Lecture Notes for Math 251: ODE and PDE. Lecture 34: 10.7 Wave Equation and Vibrations of an Elastic String

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1 ecture Notes for Math 251: ODE and PDE. ecture 3: 1.7 Wave Equation and Vibrations of an Eastic String Shawn D. Ryan Spring 212 ast Time: We studied other Heat Equation probems with various other boundary conditions. 1 The Wave Equation 1.1 Derivation of the Wave Equation Consider a competey fexibe string of ength and constant density ρ. We wi assume that the string wi ony undergo reativey sma vertica vibrations, so that points do not move from side to side. An exampe might be a pucked guitar string. Thus we can et u(x, t) be its dispacement from equiibrium at time t. The assumption of compete fexibiity means that the tension force is tangent to the string, and the string itsef provides no resistance to bending. This means the tension force ony depends on the sope of the string. Take a sma piece of string going from x to x + x. et Θ(x, t) be the ange from the horizonta of the string. Our goa is to use Newton s Second aw F = ma to describe the motion. What forces are acting on this piece of string? (a) Tension puing to the right, which has magnitude T(x + x, t) and acts at an ange of Θ(x + x, t) from the horizonta. (b) Tension puing to the eft, which has magnitude T(x, t) and acts at an ange of Θ(x, t) from the horizonta. (c) Any externa forces, which we denote by F(x, t). Initiay, we wi assume that F(x, t) =. The ength of the string is essentiay ( x) 2 + ( u) 2, so the vertica component of Newton s aw says that ρ ( x) 2 + ( u) 2 u tt (x, t) = T(x + x, t) sin(θ(x + x, t)) T(x, t) sin(θ(x, t)). (1) Dividing by x and taking the imit as x, we get ρ 1 + (u x ) 2 u tt (x, t) = T(x, t) sin(θ(x, t))]. (2) We assumed our vibrations were reativey sma. This means that Θ(x, t) is very cose to zero. As a resut, sin(θ(x, t)) tan(θ(x, t)). Moreover, tan(θ(x, t)) is just the sope of the string 1

2 u x (x, t). We concude, since Θ(x, t) is sma, that u x (x, t) is aso very sma. The above equation becomes ρu tt (x, t) = (T(x, t)u x (x, t)) x. (3) We have not used the horizonta component of Newton s aw yet. Since we assume there are ony vertica vibrations, our tiny piece of string can ony move verticay. Thus the net horizonta force is zero. T(x + x, t) cos(θ(x + x, t)) T(x, t) cos(θ(x, t)) =. () Dividing by x and taking the imit as x yieds T(x, t) cos(θ(x, t))] =. (5) Since Θ(x, t) is very cose to zero, cos(θ(x, t)) is cose to one. thus we have that T (x, t) is cose to zero. So T(x, t) is constant aong the string, and independent of x. We wi aso assume that T is independent of t. Then Equation (??) becomes the one-dimensiona wave equation where c 2 = T ρ. 1.2 The Homogeneous Dirichet Probem u tt = c 2 u xx (6) Now that we have derived the wave equation, we can use Separation of Variabes to obtain basic soutions. We wi consider homogeneous Dirichet conditions, but if we had homogeneous Neumann conditions the same techniques woud give us a soution. The wave equation is second order in t, unike the heat equation which was first order in t. We wi need to initia conditions in order to obtain a soution, one for the initia dispacement and the other for the initia speed. The reevant wave equation probem we wi study is u tt = c 2 u xx (7) u(, t) = u(, t) = (8) u(x, ) = f(x), u t (x, ) = g(x) (9) The physica interpretation of the boundary conditions is that the ends of the string are fixed in pace. They might be attached to guitar pegs. We start by assuming our soution has the form Pugging this into the equation gives u(x, t) = X(x)T(t). (1) T (t)x(x) = c 2 T(t)X (x). (11) Separating variabes, we have X X = T c 2 T 2 = λ (12)

