A particle of mass m and angular momentum ℓmoves in a central spring-like force field F=-k r. (a

A particle of mass m and angular momentum ℓmoves in a...

Key Concepts

Effective Potential Energy

In central force problems, the effective potential energy combines the actual potential energy of the central force with the contribution from the centrifugal potential due to angular momentum. This concept transforms a two-dimensional orbital problem into an equivalent one?dimensional problem in the radial coordinate, simplifying the analysis of particle motion and stability of orbits.

Circular Orbits in Central Force Problems

Circular orbits occur when the radial force balance is achieved at a specific radius where the derivative of the effective potential energy with respect to the radial coordinate equals zero. This condition ensures that the inward central force exactly balances the outward centrifugal force, resulting in a stable circular path for the particle.

Small Oscillations About Equilibrium

When a particle in a circular orbit is slightly perturbed, its radial motion can be approximated by a harmonic oscillator through a linearization process. The curvature (second derivative) of the effective potential at the equilibrium point determines the frequency of these small oscillations. This small oscillation analysis is a fundamental tool in studying the stability and dynamics of perturbed orbits.

Closed versus Open Orbits in Central Force Motion

The relationship between the period of small radial oscillations and the period of the angular motion about the center drives the global orbital behavior. If these periods are commensurate (i.e., in a rational ratio), the orbit is closed, meaning the particle retraces its path. Otherwise, the orbit is open and does not exactly repeat. This concept is central to understanding orbital resonances and is closely related to Bertrand's theorem in classical mechanics.

Transcript

00:01 In this problem, we have a mass spring system which is oscillating on a frictionless track, and we're told that the velocity of the spring will vary linearly with its displacement.

00:13 And we're asked to find first the total kinetic energy of the system, and second, the period of oscillation of the system.

00:22 So how we'll do this problem? for the first part, finding the total kinetic energy, the total kinetic energy will be the sum of the kinetic energy of the block, the mass capital m, which is just simply one -half capital m v squared, plus the kinetic energy of the spring, and it will be a little more complicated to find the k of the spring.

00:47 So what we'll do is we'll break it into a segment.

00:51 So we'll look at the segment dx and we'll find the kinetic energy for that segment and then add up all those segments.

00:59 So integrating over the whole spring to find the kinetic energy of the spring.

01:04 And then we'll have our total energy.

01:06 And for part b to find the period of oscillation, we'll need to recognize when potential energy and kinetic energy are maximums.

01:15 And use conservation of total energy to relate the two of those to each other.

01:21 And then we'll look at our position and velocity expressions for a simple harmonic oscillator.

01:28 We'll be able to get our angular frequency and our period from those relations.

01:36 Okay, so part a looking at our spring.

01:40 So we're given expressions for velocity vx.

01:45 We can say equals x over l times v where l is the maximum displacement of our spring and we're given dm so if we took in a segment of mass dm we could say that was equal to small m total mass of the spring divided by the displacement l times d x so we can say our k dx so kinetic energy for this segment dx would be equal to one so one half mv squared so our mass of that segment is dm and our velocity is vx right and so then we'll rewrite that with our given expressions for dm and vx so 1 1 .5m over l dx times x squared over l squared.

02:54 And that will become mv squared over 2l cubed times an x squared dx.

03:10 So total kinetic energy of our spring will be integrating from length l to zero this mv squared over 2l cubed and x squared dx.

03:32 And so that will give us an mv squared over 2l cubed.

03:45 Will be x cubed over 3 from an l to a 0 and that works out to little m v squared over 6 and k of our block just simply one -half capital m v squared so our k total is equal to the kinetic energy of the spring plus k of the block.

04:24 And so you see, we'll have a v squared in common there.

04:28 And we can write that as one half m plus capital m plus little m over three times v squared.

04:40 All right.

04:41 So that will be our total kinetic energy for part a.

04:47 And then part b.

04:50 So we want to find the period of oscillation.

04:55 So to get there, first we want to think about our, what we know for energy.

05:03 So we can say our total energy is going to be a constant, and it's going to be equal to the potential energy u plus the potential energy k.

05:16 And we can say a few things about umax and kmax...