Transcript Sects. 3.3 & 3.4

SHO in 2d
Sect. 3.3
• Look at particle motion in 2d (xy plane), under a
force, linear in the displacement (Hooke’s “Law”):
– 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 cosθ
Fy = -ky = -kr sinθ
– Lets work in rectangular coordinates
– Newton’s 2nd Law equations: Components of
F = -kr = mr = ma
• x & y components of Newton’s 2nd Law:
-kx = mx
-ky = my
Or:
x + (ω0)2 x = 0 y + (ω0)2 y = 0
with (ω0)2  k/m
• Solutions:
x(t) = A cos(ω0t - α)
y(t) = B cos(ω0t - β)
– The amplitudes A, B, & the phases α, β 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(ω0t - α) and
y(t) = B cos(ω0t - β). Defining δ  α - β, algebra
gives: B2x2 -2ABxy cosδ +A2y2 =A2B2sin2δ
• The general path is complicated! Some special cases:
– δ = π/2
 An ellipse: (x/A)2 + (y/B)2 = 1
• If A = B, this is a circle.
–δ=0
 A straight line: y = (B/A) x
– δ = π
 A straight line: y = -(B/A) x
– Except for special cases, the general path is an
ellipse!
Paths in the xy Plane for A = B & Various δ
• More general motion in 2d (xy plane), under a
Hooke’s “Law” force:
– Assume different force constants for x & y
Fx = -kx x
Fy = -ky y
– Newton’s 2nd Law equations:
-kx x = mx and -ky y = my
Or:
x + (ωx)2x = 0 and y + (ωy)2 y = 0
with the definitions: (ωx)2  (kx/m), (ωy)2  (ky/m)
• Solutions: x(t) = A cos(ωxt - α) y(t) = B cos(ωyt - β)
x(t)= A cos(ωxt - α) y(t)= B cos(ωyt - β)
• 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 (ωx/ωy) is a rational
fraction (if the frequencies are  “commensurable”)
– If (ωx/ωy)  a rational fraction, the curve will be
open. In this case, the mass will never pass twice through
the same point with the same velocity.
 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, ωy = (¾)ωx, α = β
• If the 2 frequencies aren’t 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, (ωx/ωy) must be a
rational fraction to infinite precision!
• The shape of the Lissajous curve strongly
depends on the phase difference δ  α - β
• Some typical Lissajous figures:
A = B; ωy = 2ωx; δ = 0, (π/3), (π/2)
Phase Diagrams
Sect. 3.4. Back to 1d!
• 1d Oscillator: Because Newton’s 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 space.
– For a 3d particle, the phase space would be 6 dimensional!
• 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 space.
• As time progresses, P = P(x,x) will move in this
plane & trace out a phase path or phase trajectory.
Different initial conditions  Different phase paths
• 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 isn’t limited to oscillators, but is clearly valid
for any particle.
– A very useful concept in statistical mech (Phys. 4302)
• Look in detail at the phase diagram for the 1d
simple harmonic oscillator:
x(t) = A sin(ω0t - δ) v(t) = x(t) = ω0A cos(ω0t - δ)
– Eliminating t from these 2 equations gives:
[x2/A2] + [x2/(A2ω02)] = 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, ω02 = (k/m)  The ellipse equation can be written:
[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 >
0, x = v decreases & if x < 0, x = v increases!
• Earlier we obtained x(t) = A sin(ω0t - δ) &
v(t) = x(t) = ω0A cos(ω0t - δ) by integrating N’s 2nd
Law Eq. (a 2nd order differential equation):
(d2x/dt2)+ (ω0)2 x = 0
• But, we can get the phase path in a simpler way.
• Can use:
(dx/dt) = x = v
&
(dx/dt) = (dv/dt) = - (ω0)2x
(1)
(2)
• Divide (2) by (1): (dx/dx) = - (ω0)2(x/x)
A 1st order differential eqtn for x(x) = v(x):
xdx = - (ω0)2xdx
• Solution: [x2/A2] + [x2/(A2ω02)] = 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 equation.