3.2.1: Driven Springs Simulation
- Page ID
- 26147
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The following simulation shows five different masses, each attached to a spring of the same stiffness. The springs are mounted on a mechanical device that shakes the springs and attached masses. You can adjust the driving frequency, \(f\) in \(\text{Hz}\), of the shaking mechanism, the amplitude of the driving force, \(F_{o}\), in Newtons, the amount of friction, \(b\) in \(\text{Ns/m}\), and the stiffness of the springs, \(κ\), measured in \(\text{N/m}\).
Simulation Questions:
- Start the simulation. Do any of the masses have a very large amplitude?
- Increase the amplitude of the driving force, \(F_{o}\). Now do any of the masses have a very large amplitude?
- Reset the simulation and change the driving frequency, \(\text{f}\), to \(0.5\text{ Hz}\). Wait a few seconds. What do you see now?
- Why is the oscillation of mass number five much larger than the other ones now?
- Reset the simulation and change the driving frequency to \(2.0\text{ Hz}\). What do you notice now?
- See if you can determine the resonance frequency of the center mass by trial and error. What is its resonance frequency?
- In the previous simulation we saw that the natural frequency, written as \(f_{o}\) is given by the stiffness of the spring, \(κ\), and the mass; \(f_{o} = (κ/m)^{1/2}/(2π)\). In this simulation the large mass is \(10\text{ kg}\) and the spring constant is initially \(100\text{ N/m}\) so \(f_{o}= 0.5\text{ Hz}\). This is why it has a large oscillation when driven at \(0.5\text{ Hz}\); it will resonate with a driving frequency equal to its natural frequency. The center mass (number three) is \(2.5\text{ kg}\) so the natural frequency is \(1.0\text{ Hz}\). Did you find a resonance frequency of \(1.0\text{ Hz}\) for this mass?
- Mass number two is \(1.25\text{ kg}\). Calculate its natural frequency. Verify your result by trying it out in the simulation.
- Reset the simulation so that none of the masses are in resonance. Why doesn't the driving amplitude have much of an effect on the oscillation of the masses?
- Change the spring stiffness, \(κ\) to \(150\text{ N/m}\). (This changes the stiffness of all the springs.) Do the masses have the same resonance frequencies? Explain.
- Use the formula for the natural frequency to calculate the natural frequency of the \(10\text{ kg}\) mass (mass number five) with a spring constant of \(150\text{ N/m}\). Verify your result using the simulation.