Question

Problem 2. A string of a guitar is fixed at the two ends, x = 0 and r = a. The string is set in motion with initial position f(x) = (h/a)., 0 <r <a, where h > 0, and then it is released with no initial velocity. The displacement u of the string is described by the PDE au 1 au ar2 2 212 0<x<a, t> 0. (i) State the boundary value initial value problem that u satisfies. (ii) Find the solution u(x, t) and express it as an infinite series. (In other words, find the solution that is given by the method of separation of variables. You do not have to apply the method of separation of variables from scratch. Just compute the solution.) (iii) You are given the frequencies of vibration (in Hertz) of the guitar strings: Note Rounded up String 6 82 ہی ایم Frequency (Hz) 82.407 110.000 146.832 195.998 246.942 329.628 110 147 196 247 330 в E If the length of the string is 1.25 meters and c= 500, which note (approximately) does the vibrating string sound like?

Answer 1

• ( tired at tre x=a the two two ends x=o ends Boundary conditions - 4(0,t) = 0 (b) (a,0)=0 a = length of the stoing. litial conditions 10 () 20 (2000 initial velocity) (d) {(2,0) -(1.) (imisind poi horo) - (1) displacement of string is Let u=X) TH) U-X.T. bu -x"I 04 - XT" substitute in equation ④ TU x"T = *XT"

Xll į "T!! X = ke Separation constant = X"=kx T". KC²T T"_kc²7=0 x"-kx=0 by auxillary equation m²-kc²=0 m² = K C m²-k=o mk // Let k=-p2 (for real problem) (practical) m= tipe m²=-p² matip T = Clospet +Dsinplet .: X-kos px tesinpr) ee T in u Substitute x we know NOMINEROSITALASHNI u=X.T a = Acospargsinpx) (ccospet + postimp Dsinpet)

Apply boundary and initial conditions, Apply bounding condition (1) U(0,1)=0 U10,1) -(Acos pro + B sin pxo) (ccospet + D siopet) =0 = (A + Bxo) (ccospetto sinpet)=0 A cc cospet +D sinpet) = 0 = 0 A S Apply in equation ④ U12,t) = B sinpa ( cengpett D sin pet) -3 - Apply Boundary condition (6) in equation ® ula,t)=0 Ulat) - B sinpa (e cospett D sinpct) = 0 Вор в о sinpa so

pa= on .. C Substitute pin equation ③ @ M(2,4) = B sinon x lecos 27.ct to sio act in equation ④ - podly Apply initial condition (c) not attro LO du at "(sio na x) F-c (sin on Ct). nge + Dlcos pact) xlord 1 X ) (Best na ) (-e-sinovog cet D cos o .C).o. le sionga) (Dugte) - ..D AC =0 "D =0

Substitute Din eruation ④ ucard (sinna x)(c cos pact) most general solution using Super position po O 41xt) - 3 sinnt a) Bn cosom et © Be=Ba= another constant Apply IC (d) in ③ 41(7,0) - Tax cos x U12,0) = Ê sin ore. Bn Coso = - 1x ba sinna - that n is chosen must This coefficient u(90) is a .:. bn = 2 -2 fourier sine series. Julxo) sin nha da IT ha sinn a da

sin Tac 2h (tos sinon + 10+ sino n n = (a! (-13" Br= - ab (-) D equation © Bubstitute Bo in 1 (1,8) = 2 2 65" (bia ona) (1050, C2) met