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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 \(\omega_{0} < \beta < \infty\).

Short Answer

Expert verified
The expression is decreasing for \(\beta > \omega_0\).

Step by step solution

01

Analyze the expression

The expression we need to verify for decreasing behavior is \(\beta - \sqrt{\beta^2 - \omega_0^2}\). To understand if it decreases with increasing \(\beta\), we need to explore how each component changes as \(\beta\) increases.
02

Differentiate the expression

Let's differentiate the expression \(\beta - \sqrt{\beta^2 - \omega_0^2}\) with respect to \(\beta\):1. Differentiate \(\beta\) to get \(1\).2. Differentiate \(-\sqrt{\beta^2 - \omega_0^2}\) using the chain rule. The derivative of \(\sqrt{u}\) where \(u = \beta^2 - \omega_0^2\) is \(\frac{1}{2\sqrt{u}}\cdot 2\beta = \frac{\beta}{\sqrt{\beta^2 - \omega_0^2}}\).So, the derivative of the whole expression is:\[1 - \frac{\beta}{\sqrt{\beta^2 - \omega_0^2}}\]
03

Determine the sign of the derivative

We need to check the sign of the derivative \(1 - \frac{\beta}{\sqrt{\beta^2 - \omega_0^2}}\). For a function to be decreasing, this derivative must be negative:\[1 - \frac{\beta}{\sqrt{\beta^2 - \omega_0^2}} < 0\]Rewriting this, we have:\[\beta > \sqrt{\beta^2 - \omega_0^2}\]Since \(\beta > \omega_0\), this inequality holds true, confirming that our original function is decreasing.
04

Sketch the behavior

Since the derivative is negative, \(\beta - \sqrt{\beta^2 - \omega_0^2}\) is a decreasing function for \(\beta > \omega_0\). As \(\beta\) approaches infinity, the expression approaches \(0\). The sketch would have a steep decrease starting near \(\omega_0\) and gradually leveling off as it approaches zero.

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Key Concepts

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

Decay Parameter
Understanding the decay parameter is critical when studying an overdamped oscillator. The decay parameter is expressed as \( \beta - \sqrt{\beta^2 - \omega_0^2} \). This parameter describes the rate at which the oscillations decay over time. For overdamped oscillators, this decay parameter helps us determine how the system returns to its equilibrium position without oscillating.
To examine how the decay parameter behaves, particularly its dependence on \( \beta \), it's key to see that \( \beta > \omega_0 \) in an overdamped system. With increasing \( \beta \), the decay parameter tends to decrease because each increment in \( \beta \) makes the square root term larger, thereby reducing the difference in the expression.
This gives insight into the system's behavior as the resistance in the damping force changes; the stronger the damping \( \beta \), the quicker the oscillations decay and the system stabilizes.
Differentiation
Differentiation is a fundamental concept in calculus, used to determine the rate of change of a function or its slope at any given point. In the context of the decay parameter, differentiating is essential to confirm whether it decreases as \( \beta \) increases. By differentiating \( \beta - \sqrt{\beta^2 - \omega_0^2} \) with respect to \( \beta \), we are performing a calculus operation that helps us understand how the decay parameter changes with changes in \( \beta \).
Here, differentiating \( \beta \) yields 1 immediately since the derivative of a linear term with respect to itself is just the coefficient. The more complex task is the differentiation of \( -\sqrt{\beta^2 - \omega_0^2} \), which involves applying the chain rule. By the end of this differentiation process, we obtain the expression \( 1 - \frac{\beta}{\sqrt{\beta^2 - \omega_0^2}} \). This result is crucial for the next step, which involves analyzing the derivative's sign.
Chain Rule
The chain rule is a rule in calculus used for computing the derivative of the composition of two or more functions. It's particularly useful when dealing with nested functions, like \( \sqrt{\beta^2 - \omega_0^2} \) in our case.
To differentiate \( -\sqrt{\beta^2 - \omega_0^2} \), we identify an outer function, \( \sqrt{u} \), and an inner function, \( u = \beta^2 - \omega_0^2 \). First, we differentiate the outer function with respect to \( u \), yielding \( \frac{1}{2\sqrt{u}} \). Then, we differentiate the inner function, \( \beta^2 - \omega_0^2 \), with respect to \( \beta \), which results in \( 2\beta \).
Applying the chain rule, we multiply these derivatives resulting in \( \frac{\beta}{\sqrt{\beta^2 - \omega_0^2}} \). The chain rule simplifies handling complex expressions by breaking them into simpler parts tackled individually, which in turn makes it clearer to see how each part contributes to the change of the whole.
Decreasing Function Analysis
Analyzing whether a function is decreasing involves studying the sign of its derivative. For the decay parameter \( \beta - \sqrt{\beta^2 - \omega_0^2} \), once differentiated, we end up with \( 1 - \frac{\beta}{\sqrt{\beta^2 - \omega_0^2}} \). To determine if this indicates a decrease, this derivative must be less than zero.
Thus, we need to verify that \( \beta > \sqrt{\beta^2 - \omega_0^2} \) is true. Given the constraint that \( \beta > \omega_0 \), this inequality holds. Therefore, the function is indeed decreasing for values of \( \beta \) greater than \( \omega_0 \).
In practical terms, as the value of \( \beta \) increases, the decay parameter decreases, meaning the rate of decay becomes less over time. This explains the slower approach to stability as damping increases, leading to a smoother transition to equilibrium.

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Most popular questions from this chapter

A massless spring is hanging vertically and unloaded, from the ceiling. A mass is attached to the bottom end and released. How close to its final resting position is the mass after 1 second, given that it finally comes to rest 0.5 meters below the point of release and that the motion is critically damped?

An unusual pendulum is made by fixing a string to a horizontal cylinder of radius \(R\), wrapping the string several times around the cylinder, and then tying a mass \(m\) to the loose end. In equilibrium the mass hangs a distance \(l_{\mathrm{o}}\) vertically below the edge of the cylinder. Find the potential energy if the pendulum has swung to an angle \(\phi\) from the vertical. Show that for small angles, it can be written in the Hooke's law form \(U=\frac{1}{2} k \phi^{2} .\) Comment on the value of \(k\).

We know that if the driving frequency \(\omega\) is varied, the maximum response \(\left(A^{2}\right)\) of a driven damped oscillator occurs at \(\omega \approx \omega_{\mathrm{o}}\) (if the natural frequency is \(\omega_{\mathrm{o}}\) and the damping constant \(\beta \ll\) \(\omega_{\mathrm{o}}\) ). Show that \(A^{2}\) is equal to half its maximum value when \(\omega \approx \omega_{\mathrm{o}} \pm \beta,\) so that the full width at half maximum is just \(2 \beta\). [Hint: Be careful with your approximations. For instance, it's fine to say \(\left.\omega+\omega_{\mathrm{o}} \approx 2 \omega_{\mathrm{o}}, \text { but you certainly mustn't say } \omega-\omega_{\mathrm{o}} \approx 0 .\right]\)

(a) If a mass \(m=0.2 \mathrm{kg}\) is tied to one end of a spring whose force constant \(k=80 \mathrm{N} / \mathrm{m}\) and whose other end is held fixed, what are the angular frequency \(\omega\), the frequency \(f\), and the period \(\tau\) of its oscillations? (b) If the initial position and velocity are \(x_{\mathrm{o}}=0\) and \(v_{\mathrm{o}}=40 \mathrm{m} / \mathrm{s},\) what are the constants \(A\) and \(\delta\) in the expression \(x(t)=A \cos (\omega t-\delta) ?\)

(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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