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Two particles of equal masses \(m_{1}=m_{2}\) move on a frictionless horizontal surface in the vicinity of a fixed force center, with potential energies \(U_{1}=\frac{1}{2} k r_{1}^{2}\) and \(U_{2}=\frac{1}{2} k r_{2}^{2} .\) In addition, they interact with each other via a potential energy \(U_{12}=\frac{1}{2} \alpha k r^{2},\) where \(r\) is the distance between them and \(\alpha\) and \(k\) are positive constants. (a) Find the Lagrangian in terms of the CM position \(\mathbf{R}\) and the relative position \(\mathbf{r}=\mathbf{r}_{1}-\mathbf{r}_{2} \cdot(\mathbf{b})\) Write down and solve the Lagrange equations for the \(\mathrm{CM}\) and relative coordinates \(X, Y\) and \(x, y .\) Describe the motion.

Short Answer

Expert verified
The CM undergoes harmonic motion with frequency \( \omega = \sqrt{\frac{k}{m}} \), while the relative motion also oscillates harmonically with \( \omega' = \sqrt{\frac{k(1+\alpha)}{m}} \).

Step by step solution

01

Define the Positions

The positions of the two particles can be written as \( \mathbf{r}_1 \) and \( \mathbf{r}_2 \). The center of mass (CM) position, \( \mathbf{R} \), is given by \( \mathbf{R} = \frac{1}{2}(\mathbf{r}_1 + \mathbf{r}_2) \). The relative position, \( \mathbf{r} \), is \( \mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2 \). These definitions will allow us to express the system in terms of \( \mathbf{R} \) and \( \mathbf{r} \).
02

Express Kinetic Energy

The kinetic energy of the system is the sum of the kinetic energies of the two particles. If \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are their velocities, the kinetic energy \( T \) is \( T = \frac{1}{2}m \mathbf{v}_1^2 + \frac{1}{2}m \mathbf{v}_2^2 \). Using \( \mathbf{v}_1 = \mathbf{V} + \frac{\mathbf{v}}{2} \) and \( \mathbf{v}_2 = \mathbf{V} - \frac{\mathbf{v}}{2} \) where \( \mathbf{V} = \dot{\mathbf{R}} \) and \( \mathbf{v} = \dot{\mathbf{r}} \), we find that \( T = m \mathbf{V}^2 + \frac{1}{4}m \mathbf{v}^2 \).
03

Express Potential Energy

The total potential energy \( U \) is given by \( U = U_1 + U_2 + U_{12} \). Substituting the given functions of potential energy, \( U = \frac{1}{2}k r_1^2 + \frac{1}{2}k r_2^2 + \frac{1}{2}\alpha k r^2 \). To express this in terms of \( \mathbf{R} \) and \( \mathbf{r} \), note \( r_1^2 + r_2^2 = 2R^2 + \frac{r^2}{2} \). Therefore, \( U = k R^2 + \frac{k(1+\alpha)}{2}r^2 \).
04

Write the Lagrangian

The Lagrangian \( \mathcal{L} \) is the difference between the kinetic and potential energies, \( \mathcal{L} = T - U \). Substituting the expressions for \( T \) and \( U \), we get \( \mathcal{L} = m \mathbf{V}^2 + \frac{1}{4} m \mathbf{v}^2 - k R^2 - \frac{k(1+\alpha)}{2}r^2 \).
05

Write Lagrange Equations

For the CM coordinates \( X, Y \), we have Lagrange's equations \( m \ddot{X} = -kX \) and \( m \ddot{Y} = -kY \). For the relative coordinates \( x, y \), the equations are \( \frac{1}{2}m \ddot{x} = -\frac{k(1+\alpha)}{2}x \) and \( \frac{1}{2}m \ddot{y} = -\frac{k(1+\alpha)}{2}y \).
06

