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  F x = -kx = -kr cosθ F y = -ky = -kr sinθ –Lets work in rectangular coordinates –Newton’s 2 nd Law equations: Components of F = -kr = mr = ma

x & y components of Newton’s 2 nd Law: -kx = mx -ky = my Or:x + (ω 0 ) 2 x = 0 y + (ω 0 ) 2 y = 0 with (ω 0 ) 2  k/m Solutions: x(t) = A cos(ω 0 t - α) y(t) = B cos(ω 0 t - 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(ω 0 t - α) and y(t) = B cos(ω 0 t - β). Defining δ  α - β, algebra gives: B 2 x 2 -2ABxy cosδ +A 2 y 2 =A 2 B 2 sin 2 δ 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 F x = -k x x F y = -k y y –Newton’s 2 nd Law equations: -k x x = mxand -k y y = my Or: x + (ω x ) 2 x = 0 and y + (ω y ) 2 y = 0 with the definitions: (ω x ) 2  (k x /m), (ω y ) 2  (k y /m) Solutions: x(t) = A cos(ω x t - α) y(t) = B cos(ω y t - β)

x(t)= A cos(ω x t - α) y(t)= B cos(ω y t - 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 (ω x /ω y ) is a rational fraction ( if the frequencies are  “commensurable”) –If (ω x /ω 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 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, ω y = (¾)ω x, 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, (ω x /ω y ) 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; ω y = 2ω x ; δ = 0, (π/3), (π/2)

Phase Diagrams Sect Back to 1d! 1d Oscillator: Because Newton’s 2 nd Law eqtn of motion is a 2 nd order diff. eqtn, the state of motion of the oscillator is completely specified if two quantities are given at an initial time t 0 : x(t 0 ), v(t 0 ) = x(t 0 ).  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(ω 0 t - δ) v(t) = x(t) = ω 0 A cos(ω 0 t - δ) –Eliminating t from these 2 equations gives: [x 2 /A 2 ] + [x 2 /(A 2 ω 0 2 )] = 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)kA 2. Also, ω 0 2 = (k/m)  The ellipse equation can be written: [x 2 /(2E/k)] + [x 2 /(2E/m)] = 1

Writing the elliptical phase path as [x 2 /(2E/k)] + [x 2 /(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(t 0 ), v(t 0 ), the motion could proceed on different phase paths. This is impossible, since the solutions to a linear 2 nd 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(ω 0 t - δ) & v(t) = x(t) = ω 0 A cos(ω 0 t - δ) by integrating N’s 2 nd Law Eq. (a 2 nd order differential equation): (d 2 x/dt 2 )+ (ω 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) = - (ω 0 ) 2 x (2) Divide (2) by (1): (dx/dx) = - (ω 0 ) 2 (x/x) A 1 st order differential eqtn for x(x) = v(x): xdx = - (ω 0 ) 2 xdx Solution: [ x 2 /A 2 ] + [x 2 /(A 2 ω 0 2 )] = 1 as before! For the SHO, we can easily use either the 2 nd order differential eqtn or the 1 st order one just described. For motion in more complicated situations, its sometimes easier to find the solution to the path directly from a 1 st order equation.