MAS 315 Waves 1 of 7 Answers to Example Sheet 3. NB Questions 1 and 2 are relevant to resonance - see S2 Q7

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1 MAS 35 Waves o 7 Answers to Example Sheet 3 NB Questions and are relevant to resonance - see S Q7. The CF is A cosnt + B sin nt (i ω n: Try PI y = C cosωt. OK provided C( ω + n = C = GS is y = A cosnt + B sin nt + (n ω cosωt y(0 = ẏ(0 = 0 B = 0, A = (n ω y = n ω [cosωt cosnt] (n ω (ii ω = n: Try PI y = Dt sinnt (Y or MAS ẏ = D sin nt + Dnt cosnt, ÿ = Dn cosnt Dn t sin nt. Satisies ODE Dn cosnt Dn t sin nt + n sin nt = cos nt D = GS is y = A cosnt + B sin nt + t sin nt The answer to (i, viz. y(0 = ẏ(0 = 0 A = B = 0 y = y = t sin nt (cosnt cosωt (n ω is not deined at ω = n but does it have a limit as ω n? The answer is YES. The value o the limit can be ound by l Hopital s Rule or in the ollowing way: Since Hence and y = [ sin (n ωt sin (n + ωt] (n ω(n + ω lim θ 0 ( sin θ θ =, lim θ 0 lim ω n { sin tθ tθ { sin (n ωt = t (n ω sin nt { sin (n ωt (n ω { sin tθ = lim t = t θ 0 tθ y t sin nt as required, viz (ii NB the answer to (ii, when ( the orcing is at the natural requency ω, is an oscillation t at a growing amplitude. In practice, there is always damping (or riction, represented by the term λẏ in the ODE in Q.

2 MAS 35 Waves o 7 Answers to Example Sheet 3. The CF is A e m t +B e m t where m, m are the roots o m +λm+n = 0 m = λ±iσ where σ = n λ [NB λ << n σ is real]. Thus CF is e λt A cosσt + B sin σt The PI is Re{C iωt where C{ ω + λiω + n = C = {(n ω + λiω. Now we can write (n ω + λiω = Re iα where C = R e iα PI is cos(ωt α R R = (n ω + 4λ ω R cos α = (n ω R sin α = λω Thus GS is y = e λt {A cosσt + B sin σt + cos(ωt α R y(0 = ẏ(0 = 0 (ater some algebra A = R y = R {cos(ωt α e λt [ cos α cosσt + cos α, B = (λ cosα + ω sin α σr ( λ cosα + ω sin α (i (a λ = 0, ω n R = (n ω, α = 0, σ = n result in Q(i above (b ω = n, λ 0 α = π, R = λn, σ = (n λ { y = n sin nt e λt sin(n λ λn (n λ t As λ 0, this can be simpliied to y λn {sin nt { e λt sin nt ( e λt λ sin nt [ ( λt + tλ t...] λ t sin nt result in Q(ii above (ii For large λt, y R cos(ωt α = ρ cos(ωt α. Here σ ] sin σt ρ = {(ω n + 4λ ω

3 MAS 35 Waves 3 o 7 Answers to Example Sheet 3 ρ λ (n λ n λ small ρ max is large. but λ 0 no singularity ω 3. With ρ = ρ 0 ( + s = ρ 0 + ρ 0 s, u = φ x, v = φ y, w = φ z, (3.3 ρ 0 s t = ρ 0 (φ xx + φ yy + φ zz s t = (φ xx + φ yy + φ zz (A (3. x (φ t = c s x x (c s + φ t = 0. Likewise y (c s + φ t = z (c s + φ t = 0. Thus c s + φ t = F(t. As x, there is no disturbance and conditions are steady. Thus F(t = 0 so c s = φ t (B Eliminating s between (A and (B c s t = φ tt = c (φ xx + φ yy + φ zz φ tt = c (φ xx + φ yy + φ zz (N.B. (φ xx + φ yy + φ zz is oten written φ ( delsquared φ and is known as the Laplacian o φ. 3

