration of the car at its maximum?

The uniform solid cylinder of mass \(m\) and radius \(r\) rolls without slipping during its oscillation on the circular surface of radius \(R .\) If the motion is confined to small amplitudes \(\theta=\theta_{0},\) determine the period \(\tau\) of the oscillations. Also determine the angular velocity \(\omega\) of the cylinder as it crosses the vertical centerline. (Caution: Do not confuse \(\omega\) with \(\dot{\theta}\) or with \(\omega_{n}\) as used in the defining equations. Note also that \(\theta\) is not the angular displacement of the cylinder.)

The spoked wheel of radius \(r,\) mass \(m,\) and centroidal radius of gyration \(\bar{k}\) rolls without slipping on the incline. Determine the natural frequency of oscillation and explore the limiting cases of \(\bar{k}=0\) and \(\bar{k}=r\)

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