Two particles with masses m1 and m2 are joined by a light...

Two particles with masses $m_{1}$ and $m_{2}$ are joined by a light spring of force constant $k .$ They vibrate along a straight line with their center of mass fixed. (a) Show that the total energy $$\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}+\frac{1}{2} k x^{2}$$ can be written as $\frac{1}{2} \mu u^{2}+\frac{1}{2} k x^{2},$ where $u=\left|u_{1}\right|+\left|u_{2}\right|$ is the relative speed of the particles and $\mu=m_{1} m_{2} /\left(m_{1}+m_{2}\right)$ is the reduced mass of the system. This result demonstrates that the pair of freely vibrating particles can be precisely modeled as a single particle vibrating on one end of a spring that has its other end fixed. (b) Differentiate the equation $$\frac{1}{2} \mu u^{2}+\frac{1}{2} k x^{2}=$ $ constant with respect to $x$ . Proceed to show that the system executes simple harmonic motion. (c) Find its frequency.

Two particles with masses $m_{1}$ and $m_{2}$ are joined by a light spring of force constant $k .$ They vibrate along a straight line with their center of mass fixed. (a) Show that the total energy
$$\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}+\frac{1}{2} k x^{2}$$
can be written as $\frac{1}{2} \mu u^{2}+\frac{1}{2} k x^{2},$ where $u=\left|u_{1}\right|+\left|u_{2}\right|$ is the relative speed of the particles and $\mu=m_{1} m_{2} /\left(m_{1}+m_{2}\right)$ is the reduced mass of the system. This result demonstrates that the pair of freely vibrating particles can be precisely modeled as a single particle vibrating on one end of a spring that has its other end fixed. (b) Differentiate the equation
$$\frac{1}{2} \mu u^{2}+\frac{1}{2} k x^{2}=$ $
constant with respect to $x$ . Proceed to show that the system executes simple harmonic motion. (c) Find its frequency.

Key Concepts

Reduced Mass

In the analysis of two-body problems, the reduced mass is a crucial concept that enables one to simplify the system’s dynamics by replacing it with a single effective mass. It is defined as the product of the two masses divided by their sum and plays a key role when dealing with interactions like the spring force or gravitational force between the particles.

Energy Partitioning

This concept distinguishes between kinetic energy and potential energy in a physical system. In the context of coupled oscillators, kinetic energy is associated with the motion of each mass, and potential energy is related to the distortion of the spring. Partitioning the total energy into these two components is essential for forming an equivalent single-particle model.

Coordinate Transformation to a Single Particle Model

Transforming the equations of motion using relative coordinates and the reduced mass allows for the re-formulation of a two-body interaction as an equivalent one-body problem. This transformation simplifies the analysis, as the dynamics can then be studied using the familiar framework of a single particle oscillating in a potential well.

Simple Harmonic Motion (SHM)

SHM describes a system in which the restoring force is directly proportional to the displacement from equilibrium, resulting in sinusoidal oscillations. By differentiating the energy expression and comparing it with the standard form of a simple harmonic oscillator, one can confirm that the model follows SHM dynamics.

Frequency of Oscillation

The frequency of oscillation in a simple harmonic oscillator is determined by the stiffness of the spring (force constant) and the effective inertial parameter (mass or reduced mass). Computing the angular frequency from these parameters provides a direct link between the physical properties of the system and its temporal oscillatory behavior.

Transcript

00:01 Assuming that the center of mass does not move, we have for the two particles that the total momentum will be zero.

00:08 So the two particles will either be approaching each other, moving away from each other, or will both be at rest, if only momentarily.

00:16 So u1 is the velocity of one particle, and u2 is the velocity of the other particle, and m1 and m2 are their masses.

00:24 So we then have that from this relation, we can obtain one velocity in terms of the other.

00:32 For example, u2 will equal negative m1 over m2 times u1.

00:39 Or the magnitude of u2 simply equals m1 over m2 times the magnitude of u1.

00:46 The total velocity will then be the magnitudes, really the speed, the total speed, will be the magnitudes added together.

00:56 And when you substitute in the magnitude of u2 in terms of the magnitude of u1, you get the following relation solely in terms of the magnitude u1, which can be rewritten in the following way.

01:18 So now you can rewrite that, and now we see that u is equal to the sum of the masses divided by m2 times the magnitude of u1.

01:30 That can be then multiplied and rewritten so that the magnitude u1 is written as mass 2 times u divided by the sum of the masses.

01:44 Also, we have some more relations we can obtain.

01:50 So, for example, from the total momentum equaling 0, we have that we can work the other way.

01:58 We get that the total speed can be rewritten in terms of the magnitude of velocity u2 as well, instead of being rewritten in terms of the magnitude of velocity u1.

02:28 And this is analogous to how it was rewritten in terms of u1.

02:33 So this time, it'll be, you will be written as the sum of the masses divided by m1, instead of being divided by m2, times the magnitude of u2.

02:47 And then the speed for u2 would be equal to m1 times the speed u u divided by the total of the masses, the sum of the masses.

03:02 So we can use these two relations, these two equations, or expressions for the speeds u1 and u2, to get an equation for the kinetic energy and rewrite it.

03:16 So we substitute them into the squares of u1 and u2.

03:21 In the kinetic energy.

03:24 And the result that we get is the following.

03:35 So we see in this first step, we substitute looking at the expressions for u1 and u2.

03:42 And what we have is m2 and m2 and m2 is m2 and m1 squared in the numerator as well as u squared.

03:50 And then we have the sum of the mass is squared in the denominator.

03:56 And of course the potential term is unaffected.

03:59 It only depends on x squared and x is the relative position between the two particles.

04:06 So continuing with this, we'll be able to do some factoring between these two terms and the kinetic energy...

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