3 where λ is a constant. This gives a pair of ODEs The boundary conditions transform into T + c 2 λt = (13) X + λx =. (1) u(, t) = X()T(t) = X() = (15) u(, t) = X()T(t) = X() =. (16) This is the same boundary vaue probem that we saw for the heat equation and thus the eigenvaues and eigenfunctions are λ n = ( nπ) 2 (17) X n (x) = sin ( nπx) (18) for n = 1, 2,... The first ODE (??) is then ) 2T =, (19) and since the coefficient of T is ceary positive this has a genera soution T n (t) = A n cos ( nπct ) + Bn sin ( nπct). (2) There is no reason to think either of these are zero, so we end up with separated soutions u n (x, t) = A n cos( nπct ) + B n sin( nπct ] ) sin( nπx ) (21) and the genera soution is u(x, t) = A n cos( nπct ) + B n sin( nπct ] ) sin( nπx ). (22) We can directy appy our first initia condition, but to appy the second we wi need to differentiate with respect to t. This gives us u t (x, t) = nπc A n sin( nπct ) + nπc B n cos( nπct ] ) sin( nπx ) (23) Pugging in the initia condition then yieds the pair of equations u(x, ) = f(x) = A n sin( nπx ) (2) nπc u t (x, ) = g(x) = B n sin( nπx ). (25) 3

4 These are both Fourier Sine series. The first is directy the Fourier Since series for f(x) on (, ). The second equation is the Fourier Sine series for g(x) on (, ) with a sighty messy coefficient. The Euer-Fourier formuas then te us that 1.3 Exampes A n nπc B n A n = 2 = 2 = 2 B n = 2 nπc f(x) sin( nπx )dx (26) g(x) sin( nπx )dx (27) f(x) sin( nπx )dx (28) g(x) sin( nπx )dx. (29) Exampe 1. Find the soution (dispacement u(x, t)) for the probem of an eastic string of ength whose ends are hed fixed. The string has no initia veocity (u t (x, ) = ) from an initia position x x u(x, ) = f(x) = 1 < x < 3 (3) ( x) 3 x By the formuas above we see if we separate variabes we have the foowing equation for T with the genera soution )2 T = (31) T n (t) = A n cos( nπct ) + B n sin( nπct ). (32) since the initia speed is zero, we find T () = and thus B n =. Therefore the genera soution is u(x, t) = A n cos( nπct ) sin(nπx ). (33) where the coefficients are the Fourier Sine coefficients of f(x). So A n = 2 = 2 / f(x) sin( nπx )dx (3) x 3/ sin(nπx )dx + sin( nπx / )dx + x sin( nπx ] 3/ )dx (35) = 8 sin(nπ ) + sin(3nπ ) n 2 π 2 (36)

5 Thus the dispacement of the string wi be u(x, t) = 8 sin( nπ) + sin(3nπ) cos( nπct π 2 π 2 ) sin(nπx ). (37) Exampe 2. Find the soution (dispacement u(x, t)) for the probem of an eastic string of ength whose ends are hed fixed. The string has no initia veocity (u t (x, ) = ) from an initia position u(x, ) = f(x) = 8x( x)2 3 (38) By the formuas above we see if we separate variabes we have the foowing equation for T with the genera soution )2 T = (39) T n (t) = A n cos( nπct ) + B n sin( nπct ). () since the initia speed is zero, we find T () = and thus B n =. Therefore the genera soution is u(x, t) = A n cos( nπct ) sin(nπx ). (1) where the coefficients are the Fourier Sine coefficients of f(x). So A n = 2 = 2 Thus the dispacement of the string wi be u(x, t) = 32 π 3 f(x) sin( nπx )dx (2) 8x( x) 2 3 sin( nπx )dx (3) = cos(nπ) n 3 π 3 Integrate By Parts () Exampe 3. Probem 12 is a great exercise. HW 1.7 # a, 5a, 7a, 8a, cos(nπ) n 3 Go through the Separation of Variabes it wi be important for Exams. cos( nπct ) sin(nπx ). (5) 5

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