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Consider a cart on a spring with natural frequency \(\omega_{\mathrm{o}}=2 \pi,\) which is released from rest at \(x_{\mathrm{o}}=1\) and \(t=0 .\) Using appropriate graphing software, plot the position \(x(t)\) for \(0 < t < 2\) and for damping constants \(\beta=0,1,2,4,6,2 \pi, 10,\) and \(20 .\) [Remember that \(x(t)\) is given by different formulas for \(\left.\beta<\omega_{\mathrm{o}}, \beta=\omega_{\mathrm{o}}, \text { and } \beta > \omega_{\mathrm{o}} .\right]\)

Short Answer

Expert verified
Plot position \( x(t) \) for each provided \( \beta \); use formulas based on damping level.

Step by step solution

01

Understand the Problem

We have a damped harmonic oscillator with a natural frequency of \( \omega_{\mathrm{o}} = 2\pi \). We are asked to plot the position \( x(t) \) for different damping constants (\( \beta \)) and for a specific time interval from \( t = 0 \) to \( t = 2 \). We should apply the appropriate formula for \( x(t) \) based on whether \( \beta < \omega_{\mathrm{o}} \), \( \beta = \omega_{\mathrm{o}} \), or \( \beta > \omega_{\mathrm{o}} \).
02

Formulas for Position x(t)

The position \( x(t) \) for a damped harmonic oscillator is given by different formulas based on the damping constant:- For \( \beta < \omega_{\mathrm{o}} \) (underdamped), \( x(t) = x_{\mathrm{o}} e^{-\beta t} \cos(\sqrt{\omega_{\mathrm{o}}^2 - \beta^2} \cdot t) \).- For \( \beta = \omega_{\mathrm{o}} \) (critically damped), \( x(t) = x_{\mathrm{o}} (1 + \omega_{\mathrm{o}} t) e^{-\omega_{\mathrm{o}} t} \).- For \( \beta > \omega_{\mathrm{o}} \) (overdamped), \( x(t) = x_{\mathrm{o}} e^{-\beta t} \cosh(\sqrt{\beta^2 - \omega_{\mathrm{o}}^2} \cdot t) \).
03

Determine the Damping Scenarios

Calculate \( \omega_{\mathrm{o}} \) using the given value \( \omega_{\mathrm{o}} = 2\pi \), which gives us approximately 6.28. We'll evaluate different \( \beta \) values: 0, 1, 2, 4, 6, 2\pi (approx. 6.28), 10, and 20. Some of these will correspond to underdamped, critically damped, and overdamped cases.
04

Apply the Formulas

For each damping constant \( \beta \):- If \( \beta < 6.28 \) (0, 1, 2, 4, 6): use the formula for underdamped motion.- If \( \beta = 6.28 \): use the critical damping formula.- If \( \beta > 6.28 \) (10, 20): use the formula for overdamped motion.Calculate \( x(t) \) values from \( t = 0 \) to \( t = 2 \).
05

Graph the Results

Use graphing software to plot each scenario. Create a graph showing \( x(t) \) on the y-axis and \( t \) on the x-axis, with separate curves for each \( \beta \) value. Ensure that each curve is clearly labeled to show the different damping constants.

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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, often denoted as \( \omega_{\mathrm{o}} \), is a fundamental characteristic of oscillatory systems like springs. It represents the frequency at which a system oscillates when not subjected to any external force or damping. In our exercise, this frequency is given as \( \omega_{\mathrm{o}} = 2\pi \), which simplifies calculations since it can convert easily between cycles and angular measurements.
Understanding natural frequency helps to identify how fast a system tends to oscillate naturally. In this scenario, it's useful for determining the type of damping effect by comparing it to the damping constant \( \beta \). Whether the system is underdamped, critically damped, or overdamped will depend on the relationship between \( \beta \) and \( \omega_{\mathrm{o}} \).
It's crucial because it helps us use the right formula for calculating position \( x(t) \), which changes based on what type of damping occurs.
Damping Constant
The damping constant, symbolized by \( \beta \), measures how quickly an oscillating system loses energy over time. It's a key player in defining the behavior of damped harmonic oscillators. In the given exercise, we're asked to consider various values of \( \beta \) like 0, 1, 2, 4, 6, 10, and 20.
A higher \( \beta \) value indicates stronger damping, meaning the system will return to equilibrium more quickly. Conversely, a lower \( \beta \) leads to a slower dissipation of energy, allowing oscillations to continue longer. This constant guides us to understand whether the system exhibits underdamped, critically damped, or overdamped behavior.
  • Underdamped: Oscillates with decreasing amplitude.
  • Critically damped: Returns to equilibrium as quickly as possible without oscillating.
  • Overdamped: Returns to equilibrium without oscillating, slower than critically damped.
Each of these behaviors requires a specific formula to measure \( x(t) \) accurately over time.
Underdamped, Critically Damped, and Overdamped Cases
These terms describe how a damped harmonic oscillator behaves based on the relationship between the damping constant \( \beta \) and the natural frequency \( \omega_{\mathrm{o}} \).
Underdamped:
When \( \beta < \omega_{\mathrm{o}} \), the system is underdamped. It will oscillate, but the amplitude decreases over time. The position function is \( x(t) = x_{\mathrm{o}} e^{-\beta t} \cos(\sqrt{\omega_{\mathrm{o}}^2 - \beta^2} \cdot t) \). This reflects oscillatory motion that gradually diminishes.
Critically Damped:
When \( \beta = \omega_{\mathrm{o}} \), the system is critically damped. Here, the system returns to equilibrium as fast as possible without oscillations. The mathematical expression is \( x(t) = x_{\mathrm{o}} (1 + \omega_{\mathrm{o}} t) e^{-\omega_{\mathrm{o}} t} \).
Overdamped:
When \( \beta > \omega_{\mathrm{o}} \), the system is overdamped. It returns to equilibrium without oscillating. The relevant formula is \( x(t) = x_{\mathrm{o}} e^{-\beta t} \cosh(\sqrt{\beta^2 - \omega_{\mathrm{o}}^2} \cdot t) \). This signifies a slower response compared to the critically damped case.
Graphing of Position Over Time
Graphing the position \( x(t) \) over time is essential to visualize how a damped harmonic oscillator behaves under different damping scenarios. In our exercise, we plot \( x(t) \) from \( t = 0 \) to \( t = 2 \) for various \( \beta \) values.
Using a graph helps us compare how quickly the oscillations decay depending on the damping constant. Here's what to expect:
  • For a small \( \beta \), the oscillation is more evident, decaying slowly.
  • At critical damping, the graph shows a smooth return to equilibrium without oscillating.
  • In overdamped conditions, the return to equilibrium is visible but slower, with no oscillations.
This graphical representation simplifies understanding how different damping constants affect motion. By visually interpreting the data, students can grasp how energy dissipation varies in each scenario. It’s a powerful way of connecting theoretical formulas with real-world behavior.

