1 String with massive end-points Πρόβλημα 5.11:Θεωρείστε μια χορδή μήκους, τάσης T, με δύο σημειακά σωματίδια στα άκρα της, το ένα μάζας m, και το άλλο μάζας m. α) Μελετώντας την κίνηση των άκρων βρείτε τις οριακές συνθήκες που πρέπει να επιβληθούν στην κυμματική εξίσωση. β) Βρείτε τις ιδιοσυχνότητες της χορδής. γ) Υπολογίστε την ολική ενέργεια της χορδής και δείξατε ότι διατηρείται. δ) Βρείτε την συνάρτηση μετατόπισης της χορδής όταν οι αρχικές συνθήκες είναι u(x, ) =. u(x, ) =, u t (x, ) = f(x) (1.1) Solution:(a) On every mass at the end points, there is a force coming from the string acting: according to the picture above, this force is T sin ϕ T u x so that Newton s law at the two endpoints reads T u x (, t) = m u tt (, t), T u x (, t) = m u tt (, t) (1.2) where we took into account the fact that that the signs of the force at the two end-points are opposite. These conditions can be rewritten as follows T u x (, t) = m c 2 u xx (, t), T u x (, t) = m c 2 u xx (, t) (1.3) using the wave equation and using c 2 = T ρ they simplify further to u x (, t) = m ρ u xx(, t), u x (, t) = m ρ u xx(, t) (1.4) From now on we define (b) We are now separating variables with κ, m, ρ (1.5) X(x) = A sin(kx) + B cos(kx), T (t) = C sin(ωt) + D cos(ωt), ω = kc (1.6) 1
The boundary conditions (1.29) imply and using (1.27) we obtain X () = κ X (), X () = κ X () (1.7) ka = k 2 κ B, k(a cos(k) B sin(k)) = k 2 κ (A sin(k) + B cos(k)) (1.8) From the first equation we obtain while substituting it into the second we obtain A = kκ B (1.9) kκ + tan(k) = kκ (kκ tan(k) 1) (1.1) It can also be rewritten as (κ + κ )k = tan (k) (1.11) 1 k 2 κ κ Define [ tan ϕ (k) = kκ, tan ϕ (k) = kκ, ϕ,, π ] (1.12) 2 Note that when m = m =, then ϕ = ϕ =. On the other hand, when m = m +, ϕ = ϕ = π 2 Then equation (1.1) can be written as tan ϕ + tan ϕ 1 tan ϕ tan ϕ = tan(k) tan(ϕ + ϕ ) = tan( k) (1.13) This is an eigenvalue equation for k with solutions ϕ + ϕ = k nπ k n = nπ ϕ (k) + ϕ (k) with [ ] k(κ + κ ) θ(k) ϕ + ϕ = arctan 1 k 2 κ κ From this we can now compute the eigenfrequencies θ(k) + nπ (1.14) (1.15) ω n = ck n (1.16) For each n we will have a solution k n. The solutions can be visualized by plotting the two sides of equation (1.1) on the same plot. This can be seen in the figures below. 2
In this figure above we plot the two sides of (1.1) for κ = κ = over two periods. The horizontal axis is k 2π. There is one intersection point per each period of length π. This is the same picture as above but over 6 π-periods. In this figure we plot again the two sides of (1.1) but for κ =, κ = 2. The eigenfunctions are Their inner product is dx y n (x)y m (x) = 1 2 y n = cos(k n x) k n κ sin(k n x) cos (k n x + ϕ (k n )) (1.17) [ sin (ϕ (k m ) ϕ (k n )) + ( 1) m n sin (ϕ (k m ) ϕ (k n )) k m k n + (1.18) + sin (ϕ (k m ) + ϕ (k n )) + ( 1) m+n sin (ϕ (k m ) + ϕ (k n )) k m + k n where we have used (1.14). Using also ] sin ϕ = kκ 1 + k2 κ 2, cos ϕ = 1 1 + k2 κ 2 (1.19) 3
