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An experimental package and its support structure, which are to be placed on board the International Space Station, act as an underdamped spring-mass system with a force constant of \(2.1 \times 10^{5} \mathrm{N} / \mathrm{m}\) and mass 108 \(\mathrm{kg}\) . A NASA requirement is that resonance for forced oscillations not occur for any frequency below 35 \(\mathrm{Hz}\) . Does this package meet the requirement?

Short Answer

Expert verified
No, the package does not meet the requirement as resonance can occur below 35 Hz.

Step by step solution

01

Understanding Resonant Frequency

The resonant frequency for a damped spring-mass system is the frequency at which maximum amplitude oscillations occur due to an external force. It is given by the formula \( f_r = \frac{1}{2\pi} \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}} \), where \( k \) is the force constant, \( m \) is the mass, and \( b \) is the damping coefficient. In an underdamped system, resonance is largely determined by the natural frequency \( \omega_n = \sqrt{\frac{k}{m}} \). We assume damping is low enough that it does not significantly affect the resonant frequency.
02

Calculating Natural Frequency

The natural frequency \( \omega_n \) for an undamped system can be calculated using \( \omega_n = \sqrt{\frac{k}{m}} \). Here, \( k = 2.1 \times 10^5 \, \mathrm{N/m} \) and \( m = 108 \, \mathrm{kg} \). Substituting the values, we get \( \omega_n = \sqrt{\frac{2.1 \times 10^5}{108}} \).
03

Simplifying Natural Frequency Calculation

Calculate \( \omega_n = \sqrt{1944.444} \approx 44.1 \, \mathrm{rad/s} \). To convert this to hertz (Hz), use the relation \( f = \frac{\omega_n}{2\pi} \). Therefore, \( f_n = \frac{44.1}{2\pi} \approx 7.02 \, \mathrm{Hz} \).
04

Comparing with NASA Requirement

NASA specifies that resonance should not occur for frequencies below 35 Hz. The calculated natural frequency is approximately 7.02 Hz. Since 7.02 Hz is less than 35 Hz, the system’s resonant frequency is below the required threshold. Thus, the package might experience resonant conditions within the restricted frequency range.

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Key Concepts

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

Spring-Mass System
In a spring-mass system, we have a mass attached to a spring that can oscillate when disturbed. This system is an excellent way to understand oscillations as the spring force is proportional to the displacement, following Hooke's Law. The spring force constant, denoted as \( k \), determines the stiffness of the spring. It’s given in Newtons per meter (N/m), indicating how much force is needed to extend or compress the spring by one meter.
A practical spring-mass system is the setup used for the package on the International Space Station (ISS). Here, the force constant \( k \) is \( 2.1 \times 10^{5} \, \text{N/m} \) and the mass \( m \) is \( 108 \, \text{kg} \). This establishes the parameters for determining the system's oscillatory behavior, specifically in evaluating its resonant frequency.
  • The force constant \( k \) influences the system's stiffness.
  • The mass \( m \) affects how the system responds to external forces.
Understanding the interplay between these elements helps predict how the system behaves when subjected to external vibrations or forces, both critical factors in preventing undesirable resonance on the ISS.
Damping
Damping refers to the mechanism by which oscillations in a system are gradually reduced or eliminated due to energy loss. In real-world applications like the ISS system, damping is crucial as it prevents excessive oscillations that could be harmful.

In a damped spring-mass system, as is the case for the experimental package, energy dissipation occurs through various forms, such as friction or air resistance. This energy loss is determined by a damping coefficient, \( b \). An underdamped system is one where the damping is not sufficient to prevent oscillations, but it does reduce them over time.

  • Underdamping results in oscillations that decrease in amplitude over time.
  • Critical damping occurs when the system returns to equilibrium as quickly as possible without oscillating.
  • Overdamping slows down the return to equilibrium but avoids oscillations.
For the package, damping ensures that resonance does not build excessively, which could destabilize systems or structures in the ISS. It is, however, assumed to be low enough to not significantly alter the natural resonant frequency for simplicity.
Natural Frequency
The natural frequency of a spring-mass system is the rate at which it tends to oscillate in the absence of any external forces or damping. We denote this frequency as \( \omega_n \), and it is a fundamental property of the system based on the mass and spring constant. It is given by the formula:

\[ \omega_n = \sqrt{\frac{k}{m}} \]

Here, \( k \) is the spring force constant, and \( m \) is the mass. For the ISS package, the calculation of the natural frequency \( \omega_n \) using the given \( k = 2.1 \times 10^5 \, \text{N/m} \) and \( m = 108 \, \text{kg} \) yields \(\omega_n \approx 44.1 \, \text{rad/s}\). To convert it to hertz (Hz), we use the relationship:

\[ f_n = \frac{\omega_n}{2\pi} \approx 7.02 \, \text{Hz} \]

  • \( \omega_n \) in radians per second (rad/s) shows the angular frequency.
  • Conversion to hertz gives the frequency in cycles per second, a more standard measure.
This tells us that the package will readily oscillate at about 7.02 Hz naturally when undisturbed by external forces. However, NASA's requirements include preventing resonance at frequencies below 35 Hz. Thus, it's crucial to evaluate and adjust systems accordingly to avoid resonance-related issues on the ISS.

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Most popular questions from this chapter

A 0.500 \(\mathrm{kg}\) mass on a spring has velocity as a function of time given by \(v_{x}(t)=(3.60 \mathrm{cm} / \mathrm{s}) \sin \left[\left(4.71 \mathrm{s}^{-1}\right) t-\pi / 2\right]\) . What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?

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A uniform rod of length \(L\) oscillates through small angles about a point a distance \(x\) from its center. (a) Prove that its angular frequency is \(\sqrt{g x /\left[\left(L^{2} / 12\right)+x^{2}\right] .(b)}\) Show that its maximum angular frequency occurs when \(x=L / \sqrt{12}\) . (c) What is the length of the rod if the maximum angular frequency is 2\(\pi \mathrm{rad} / \mathrm{s} ?\)

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