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Silly String. Starring. Zeb. Jeff. Mike. Bubba. Introduction. Descriptions Statements. A string can be defined as a rigid body whose dimensions are small when compared with its length .
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Silly String Starring
Introduction • Descriptions • Statements
A string can be defined as a rigidbody whose dimensions are small when compared with its length.
The string in our model will be stretched between two fixed pegs that are separated by a distance of length L. Peg 1 Peg 2 L
Tension (T0) will be the force of the two pegs pulling on the string. For our model, we will assume near constant tension.
Density can be defined as the ratio of the mass of an object to its volume. For a string, density is mass per unit length. In our model, we will also assume near constant density for the string.
Derivation of the Wave Equation • Basic modeling assumptions • Review of Newton’s Law • Calculus prereqs • Equational Derivation
Model Assumptions L Transverse: Vibration perpendicular to the X-axis
Model Assumptions • Density is assumed constant = 1 • Initial Deformation is small
Model Assumptions L T = 1 Tension T is constant and tangent to the curve of the string
Newton’s Second Law F = ma
y Angle of Inclination j i |T| T x [ ] ( ) ( ) T 1 = dy dy dy + ( ) ² ( ) ² 1 + dx dx dx 1 + Calculus Prerequisites y = f(x)
u s x x x+x Equational Derivation
u Horizontal Forces s s x x x+x Vertical Forces Equational Derivation
Vertical Forces: [ ] (x + x, t) - T u u u u x x x x ( ) 2 1 + ( ) 2 (x + x, t) (x, t) 1 + (x, t) Get smaller and go to zero
Vertical Forces: [ ] - (x + x, t) T 1 1 u u x x (x, t) ²u = (s) (x,t) t²
Vertical Forces: Net Force - u u (x + x, t) (x, t) x x ²u Mass = s (x,t) t² Acceleration
Vertical Forces = s (x,t) x - u u (x + x, t) (x, t) x x x ²u t²
Vertical Forces ²u = (x,t) (x, t) x² ²u t² One dimensional wave equation
Solution to the Wave Equation • Partial Differential equations • Multivariate Chain rule • D’Alemberts Solution • Infinite String Case • Finite String Case • Connections with Fourier Analysis
²y ²y y y ²y 0 A + B + C + D + E + F y = x² t² x xx t 2nd Order Homogeneous Partial Differential Equation
Classification of P.D.E. types • = B² - AC • Hyperbolic > 0 • Parabolic = 0 • Elliptic < 0
Boundary Value Problem • Finite String Problem • Fixed Ends with 0 < x < l • [u] = 0 and [u] = 0 X = 0 X = l
Cauchy Problem • Infinite String Problem • Initial Conditions • [u] = (x) and [du/dt] (x) = t=0 0 t=0 l
Multi-Variable Chain Rule example f(x,y) = xy² + x² g(x,y) = y sin(x) h(x) = e F(x,y) = f(g(x,y),h(x)) x
Let u = g(x,y) v = h(x) So F = f(u,v) = uv² + u²
Variable Dependency Diagram g x f u x u g F y f h v x v y
F f g f u = + x u x v x x = ((v² + 2u)(y cos(x)) + (2uv)e ) = x (e )² + 2y sin(x) (y cos(x)) + 2(y sin(x) e e ) x x
Multi-Variable Chain Rule for Second Derivative Very Similar to that of the first derivative
Our Partial Differential Equation ξ = x – t η = x + t So ξ + η = 2x x = (ξ + η)/2 And - ξ + η = 2t t = (η – ξ)/2
Using Multi-Variable Chain Rule u u u = + t ξ η u [ ] - u ²u = + η t² t ξ
Using Algebra to reduce the equation ²u ²u ²u 2 ²u - + = η² t² ξ² ηξ
u u u = + x ξ η u u ²u [ ] = + x² x ξ η
Using Algebra to simplify ²u 2 ²u ²u ²u = + + x² ξ² ηξ η²
Substitute What we just found ²u ²u = t² x²
²u 2²u ²u ²u 2 ²u ²u + = + + ξ² ηξ η² ξ² ηξ η² 2 ²u 2 ²u = ηξ ηξ 4 ²u = 0 ηξ
We finally come up with ²u = 0 ηξ When u = u(ξ, η) ξ = x - t η = x + t
Sorry, we don’t sell to strings here Can I get a Beer? A String walks into a bar
Again we can’t serve you because you are a string Can I get a beer? I’m afraid not!
Integrating with respect to Ada Next integrating with respect to Xi Relabeling in more conventional notation gives Then unsubstituting
Infinite String Solution Which is a cauchy problem Reasonable initial conditions