PHY293 Oscillations Lecture #13

Transcription

1 PHY293 Oscillations Lecture #13 October 8, Third problem set due in drop-boxes next Tuesday (October 12) at 17: 2. Looked at animations of longitudinal/sound waves at: drussell/demos/waves/wavemotion.html wave.html Begin Lecture material 1. Initial Value Problems with Waves Saw last time the full standing wave solution: y(x, t) = Σ n=1[c n sin(k n x) cos(ω n t) + D n sin(k n x) sin(ω n t) This is more convenient than the cos(ω n t + α n ) version as initial conditions are easily related to C n from y(x, ) s(x) D n from ẏ(x, ) v(x) Which gives: C m = sin(k mx)s(x)dx sin2 (k m x)dx This is the continuum analog of the expression for eigenvector coefficients for discrete oscillators: B n = ξ n x / ξ n 2 Similarly we can find the D m from: Example D m = 1 ω m sin(k mx)v(x)dx sin2 (k m x)dx Consider a string that starts out in the following shape s(x) = 4 sin(3πx/l) Sort of a weird initial shape for the string, but makes the coefficient math easier to solve In this case assume v(x) = ie. the string is motionless in this shape at t = but still under tension so it will be pulled back towards y = along its whole length Notice that this shape of the string already crosses y = at four places x = (, L/3, 2L/3, L) In particular y = at both ends of the string, so our usual solution will work: y(x, t) = Σ n=1[c n sin(k n x) cos(ω n t) + D n sin(k n x) cos(ω n t) It is just a matter of doing some integrals with the initial conditions to find the coefficients C n and D n From v(x) = immediately conclude that all D n = since all integrals involve integrating over v(x) sin(k n x) So we just have to look at the integrals for the C n : C n = sin(k nx)s(x)dx sin2 (k n x)dx = 2 L 4 sin(k n x) sin(3πx/l)dx But this integral vanishes unless k n = 3π/L k 3, when it will be equal to L 2 So the only coefficient that will enter in this solution is C 3 = 4 giving: y(x, t) = 4 sin(3πx/l) cos(ω 3 t) ω 3 = 3πc/L

2 We introduced the concept of the frequency spectrum for this solution, but it was a pretty simple in this case. It is just a plot of C n versus ω in this case the whole graph is except at ω 3 = 3πc/L where we plot a 4. In subsequent examples we ll see less trivial frequency spectra are necessary to describe more complicated initial conditions 2. The Plucked String In this case use: s(x) = 2x/L ( x L/2) and s(x) = 2 2x/L (L/2 < x L) This describes a string, fixed at y(, t) = y(l, t) = (ie. at both ends) drawn aside a distance y(l/2, t) = 1 in the middle much as a guitar player, or cellist, plucking a string might do Here we will also take v(x) = ie. the string stationary at t =. This string satisfies our usual boundary conditions (fixed at x = and x = L) so again use our usual solution: Again the D n = for all n as v(x) = To find the full solution we just have to do some integrals: y(x, t) = Σ n=1[c n sin(k n x) cos(ω n t) + D n sin(k n x) cos(ω n t) C m = sin(k mx)s(x)dx sin2 (k m x)dx Notice that we will always get sin 2 k m xdx in the denominator of this expression This will always be L/2 for k n = nπ/l so we can simplify our expression for the C m to: C m = 2 L If we substitute the string s initial position in this case, s(x) we get C m = 2 L Ł/2 sin(k m x)s(x)dx 2x L sin(k mx)dx + 2 (2 2x/L) sin(k m x)dx L L/2 To do integrals of the type x sin(x)dx we have to use integration by parts consider: d(uv) = (du)v + udv d(uv) = vdu + udv thus [uv b a = vdu + udv a a udv = vdu + [uv b a In our case substitute θ = mπx L mπ and dθ = dx to get: L I = ( ) 2 ( ) 2 2 L mπ/2 L mπ θ sin(θ)dθ To do integration by parts take: u = θ and dv = sin(θ)dθ This gives du = dθ and v = cos θ Thus: θ sin θdθ = a a cos θdθ + [ θ cos θ b a

3 With a = and b = mπ/2 this gives: I = 4 [ m 2 π 2 θ sin θdθ = 4 [ m 2 π 2 cos θdθ + [ θ cos θ mπ/2 = 4 [ m 2 π 2 [sin θ mπ/2 + [ θ cos θ mπ/2 = 4 [ m 2 π 2 sin(mπ/2) + [( mπ/2) cos(mπ/2) + = 4 [ m 2 π 2 sin(mπ/2) mπ/2 cos(mπ/2) = 4 sin(mπ/2) 2mπ cos(mπ/2) m 2 π 2 Plugging this back into our expression for C m (and adding the second integral that is to be done) we get: 4 sin(mπ/2) 2mπ cos(mπ/2) C m = m 2 π 2 = 8 m 2 π 2 sin(mπ/2) = 8 [1,, 1/9,, 1/25,, 1/49,... π2 + 4 sin(mπ/2) + 2mπ cos(mπ/2) m 2 π 2 The terms in square brackets are the Fourier Coefficients for the expansion of a triangle wave We can plug the first few terms back into our full expression for y(x, t) y(x, t) = ΣC m sin(mπx/l) cos(mπct/l) = 8 [sin(πx/l) cos(πct/l) 1/9 sin(3πx/l) cos(3πct/l) + 1/25 sin(5πx/l) cos(5πct/l)... π2 Plucking a string in the middle gives a mixture of standing waves with a dominant frequency ω 1 = cπ/l and a series of over-tones at 3ω 1, 5ω 1, 7ω 1 etc. We also plotted the frequency spectrum for this solution Every other term vanishes (for n = 2, 4, 6,...) and the terms that survive alternate in sign and fall off like 1/m 2. We can do Fourier Coefficient determinations for any initial shape. Showed want the resulting functions look for a finite number of terms for a square-wave and sawtooth initial shapes (see plots attached at the end of these notes)

4 I(x) = 8 π2[sin(πx/l) 1/9sin(3πx/L) + 1/25sin(5πx/L) +...

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