Solve the Equations

The CM motion \( X(t), Y(t) \) are simple harmonic oscillations given by \( X(t) = A_x \cos(\sqrt{\frac{k}{m}} t + \phi_x) \) and \( Y(t) = A_y \cos(\sqrt{\frac{k}{m}} t + \phi_y) \). The relative motion \( x(t), y(t) \) are also harmonic oscillations but with a different frequency, \( x(t) = A_x' \cos(\sqrt{\frac{k(1+\alpha)}{m}} t + \phi_x') \) and \( y(t) = A_y' \cos(\sqrt{\frac{k(1+\alpha)}{m}} t + \phi_y') \).
07

Conclusion: Describe the Motion

The CM moves in a simple harmonic motion with frequency \( \omega = \sqrt{\frac{k}{m}} \), while the relative motion between the particles is also harmonic but with a frequency \( \omega' = \sqrt{\frac{k(1+\alpha)}{m}} \). The motions are decoupled and oscillate independently.

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Key Concepts

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

Center of Mass Motion
Understanding the center of mass (CM) motion is essential in analyzing this type of problem. The CM represents the average location of all the mass in a system, acting as if all the system's mass were concentrated at that point. The formula for the CM position \( \mathbf{R} \) indicates that it's simply the midpoint when two particles have equal mass:
\[ \mathbf{R} = \frac{1}{2}(\mathbf{r}_1 + \mathbf{r}_2) \]
In the context of Lagrangian mechanics, focusing on the CM helps simplify complex systems into more manageable equations. This description is helpful because it telegraphs the system's overall motion without getting bogged down in the details of individual particle paths.
For two interacting particles, the CM coordinates \( X \) and \( Y \) move independently of their separations. The motion characterized by these coordinates follows simple harmonic motion (SHM), which is a type of periodic motion where the restoring force is directly proportional to the displacement. This means the CM's motion can be described by functions involving cosine and sine, such as:
\[X(t) = A_x \cos\left(\sqrt{\frac{k}{m}} t + \phi_x\right)\]Similarly for \( Y(t) \), capturing the graceful back-and-forth swaying motion typical of SHM.
  • The significance of the CM motion in our exercise lies in its independent simple harmonic behavior that simplifies problem analysis.
  • This simplified motion can be analyzed separately from the internal motions of the system's components.
Simple Harmonic Oscillator
A simple harmonic oscillator is a cornerstone concept in physics, representing systems where the force is proportional to the displacement from an equilibrium position. The hallmark of such oscillators is periodicity, where the motion repeats itself after a period. They appear in various forms like pendulums, springs, and even in our exercise with the CM and relative motions.
In the given problem, both the CM and relative motions described
obey simple harmonic motion. The CM follows:
\( \omega = \sqrt{\frac{k}{m}} \),
a natural frequency determined by the stiffness \( k \) and mass \( m \).The relative motion between particles is also simple harmonic but with its frequency affected by an additional parameter \( \alpha \), indicating the interaction between the particles:
\[ \omega' = \sqrt{\frac{k(1+\alpha)}{m}} \]This is due to the potential energy involving a term with \( \alpha \), contributing to the overall system's restoring force. Simple harmonic oscillators are beneficial for modeling real-world systems due to their predictability and symmetry in motion.
  • Simple harmonic oscillators provide an idealized model of motion with predictability that can be applied widely across physics tasks.
  • The characteristics like frequency and amplitude offer insights into the system properties and external influences.
Potential Energy
Potential energy is a fundamental concept, often represented as the stored energy in a system due to its position or configuration. In the context of our exercise, the potential energy consists of multiple components—each affecting how the particles interact and move.
For each particle near a fixed force center, the potential energies are denoted by:\[ U_1 = \frac{1}{2} k r_1^2 \quad and \quad U_2 = \frac{1}{2} k r_2^2 \]These equations suggest a dependence on the square of the distance from a fixed point, a characteristic feature of systems with harmonic potentials like springs. The potential energy \( U_{12} = \frac{1}{2} \alpha k r^2 \) reflects interactions between the two particles and contributes an additional binding force depending on their separation \( r \):
\[U = kR^2 + \frac{k(1+\alpha)}{2}r^2\]This formulation shows us how changes in \( \mathbf{R} \) and \( \mathbf{r} \) translate into energy stored in the system, influencing motion. Studying potential energy provides insight into the forces at play and how energy is conserved or transformed in motion.
  • Potential energy serves as an indicator of the stability and predictability of a system's motion.
  • Understanding its formulation helps in predicting changes in movement and interactions between elements of the system.