and similarly for ϕ we obtain dx y n (x)y m (x) = κ + ( 1) m+n κ 1 + k2 κ 2 1 + k2 κ 2 (1.2) κ + (1 + κ2 k 2 m)(1 + κ 2 k 2 n) (κ κ k 2 m 1)(κ κ k 2 n 1) κ cos(k m ) cos(k n ) γ) The Energy: Consider a one-dimensional vibrating string of equilibrium length,. Consider an infinitesimal portion of length dx. The kinetic energy of that portion is given by de kinetic = 1 2 dm u2 = 1 2 ρ u2 dx (1.21) where u(x, t) is the transverse velocity of the infinitesimal portion. The potential energy of this infinitesimal portion is T dl where T is the tension and l is the change in length from the equilibrium one. From figure 1, the length of the portion is dx2 + du 2 while the equilibrium length is dx. Therefore, dl = dx 2 + du 2 dx = 1 + ( ) 2 du 1 dx dx [ 1 + 1 2 u2 x + O ( ] ) u 4 x 1 dx 1 2 u2 x (1.22) where we assumed as usual small deformations so that we can neglect the O (u 4 x) terms. Therefore the potential energy is de pot = 1 2 T u2 x dx (1.23) Summing the energies of the various pieces of the string we obtain the total energy E string = 1 2 dx [ ] ρ u 2 + T u 2 x To this we must add the energy of the two endpoints: (1.24) E mass = 1 2 m u(x = ) 2 + 1 2 m u(x = ) 2 (1.25) and the total energy is E total = E string + E mass (1.26) Consider now the variation of this energy as a function of time Ė string = 1 2 dx [ ρ u 2 + T u 2 t x] = dx [ρ uü + T u x u x ] = (1.27) = T dx [ uu xx + u x u x ] where in the last step we used the wave equation ü = c 2 u xx ü = T ρ u xx. (1.28) 4
We rewrite now the integrand of (1.27) as a total derivative Ė string = T dx x (u x u) = T u x u Consider now the energy variation of the end-point energy x= x= (1.29) Ė mass = m u(x = )ü(x = ) + m u(x = )ü(x = ) = (1.3) = m c 2 u(x = )u xx (x = ) + m c 2 u(x = )u xx (x = ) where we have used the wave equation. We now use the boundary conditions (1.29) to rewrite this as [ ] x= x= x= Ė mass = T uu x uu x = T uu x (1.31) We then obtain d dt (E string + E mass ) = (1.32) Not surprisingly the total energy is conserved, as there is no external force acting on the system. However the energy of the string or the end-point masses is not conserved individually. x= d) The general solution can now be written as We may rewrite u(x, t) = (A n sin(ω n t) + B n cos(ω n t)) [cos(k n x) k n κ sin(k n x)] (1.33) n= cos(k n x) kκ sin(k n x) = cos(k n x) tan(ϕ ) sin(k n x) = cos(ϕ ) cos(k n x) sin(ϕ ) sin(k n x) cos ϕ = = cos(k nx + ϕ ) cos(ϕ ) Imposing u(x, ) = we obtain B n = so that finally the solution is u(x, t) = (1.34) A n sin(ω n t) cos(k n x + ϕ (k n )) (1.35) n= To impose the Initial condition u t (x, ) = f(x) we compute u t (x, ) = ω n A n cos(k n x + ϕ ) = f(x) (1.36) To solve for A n we must expand cos(k n x + ϕ ) in an orthogonal basis cos(k n x + ϕ ) = n= m= [ C nm cos 2nπx + D nm sin 2nπx ] 5 (1.37)
with Therefore C nm = 2 D nm = 2 u t (x, ) = We invert as C n = 1 dx cos(k n x + ϕ ) = 1 k n sin(k n + ϕ ) = ( 1)n sin(ϕ ) k n (1.38) dx cos(k n x + ϕ ) cos 2mπx = 2k n(sin(k n + ϕ ) sin(ϕ )) (k n ) 2 4(mπ) 2 = (1.39) = 2k n (k n ) 2 4(mπ) 2 (( 1)n sin(ϕ ) sin(ϕ )) dx cos(k n x + ϕ ) sin 2mπx = 4mπ(cos(k n + ϕ ) cos(ϕ )) (k n ) 2 4(mπ) 2 = (1.4) n= m= ω n A n C nm = 2 n= = 4mπ (k n ) 2 4(mπ) 2 (( 1)n cos(ϕ ) cos(ϕ )) ω n A n C nm cos 2mπx + n= m=1 ω n A n D nm sin 2mπx dx cos 2mπx f(x), ω n A n D nm = 2 n= = f(x) (1.41) dx sin 2mπx f(x) (1.42) u(x, t) = n= m= A n C nm sin(ω n t) cos 2mπx + n= m=1 A n D nm sin(ω n t) sin 2mπx (1.43) 6