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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?

Short Answer

Expert verified
Potential energy: \( U(x) = \frac{F_0}{\alpha} \cosh(\alpha x) \). For small oscillations: \( U(x) \approx U(0) + \frac{F_0 \alpha x^2}{2} \). Angular frequency: \( \omega = \sqrt{\frac{F_0 \alpha}{m}} \).

Step by step solution

01

Understand the relationship between force and potential energy

The potential energy function, \( U(x) \), is related to the force \( F(x) \) by the negative derivative with respect to position. Mathematically, this is expressed as: \[ F(x) = -\frac{dU}{dx} \]Given that \( F = -F_0 \sinh(\alpha x) \), we substitute to set up an equation for \( U(x) \): \[ -F_0 \sinh(\alpha x) = -\frac{dU}{dx} \]
02

Integrate to find the potential energy function

To find \( U(x) \), integrate the expression for \( F(x) \):\[ \frac{dU}{dx} = F_0 \sinh(\alpha x) \]Integrating both sides with respect to \( x \), we get:\[ U(x) = \int F_0 \sinh(\alpha x) \, dx = \frac{F_0}{\alpha} \cosh(\alpha x) + C \]Where \( C \) is the integration constant. For simplicity, let's assume \( C = 0 \), as the reference potential can be set to any constant.
03

Approximate potential energy for small oscillations

For small oscillations around \( x = 0 \), the hyperbolic cosine can be approximated by its Taylor expansion:\[ \cosh(x) \approx 1 + \frac{x^2}{2} \]Thus, \( \cosh(\alpha x) \approx 1 + \frac{(\alpha x)^2}{2} \).So, the potential energy function for small \( x \) is:\[ U(x) \approx \frac{F_0}{\alpha} \left( 1 + \frac{\alpha^2 x^2}{2} \right) \approx U(0) + \frac{F_0 \alpha x^2}{2} \]
04

Find the angular frequency of oscillations

The potential can be written in the form of a harmonic potential \( \frac{1}{2} kx^2 \), where \( k = F_0 \alpha \).The angular frequency \( \omega \) for a harmonic oscillator is given by:\[ \omega = \sqrt{\frac{k}{m}} \]Substitute \( k = F_0 \alpha \) to get:\[ \omega = \sqrt{\frac{F_0 \alpha}{m}} \]

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Key Concepts

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

Force and Potential Energy
Force and potential energy are fundamentally related in classical mechanics. Here, the force on an object at position \(x\) is given by \( F = -F_0 \sinh(\alpha x) \). The negative sign indicates that the force is a restoring force, which animates motion back to equilibrium.

The relationship between force and potential energy \(U(x)\) can be understood through the equation \( F = -\frac{dU}{dx} \). This implies that the force is the derivative of the potential energy concerning displacement, but with a negative sign. This relationship allows us to determine the potential energy from a known force.
  • The force \(F\) is a measure of how much the potential energy changes with position.
  • If \(F\) is negative, the potential energy increases with increasing \(x\), signaling a stable, bound state.
By integrating \( F = -F_0 \sinh(\alpha x) \), we find the potential energy function, \( U(x) \), to be \( U(x) = \frac{F_0}{\alpha} \cosh(\alpha x) + C \). The constant \(C\) is often set to zero, simplifying further calculations.
Small Oscillations
In many physics problems, especially those dealing with small oscillations, the system's behavior can be significantly simplified. For small displacements from equilibrium, complex forces and potentials can often be approximated as linear or quadratic.

In this problem, the potential energy for small oscillations near \(x=0\) requires an approximation of \(\cosh(\alpha x)\) using the Taylor series expansion: \(\cosh(x) \approx 1 + \frac{x^2}{2}\) for very small \(x\). Applying this approximation, \(\cosh(\alpha x) \approx 1 + \frac{(\alpha x)^2}{2}\), simplifies the potential energy to:
  • \( U(x) \approx U(0) + \frac{F_0 \alpha x^2}{2} \)
This is similar to a quadratic potential, typical for simple harmonic oscillators.
The assumption of small oscillations is significant because it leads to simpler dynamics described by linear differential equations, making predictions and calculations more manageable.
Angular Frequency
Angular frequency \(\omega\) is a key concept when analyzing oscillatory motion. It describes how many radians an oscillator completes per unit time, similar to how frequency describes oscillations per second.

