Ch. 3. Sect. 3.3: Harmonic Oscillations in Two Dimensions. Sect. 3.4: Phase Diagrams. - PowerPoint PPT Presentation


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Ch. 3. Sect. 3.3: Harmonic Oscillations in Two Dimensions. Sect. 3.4: Phase Diagrams.


Title: Lecture 12 Subject: Ch. 3. Sect. 3.3: Harmonic Oscillations in Two Dimensions. Sect. 3.4: Phase Diagrams. Author: Charles W. Myles Last modified by – PowerPoint PPT presentation

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Title: Ch. 3. Sect. 3.3: Harmonic Oscillations in Two Dimensions. Sect. 3.4: Phase Diagrams.

(No Transcript)
SHO in 2d Sect. 3.3
  • Look at particle motion in 2d (xy plane), under a
    force, linear in the displacement (Hookes
  • For simplicity, assume the force constants the
    are same in the x y directions.
  • Can do the problem in plane polar coordinates or
    in rectangular coordinates
  • F -kr ? Fx -kx -kr cos?
  • Fy -ky -kr sin?
  • Lets work in rectangular coordinates
  • Newtons 2nd Law equations Components of
  • F -kr mr ma

  • x y components of Newtons 2nd Law
  • -kx mx -ky my
  • Or x (?0)2 x 0 y (?0)2 y 0
  • with (?0)2 ? k/m
  • Solutions x(t) A cos(?0t - a)
  • y(t) B cos(?0t - ß)
  • The amplitudes A, B, the phases a, ß are
    determined (similar to 1d) by initial conditions.
  • The motion is simple harmonic in each of the 2 d.
    Both oscillations at the same frequency but (in
    general) different amplitudes.

  • The equation for the path in the xy plane is
    obtained by eliminating t in x(t) A cos(?0t -
    a) and
  • y(t) B cos(?0t - ß). Defining d ? a - ß,
    algebra gives B2x2 -2ABxy cosd A2y2
  • The general path is complicated! Some special
  • d ?p/2 ? An ellipse (x/A)2 (y/B)2 1
  • If A B, this is a circle.
  • d 0 ? A straight line y (B/A) x
  • d ?p ? A straight line y -(B/A) x
  • Except for special cases, the general path is an

Paths in the xy Plane for A B Various d
  • More general motion in 2d (xy plane), under a
    Hookes Law force
  • Assume different force constants for x y
  • Fx -kx x Fy -ky y
  • Newtons 2nd Law equations
  • -kx x mx and -ky y my
  • Or x (?x)2x 0 and y (?y)2 y 0
  • with the definitions (?x)2 ? (kx/m), (?y)2 ?
  • Solutions x(t) A cos(?xt - a) y(t) B
    cos(?yt - ß)

  • x(t) A cos(?xt - a) y(t) B cos(?yt - ß)
  • Path in the xy plane is no longer an ellipse, but
    a Lissajous curve.
  • If the motion repeats itself in regular time
    intervals, this curve is closed.
  • This will happen only if (?x/?y) is a rational
    fraction (if the frequencies are ?
  • If (?x/?y) ? a rational fraction, the curve will
    be open. In this case, the mass will never pass
    twice through the same point with the same
  • ? An infinitesimal change in initial conditions
    can result in qualitatively different motion (a
    possible sign of chaos, discussed in Ch. 4!)

  • A typical Lissajous figure A B, ?y (¾)?x, a

  • If the 2 frequencies arent commensurable (if
    their ratio deviates from a rational fraction by
    even an infinitesimal amount)
  • The path in the xy plane will not be closed it
    will eventually fill a rectangle of size
  • 2A ? 2B
  • For the path to be closed, (?x/?y) must be a
    rational fraction to infinite precision!
  • The shape of the Lissajous curve strongly depends
    on the phase difference d ? a - ß

  • Some typical Lissajous figures
  • A B ?y 2?x d 0, (p/3), (p/2)

Phase Diagrams Sect. 3.4. Back to 1d!
  • 1d Oscillator Because Newtons 2nd Law eqtn of
    motion is a 2nd order diff. eqtn, the state of
    motion of the oscillator is completely specified
    if two quantities are given at an initial time
    t0 x(t0), v(t0) x(t0).
  • ? Its convenient useful to consider x(t) x(t)
    v(t) to be coordinates in an abstract 2d phase
  • For a 3d particle, the phase space would be 6
  • At any time t, the 1d oscillator motion is
    completely specified by specifying a point in
    this 2d phase space (the x - v or x - x plane).

  • At time t, the oscillator motion is specified by
    specifying a point P P(x,x) in this phase
  • As time progresses, P P(x,x) will move in this
    plane trace out a phase path or phase
  • Different initial conditions ? Different phase
  • The totality of all possible phase paths of the
    particle ? Phase Portrait or Phase Diagram of the
    particle. Studying such diagrams gives insight
    into the physics of the particle motion.
  • This concept isnt limited to oscillators, but is
    clearly valid for any particle.
  • A very useful concept in statistical mech (Phys.

  • Look in detail at the phase diagram for the 1d
    simple harmonic oscillator
  • x(t) A sin(?0t - d) v(t) x(t) ?0A
    cos(?0t - d)
  • Eliminating t from these 2 equations gives
  • x2/A2 x2/(A2?02) 1
  • This is a family of ellipses in the x - x plane!
  • The phase diagram for the1d oscillator a family
    of ellipses, each of which is a separate phase
    path, for different initial conditions.

  • The phase diagram for the 1d oscillator is a
    family of ellipses, as in the Figure.
  • Note Oscillator total energy E (½)kA2.
  • Also, ?02 (k/m) ? The ellipse equation can be
  • x2/(2E/k) x2/(2E/m) 1

  • Writing the elliptical phase path as
  • x2/(2E/k) x2/(2E/m) 1
  • ? PHYSICS Each phase path (ellipse) corresponds
    to a definite total energy E of the oscillator
    (different initial conditions!).
  • No 2 phase paths of the oscillator can cross!
  • If they could cross, this would mean that for
    given initial conditions x(t0), v(t0), the motion
    could proceed on different phase paths. This is
    impossible, since the solutions to a linear 2nd
    order differential equation are unique!

  • In constructing a phase diagram Choose x as the
    x-axis x v as the y-axis.
  • The motion of a typical point P(x,x) is always
    clockwise! Because, for a harmonic oscillator,
    for x gt 0, x v decreases if x lt 0, x v
  • Earlier we obtained x(t) A sin(?0t - d)
  • v(t) x(t) ?0A cos(?0t - d) by integrating
    Ns 2nd Law Eq. (a 2nd order differential
  • (d2x/dt2) (?0)2 x 0
  • But, we can get the phase path in a simpler way.

  • Can use (dx/dt) x v (1)
  • (dx/dt) (dv/dt) - (?0)2x (2)
  • Divide (2) by (1) (dx/dx) - (?0)2(x/x)
  • A 1st order differential eqtn for x(x) v(x)
  • xdx - (?0)2xdx
  • Solution x2/A2 x2/(A2?02) 1 as before!
  • For the SHO, we can easily use either the 2nd
    order differential eqtn or the 1st order one just
    described. For motion in more complicated
    situations, its sometimes easier to find the
    solution to the path directly from a 1st order