Question

3. Find u(x, t) for the string of length L = 1 and c2 = 1 when the initial velocity is zero and the initial deflection with small k is kx'(1 - x) Answer:

Answer 1

Page 1 As we know the solution of the wave equation gru 2 je u c² ax ² 2 +2 for all tro with uco,t=0, ucl, d)=0 U (0,0) = f(x), M (8,0) = gea), ocuch is given by , a(ait) = E (By cos Con 1) + B * sin du t) sin fare nel Where. dan ch L Bn = 2 25 sen sin (nov) air & 2 S (Ax ) dr Br ex) sin cha As given initial velocity being zero gen) = 0 Bu - O
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Page: -2] given the initial deflection is kn ² (1-1) (where k is very small) soi ben) = kx? (1-x) and c ² = 1 we have with LE) du Bn - 2 25 sen sin (ren) = 2 25+ Kx*(1-x) sin (nandre 2 sin (nan) du - I sin (rak) dx] - 2K 2 u au sin (nan) da dv=sin(nan) cos (nan) n² cos (non 24 cos (nan) dn Na 0 t ( cos (na) ro + 2 n cas (nan) da MT o
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Page :3 n sin(mann sin (nan) du 2. t na - – Cos (ha) ha ht cos(nan) 2 ·cas (na) et ht ht ht na (cos (AT) - 1)) 2 cos (uk) t na 2 NT na cos (ha) 2 (cu cos (at) -- it 373 nt cos (na) 2 na 63 43 (cos (AR) - 2) (use integration by party) x² sin (nan) at 34² cos Carn) .dk n cos (nan) na NT o 1 3 n² cas (nan) an cos (na) hã NT x² sin(nant ansin (max) cos (na) ***- pada na
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Page: 4 cas (nh) 3 nsin(nanan t 16.) - ht nk nh - cos (na) 6 u cos Changt los (hang du (na)? ( NK ht ט 6 cos (ha) Cas(na) Sin 2 [- Lose (na na h NT - – cos (NA) cas (na) t 6 (A) NT o ] راما nr - -cos (na) + 6 cos (at) (na) L so; Bu = ek (0-0) (Cos (mt) -d) - cos (na) 2 - 2k 3 MT 43 coffax) 6 Cos (na) + ht (na) rk 4 cos (ah) 2 na n3n
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Page: 5 .: Bn - - Uk را 2 cos (na) tt *) +-) h²3 the solution is . unt) = Bn cos (nat) Sin (nan) hal S. lak 2 cos()+1) cos (nat) sin(max) n33 n=1 so uch, t) - S (2 2 Cos (na) +1) cos chat) Sinnan) 23 nal
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Answer 2