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An undamped oscillator has period \(\tau_{\mathrm{o}}=1.000 \mathrm{s},\) but \(\mathrm{I}\) now add a little damping so that its period changes to \(\tau_{1}=1.001\) s. What is the damping factor \(\beta\) ? By what factor will the amplitude of oscillation decrease after 10 cycles? Which effect of damping would be more noticeable, the change of period or the decrease of the amplitude?

Short Answer

Expert verified
The damping factor \(\beta\) causes more noticeable amplitude decrease than period change.

Step by step solution

01

Understand the Problem

We need to determine the damping factor, \(\beta\), for a damped oscillator given the change in its period from 1.000 s to 1.001 s. We also have to calculate how much the amplitude decreases after 10 cycles and compare which effect of damping—change in period or decrease in amplitude—is more noticeable.
02

Recall Relevant Equations

The damped oscillator's period \(\tau_1\) is related to the natural period \(\tau_0\) by the equation \(\tau_1 = \frac{\tau_0}{\sqrt{1 - (\beta/\omega_0)^2}}\). The damping factor \(\beta\) affects the amplitude decay, described by the factor \(e^{-\beta t}\) after time \(t\).
03

Calculate the Damping Factor \(\beta\)

Since \(\tau_1 = 1.001\) and \(\tau_0 = 1.000\), plug these into the equation \(1.001 = 1.000 / \sqrt{1 - (\beta/\omega_0)^2}\) and solve for \(\beta\). This yields \(\sqrt{1 - (\beta/\omega_0)^2} = 1/1.001\). Here, \(\omega_0\) is the natural angular frequency related to \(\tau_0\) by \(\omega_0 = \frac{2\pi}{\tau_0}\). So \(\omega_0 = 2\pi\). Solving gives \(\beta \approx \frac{\omega_0}{2}((1/1.001)^2 - 1)^{1/2}\).
04

Calculate the Amplitude Decrease Factor

After 10 cycles, the time elapsed is \(t = 10\tau_1 = 10.01\, \text{s}\). The amplitude decreases by the factor \(e^{-\beta t}\). Plugging \(\beta\) from the previous step and \(t = 10.01\) s gives the amplitude decrease factor.
05

Evaluate Noticeability of Effects

Compare the relative change in period and the amplitude decrease factor. A 0.1% change in period (from 1.000 s to 1.001 s) might not be as perceivable as a substantial decrease in the amplitude over 10 cycles. Analyze which change would be more significant.

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Key Concepts

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

Damping Factor
The damping factor, often represented by the symbol \( \beta \), is a crucial parameter when dealing with damped oscillators. It determines how rapidly the system loses energy over time. In simple terms, it reflects the effect of friction or resistance within the system. The damping factor causes the oscillations to gradually lose their energy until they eventually stop.

To calculate the damping factor \( \beta \), we use the relationship between the damped period \( \tau_1 \) and the natural period \( \tau_0 \). The formula is given by:
  • \( \tau_1 = \frac{\tau_0}{\sqrt{1 - (\beta/\omega_0)^2}} \)
Where \( \omega_0 \) is the natural angular frequency, calculated as \( \omega_0 = \frac{2\pi}{\tau_0} \).

By substituting the known periods into the formula and solving for \( \beta \), we gain insight into the damping behavior of the system. A small \( \beta \) indicates low damping, meaning the oscillator will take longer to come to rest, while a large \( \beta \) shows significant damping, leading to faster energy loss.
Period of Oscillation
The period of oscillation is a key aspect of both damped and undamped oscillators. It represents the time taken for the system to complete one full cycle of motion. For an undamped oscillator, the period remains constant, represented as \( \tau_0 \). However, introducing damping changes this period.

When damping is present, the period increases slightly, as energy is dissipated with every oscillation. In the specific example given, the period increases from \( \tau_0 = 1.000 \) s to \( \tau_1 = 1.001 \) s. Even a small change in period can significantly affect the timing of oscillations over long durations.

Damping slightly alters the period because the system's energy is reduced, causing it to move slower. This relationship is often negligible for small damping factors, but still a critical aspect in precision-dependent applications.
Amplitude Decay
Amplitude decay in a damped oscillator is a direct result of the damping factor \( \beta \). Over time, the amplitude — or maximum displacement from equilibrium — decreases due to energy loss in the form of heat, sound, or other forms of energy. It's like a swinging pendulum coming to a stop naturally as its swings get smaller.

The amount by which the amplitude decreases follows an exponential decay formula:
  • Amplitude decay factor: \( e^{-\beta t} \)
Here, \( t \) is the time elapsed. For instance, if we wish to determine the amplitude decrease after 10 cycles, we can calculate the total time \( t = 10 \tau_1 \), and apply this in the formula to see how much amplitude is lost.

This aspect of damped oscillation is usually more noticeable than the subtle changes in period, especially when the amplitude significantly reduces over time, making it crucial for systems where maintaining motion is essential.
Damped Oscillation
Damped oscillation refers to the motion of an oscillator that loses energy over time due to a damping force. This force could be due to air resistance, friction, or any other factor opposing motion. Unlike undamped systems, where oscillations continue indefinitely at a constant amplitude, damped oscillations gradually decrease in amplitude.

In essence, a damped oscillator's motion becomes more contained or less vigorous as time passes. It is characterized not only by a reduction in amplitude but also a slight modification in the period. The effects of damping are crucial in numerous real-world applications, such as in car shock absorbers, where they help maintain smooth ride quality by reducing the oscillations caused by bumps.

By understanding the interplay between the damping factor, period changes, and amplitude decay, one can predict and control the behavior of oscillating systems effectively. This understanding is vital for designing systems that require precise oscillation control.

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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 \(E / \Delta E_{\text {dis: }}\)

The maximum displacement of a mass oscillating about its equilibrium position is \(0.2 \mathrm{m},\) and its maximum speed is \(1.2 \mathrm{m} / \mathrm{s}\). What is the period \(\tau\) of its oscillations?

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 \(\left.\beta<\omega_{\mathrm{o}}, \beta=\omega_{\mathrm{o}}, \text { and } \beta > \omega_{\mathrm{o}} .\right]\)

The potential energy of two atoms in a molecule can sometimes be approximated by the Morse function, $$U(r)=A\left[\left(e^{(R-r) / S}-1\right)^{2}-1\right]$$ where \(r\) is the distance between the two atoms and \(A, R,\) and \(S\) are positive constants with \(S \ll R .\) Sketch this function for \(0 < r < \infty\). Find the equilibrium separation \(r_{\mathrm{o}}\), at which \(U(r)\) is minimum. Now write \(r=r_{\mathrm{o}}+x\) so that \(x\) is the displacement from equilibrium, and show that, for small displacements, \(U\) has the approximate form \(U=\) const \(+\frac{1}{2} k x^{2} .\) That is, Hooke's law applies. What is the force constant \(k ?\)

(a) Consider a cart on a spring which is critically damped. At time \(t=0\), it is sitting at its equilibrium position and is kicked in the positive direction with velocity \(v_{\mathrm{o}} .\) Find its position \(x(t)\) for all subsequent times and sketch your answer. (b) Do the same for the case that it is released from rest at position \(x=x_{\mathrm{o}} .\) In this latter case, how far is the cart from equilibrium after a time equal to \(\tau_{\mathrm{o}}=2 \pi / \omega_{\mathrm{o}}\) the period in the absence of any damping?

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