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Most popular questions from this chapter

Another interpretation of the \(Q\) of a resonance comes from the following: Consider the motion of a driven damped oscillator after any transients have died out, and suppose that it is being driven close to resonance, so you can set \(\omega=\omega_{\mathrm{o}}\). (a) Show that the oscillator's total energy (kinetic plus potential) is \(E=\frac{1}{2} m \omega^{2} A^{2} .\) (b) Show that the energy \(\Delta E_{\text {dis dissipated during one cycle by the damping force }}\) \(F_{\text {dmp }}\) is \(2 \pi m \beta \omega A^{2} .\) (Remember that the rate at which a force does work is \(F v .\) ) (c) Hence show that \(Q\) is \(2 \pi\) times the ratio \(E / \Delta E_{\text {dis: }}\)

This problem is to refresh your memory about some properties of complex numbers needed at several points in this chapter, but especially in deriving the resonance formula (5.64). (a) Prove that any complex number \(z=x+i y\) (with \(x\) and \(y\) real) can be written as \(z=r e^{i \theta}\) where \(r\) and \(\theta\) are the polar coordinates of \(z\) in the complex plane. (Remember Euler's formula.) (b) Prove that the absolute value of \(z\), defined as \(|z|=r,\) is also given by \(|z|^{2}=z z^{*},\) where \(z^{*}\) denotes the complex conjugate of \(z\) defined as \(z^{*}=x-\)iy. \(\left(\text { c) Prove that } z^{*}=r e^{-i \theta} . \text { (d) Prove that }(z w)^{*}=z^{*} w^{*} \text { and that }(1 / z)^{*}=1 / z^{*}\right.\) (e) Deduce that if \(z=a /(b+i c),\) with \(a, b,\) and \(c\) real, then \(|z|^{2}=a^{2} /\left(b^{2}+c^{2}\right)\).

The force on a mass \(m\) at position \(x\) on the \(x\) axis is \(F=-F_{0} \sinh \alpha x,\) where \(F_{0}\) and \(\alpha\) are constants. Find the potential energy \(U(x),\) and give an approximation for \(U(x)\) suitable for small oscillations. What is the angular frequency of such oscillations?

Verify that the decay parameter \(\beta-\sqrt{\beta^{2}-\omega_{0}^{2}}\) for an overdamped oscillator \(\left(\beta > \omega_{\mathrm{o}}\right) d e\) creases with increasing \(\beta\). Sketch its behavior for \(\omega_{0} < \beta < \infty\).

A massless spring has unstretched length \(l_{\mathrm{o}}\) and force constant \(k\). One end is now attached to the ceiling and a mass \(m\) is hung from the other. The equilibrium length of the spring is now \(l_{1}\). (a) Write down the condition that determines \(l_{1}\). Suppose now the spring is stretched a further distance \(x\) beyond its new equilibrium length. Show that the net force (spring plus gravity) on the mass is \(F=-k x\). That is, the net force obeys Hooke's law, when \(x\) is the distance from the equilibrium position \(-\) a very useful result, which lets us treat a mass on a vertical spring just as if it were horizontal. (b) Prove the same result by showing that the net potential energy (spring plus gravity) has the form \(U(x)=\) const \(+\frac{1}{2} k x^{2}\)

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