In the context of harmonic oscillations, we derive \(\omega\) using the effective spring constant \(k\) from the potential energy approximation. For our problem, this implies a potential energy of form \(\frac{1}{2}kx^2\). Matching this to \(U(x) \approx \frac{F_0 \alpha x^2}{2}\) suggests \(k = F_0 \alpha\).
  • Angular frequency \(\omega\) for a harmonic oscillator is given by \(\omega = \sqrt{\frac{k}{m}}\).
  • Substituting \(k\) gives \(\omega = \sqrt{\frac{F_0 \alpha}{m}}\).
This frequency is crucial for understanding how quickly and under what conditions the system returns to equilibrium during small oscillations. Understanding \(\omega\) allows us to predict the time behavior and rhythms of various physical systems in mechanics.

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Most popular questions from this chapter

Let \(f(t)\) be a periodic function with period \(\tau\). Explain clearly why the average of \(f\) over one period is not necessarily the same as the average over some other time interval. Explain why, on the other hand, the average over a long time \(T\) approaches the average over one period, as \(T \rightarrow \infty\)

When a car drives along a "washboard" road, the regular bumps cause the wheels to oscillate on the springs. (What actually oscillates is each axle assembly, comprising the axle and its two wheels.) Find the speed of my car at which this oscillation resonates, given the following information: (a) When four \(80-\mathrm{kg}\) men climb into my car, the body sinks by a couple of centimeters. Use this to estimate the spring constant \(k\) of each of the four springs. (b) If an axle assembly (axle plus two wheels) has total mass \(50 \mathrm{kg}\), what is the natural frequency of the assembly oscillating on its two springs? ( \(\mathbf{c}\) ) If the bumps on a road are \(80 \mathrm{cm}\) apart, at about what speed would these oscillations go into resonance?

In order to prove the crucial formulas (5.83)?5.85) for the Fourier coefficients \(a_{n}\) and \(b_{n},\) you must first prove the following: $$\int_{-\tau / 2}^{\tau / 2} \cos (n \omega t) \cos (m \omega t) d t=\left\\{\begin{array}{ll} \tau / 2 & \text { if } m=n \neq 0 \\ 0 & \text { if } m \neq n \end{array}\right.$$ (This integral is obviously \(\tau\) if \(m=n=0 .\) ) There is an identical result with all cosines replaced by sines, and finally $$\int_{-\tau / 2}^{\tau / 2} \cos (n \omega t) \sin (m \omega t) d t=0 \quad \text { for all integers } n \text { and } m$$ where as usual \(\omega=2 \pi / \tau\). Prove these. [Hint: Use trig identities to replace \(\cos (\theta) \cos (\phi)\) by terms like \(\cos (\theta+\phi) \text { and so on. }]\)

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 \(k\).

Suppose that you have found a particular solution \(x_{\mathrm{p}}(t)\) of the inhomogeneous equation (5.48) for a driven damped oscillator, so that \(D x_{\mathrm{p}}=f\) in the operator notation of \((5.49) .\) Suppose also that \(x(t)\) is any other solution, so that \(D x=f .\) Prove that the difference \(x-x_{\mathrm{p}}\) must satisfy the corresponding homogeneous equation, \(D\left(x-x_{\mathrm{p}}\right)=0 .\) This is an alternative proof that any solution \(x\) of the inhomogeneous equation can be written as the sum of your particular solution plus a homogeneous solution; that is, \(x=x_{\mathrm{p}}+x_{\mathrm{h}}\).

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