4 MAS 35 Waves 4 o 7 Answers to Example Sheet 3 4. r = x + y + z x (r = x r r x r x = r x φ x = Φ r r x = r Φ r x & φ x = r [ ] r Φ r x + r Φ r r = x(chain rule φ x + φ y + φ z = 3r Φ r + (x + y + z r { r Φ rr r Φ r = Φrr + r Φ r Thus c (φ xx + φ yy + φ zz = φ tt c (Φ rr + r Φ r = Φ tt (B There are many ways o doing the last part. Here is one only. Write rφ = Ψ Φ = r Ψ Φ r = r Ψ r r Ψ and Φ rr = r Ψ rr r Ψ r + r 3 Ψ. Substitute into (B to get c {r Ψ rr r Ψ r + r 3 Ψ + r Ψ r r 3 Ψ = r Ψ tt. Hence c Ψ rr = Ψ tt. This is (.6 in another notation, so the GS or Ψ is (r ct + g(r + ct by (.. But e iω/c(r ct is (r ct with (ξ = e iωξ/c and e iω/c(r + ct is g(r +ct with g(ξ = e iωξ/c. Hence Ar e iω/c(r ± ct both satisy (B [NB See S Q6.] 4

5 MAS 35 Waves 5 o 7 Answers to Example Sheet 3 5. Substitute φ = (xe i(ωt kz into φ xx + φ zz = c φ tt, obtaining ( k = ω ω + k = 0. c c Now suppose k > ω c, and write k ω c = m m = 0 = Ae me + Be mx. Now φ x = 0 = 0. Thus = 0 at x = 0 m(a B = 0 B = A, = A cosh mx. Then = 0 at x = d A(sinh mdm = 0 A = 0 φ = 0. No interest! I k = ω c, we have = 0 = Bx + A. = 0 at x = 0, d B = 0. Thus = Ae i(ωt kz and k = k 0 = ω c with A = A. I k < ω ω, write c c k = m + m = 0 = Ae imx + Be imx. Then = 0 at x = 0 B = A, = A cosmx. Then = 0 at x = d A(sin mdm = 0 md = nπ (n =,,... ( nπx ( ω Thus = A cos e i(ωt kz with k = k n = d c n π. d [NB For real waves to propagate we need real ω, k. Thus, i (or example ω is ixed, there are only a inite number o k values, ( those with n n max, where n max is the ωd largest positive integer or which n max.] πc 5

6 MAS 35 Waves 6 o 7 Answers to Example Sheet 3 6. φ = g(re i(ωt kz φ x = g r ei(ωt kz r x r = x + y r r x φ x = { (r g x e i(ωt kz { & φ x = r (r g r x + (r g r = x x = r (see answer to Q4 = { r g x r 3 g x + r g e i(ωt kz e i(ωt kz φ x + φ y = { g r g + r g e i(ωt kz = {g + r g ei(ωt kz Write c (φ xx + φ yy + φ zz = φ tt g + r g k g = ω c g Thus ( becomes g (r + r g (r + m g(r = 0 where m = ω c k mr = x dg dr = mdg dx, md g dr = m d g dx. x d g dx + xdg dx + x g = 0. Since g is bounded at x = r = 0, g J 0 (x = J 0 (mr. [See.5 in Notes.] We need g = 0 at r = a J 0 (ma = 0 ma = β n where β n is the nth zero o J 0 (x = 0 with β 0 = 0 (see sketch on handout. ( ( βn r ω Thus, inally, g(r J 0 and k = k n = a c β n a (N.B. as is Q5, or ixed ω there are only a inite number o k values. (N.B. In a real stethoscope most o the acoustic energy is carried by the mode with β n = 0. ( 6

7 MAS 35 Waves 7 o 7 Answers to Example Sheet 3 7. Look or separable solutions o the orm φ = X(xT(t. Substitute into the PDE c X T = X T c X X = T T. In the normal way, each side is a constant which we take to be ω since we seek time-harmonic solutions, i.e. X X = ω c, T ( ωx T = ω X = A cos c T cos(ωt + ǫ + B sin ( ωx c { φ = A cos ωx ωx + B sin cos(ωt + ǫ c c ( ωx φ x = 0 at x = 0 B = 0 φ = A cos cos(ωt + ǫ ( c ωl φ t = 0 at x = l cos = 0 ωl ( c c = n + π ω = ω n = πc ( n + ( ωn x and φ = φ n cos l c cos(ω n t + ǫ n N.B. I you have done Q5 or Q6 you may ask why the cross-section o the tube does not appear in this question. This is because the work in this question relates to the n = 0 case in both Q5 and Q6 or which there is no dependence on cross-sectional position or shape. 7