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Most popular questions from this chapter

Consider a particle of reduced mass \(\mu\) orbiting in a central force with \(U=k r^{n}\) where \(k n>0\). (a) Explain what the condition \(k n>0\) tells us about the force. Sketch the effective potential energy \(U_{\mathrm{eff}}\) for the cases that \(n=2,-1,\) and \(-3 .\) (b) Find the radius at which the particle (with given angular momentum \(\ell\) ) can orbit at a fixed radius. For what values of \(n\) is this circular orbit stable? Do your sketches confirm this conclusion? (c) For the stable case, show that the period of small oscillations about the circular orbit is \(\tau_{\mathrm{osc}}=\tau_{\mathrm{orb}} / \sqrt{n+2} .\) Argue that if \(\sqrt{n+2}\) is a rational number, these orbits are closed. Sketch them for the cases that \(n=2,-1,\) and 7.

Consider two particles of equal masses, \(m_{1}=m_{2},\) attached to each other by a light straight spring (force constant \(k\), natural length \(L\) ) and free to slide over a frictionless horizontal table. (a) Write down the Lagrangian in terms of the coordinates \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\), and rewrite it in terms of the \(\mathrm{CM}\) and relative positions, \(\mathbf{R}\) and \(\mathbf{r},\) using polar coordinates \((r, \phi)\) for \(\mathbf{r} .\) (b) Write down and solve the Lagrange equations for the CM coordinates \(X, Y\). (c) Write down the Lagrange equations for \(r\) and A. Solve these for the two special cases that \(r\) remains constant and that \(\phi\) remains constant. Describe the corresponding motions. In particular, show that the frequency of oscillations in the second case is \(\omega=\sqrt{2 k / m_{1}}\).

An earth satellite is observed at perigee to be \(250 \mathrm{km}\) above the earth's surface and traveling at about \(8500 \mathrm{m} / \mathrm{s}\). Find the eccentricity of its orbit and its height above the earth at apogee. [Hint: The earth's radius is \(R_{e} \approx 6.4 \times 10^{6} \mathrm{m} .\) You will also need to know \(G M_{\mathrm{e}},\) but you can find this if you remember that \(\left.G M_{\mathrm{e}} / R_{\mathrm{e}}^{2}=g .\right]\)

Two particles whose reduced mass is \(\mu\) interact via a potential energy \(U=\frac{1}{2} k r^{2},\) where \(r\) is the distance between them. (a) Make a sketch showing \(U(r),\) the centrifugal potential energy \(U_{\mathrm{cf}}(r)\) and the effective potential energy \(U_{\text {eff }}(r) .\) (Treat the angular momentum \(\ell\) as a known, fixed constant.) (b) Find the "equilibrium" separation \(r_{\mathrm{o}},\) the distance at which the two particles can circle each other with constant \(r .\left[\text { Hint: This requires that } d U_{\text {eff }} / d r \text { be zero. }\right]\left(\text { c) By making a Taylor expansion of } U_{\text {eff }}(r)\right.\) about the equilibrium point \(r_{\mathrm{o}}\) and neglecting all terms in \(\left(r-r_{\mathrm{o}}\right)^{3}\) and higher, find the frequency of small oscillations about the circular orbit if the particles are disturbed a little from the separation \(r_{\mathrm{o}}.\)

Prove that for circular orbits around a given gravitational force center (such as the sun) the speed of the orbiting body is inversely proportional to the square root of the orbital radius.

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