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Chapter 8: Problem 28

The addition of damping to an undamped springmass system causes its period to increase by 25 percent. Determine the damping ratio \(\zeta\)

Short Answer

Expert verified
The damping ratio \( \zeta \) is approximately 0.6.

Step by step solution

01

Understand the Problem

We are given an undamped spring-mass system and a damped one where the period has been increased by 25%. We need to determine the damping ratio \( \zeta \).
02

Recall the Original Period Formula

The period \( T_0 \) of an undamped spring-mass system is calculated as \( T_0 = \frac{2\pi}{\omega_n} \), where \( \omega_n \) is the natural frequency.
03

Recall the Damped Period Formula

For a damped system, the period \( T_d \) is given by \( T_d = \frac{2\pi}{\omega_d} \), where \( \omega_d = \omega_n \sqrt{1-\zeta^2} \) is the damped frequency.
04

Relate the Damped and Undamped Periods

The problem states \( T_d = 1.25T_0 \). Thus, we relate these using the formula: \[ T_d = \frac{2\pi}{\omega_n \sqrt{1-\zeta^2}} = 1.25 \cdot \frac{2\pi}{\omega_n} \]
05

Simplify and Solve for \( \zeta \)

Cancel out \( \frac{2\pi}{\omega_n} \) from both sides and set up the equation:\[ 1 = 1.25 \sqrt{1-\zeta^2} \]Square both sides and solve for \( \zeta^2 \):\[ 1 = 1.5625(1-\zeta^2) \]\[ \frac{1}{1.5625} = 1-\zeta^2 \]\[ \zeta^2 = 1 - \frac{1}{1.5625} \]Calculate \( \zeta \) from \( \zeta^2 \).
06

Calculate Specific Values

Calculate \( \frac{1}{1.5625} \approx 0.64 \). Thus, \( 1 - 0.64 \approx 0.36 \), which means \( \zeta^2 \approx 0.36 \) and \( \zeta \approx \sqrt{0.36} \approx 0.6 \).

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Key Concepts

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

Spring-Mass System
A spring-mass system is a classic physics model used to study oscillatory motion. This system consists of a mass attached to a spring and allowed to move freely back and forth under the influence of the spring's restoring force. The system can be either undamped or damped.
In an undamped spring-mass system, there is no external force slowing down the motion, which means that the system oscillates indefinitely with constant amplitude. However, in real-world situations, some form of damping, such as air resistance or internal friction, is present.
When damping is introduced, the amplitude of oscillation gradually reduces over time. The study of these damped oscillations is crucial in understanding how real-world objects behave under periodic forces. Engineers might simulate such systems to ensure the stability of buildings and bridges during vibrations, like in earthquakes or strong winds.
Damped Frequency
Damped frequency refers to the frequency at which a damped spring-mass system oscillates. Unlike the natural frequency of an undamped system, which remains constant, damped frequency changes due to the presence of damping.
For a damped system, the frequency of oscillation is given by the formula:\[ \omega_d = \omega_n \sqrt{1 - \zeta^2} \]
where:
  • \( \omega_d \) is the damped frequency,
  • \( \omega_n \) is the natural frequency,
  • \( \zeta \) is the damping ratio.
As damping increases, the damped frequency decreases, making the oscillations slower. It is a key factor in analyzing how a system responds to external forces over time.
Natural Frequency
The natural frequency of a spring-mass system is the rate at which it oscillates when not subjected to any external forces, apart from the restoring force of the spring. This frequency is determined by the mass of the object and the stiffness of the spring.
The natural frequency \( \omega_n \) is given by the formula:\[ \omega_n = \sqrt{\frac{k}{m}} \]
where:
  • \( k \) is the spring constant,
  • \( m \) is the mass attached to the spring.
Understanding natural frequency allows scientists and engineers to predict how systems react under different conditions. For instance, if an external periodic force matches the natural frequency, resonance can occur, leading to large oscillations.
Period of Oscillation
The period of oscillation is the time it takes for a system to complete one full cycle of motion. It is an important characteristic of any oscillating system because it describes the overall speed of the oscillation.
For an undamped system, the period \( T_0 \) is given by:\[ T_0 = \frac{2\pi}{\omega_n} \]
where \( \omega_n \) is the natural frequency. This formula shows that the period is inversely proportional to the natural frequency, meaning that systems with a higher natural frequency will oscillate faster, leading to a shorter period.
However, in a damped system, the period \( T_d \) is increased due to the reduction in frequency caused by damping. The new period of a damped oscillation is:\[ T_d = \frac{2\pi}{\omega_d} \]
This extended period illustrates the damping effects that slow down the motion of the spring-mass system.

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Most popular questions from this chapter

The seismic instrument is mounted on a structure which has a vertical vibration with a frequency of \(5 \mathrm{Hz}\) and a double amplitude of \(18 \mathrm{mm}\). The sensing element has a mass \(m=2 \mathrm{kg},\) and the spring stiffness is \(k=1.5 \mathrm{kN} / \mathrm{m} .\) The motion of the mass relative to the instrument base is recorded on a revolving drum and shows a double amplitude of \(24 \mathrm{mm}\) during the steady-state condition. Calculate the viscous damping constant \(c\)

Viscous damping is added to an initially undamped spring-mass system. For what value of the damping ratio \(\zeta\) will the damped natural frequency \(\omega_{d}\) be equal to 90 percent of the natural frequency of the original undamped system?

The mass of a given critically damped system is released at time \(t=0\) from the position \(x_{0}>0\) with a negative initial velocity. Determine the critical value \(\left(\dot{x}_{0}\right)_{c}\) of the initial velocity below which the mass will pass through the equilibrium position.

Derive the expression for the energy loss \(E\) over a complete steady-state cycle due to the frictional dissipation of energy in a viscously damped linear oscillator. The forcing function is \(F_{0} \sin \omega t,\) and the displacement-time relation for steady-state motion is \(x_{P}=X \sin (\omega t-\phi)\) where the amplitude \(X\) is given by Eq. \(8 / 20 .\) (Hint: The frictional energy loss during a displacement \(d x\) is \(c \dot{x} d x,\) where \(c\) is the viscous damping coefficient. Integrate this expression over a complete cycle.

Determine the value of the viscous damping coefficient \(c\) for which the system is critically damped. The cylinder mass is \(m=2 \mathrm{kg}\) and the spring constant is \(k=150 \mathrm{N} / \mathrm{m} .\) Neglect the mass and friction of the pulley.
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