ANSWERS O MA56 UORIAL 7 Quetion. a) We hall ue the following -Shifting property: Lft)) = F ) Le ct ft)) = F c) Lt 2 ) = 2 3 ue Ltn ) = n! Lt 2 e 3t ) = Le 3t t 2 ) = n+ 2 + 3) 3 b) Here u denote the Unit Step Function given by { if t < a ut a) if t > a We hall ue the following t-shifting property: Lft)) = F ) L{ft a)ut a)} = e a F ) Letft 2) = t ft) = t + 2 Lft)) = Lt + 2) = Lt) + 2L) = 2 + 2 Ltut 2)) = L{ft 2)ut 2)} = e 2 2 + 2 ) Quetion 2. a) 2 + + 26 = + 5) 2 + = + 5) 5 + 5) 2 + LetF ) = 5 2 + L 2 + + 26 ) = L F + 5)) = L F 5)) = e 5t L F )) ue -hifting = e 5t L 2 + 5 2 + ) = e 5t {L 2 + ) 5L 2 + )} = e 5t cot 5 in t) b) Let F ) = +2 3
= 3 + 2 2 L F )) = t2 2 + 2t ue Ltn ) = n! n+ ) Let ft) = t2 2 + 2t Uing t-hifting, L 2 + 2 e ) = L e 2 F )) 3 = ft 2)ut 2) t 2)2 = { 2 = 2 t2 4)ut 2) = 2 t2 2)ut 2) + 2t 2)}ut 2) Quetion 3. a) Let Lyt)) = Y ) We hall ue Ly t)) = Y ) y). We have Ly ) = Ltut 2)) Y ) 4 = e 2 2 + 2 ) Y ) = e 2 + 2 3 ) + 4 yt) = L Y )) = L {e 2 + 2 3 )} + 4L ) = 2 t2 2)ut 2) + 4 by a previou quetion.) b) We hall ue Ly ) = 2 Y y) y ) We have
Ly 2y ) = L4) 2 Y y) y ) 2{Y y)} = 4 2 Y 2Y + 2 = 4 2 2)Y = 4 + 2 = 4 + 2 2 Y = 2 2 2) 2 2) = 2 2 2 y = L 2 2 2 ) = e 2t 2t Quetion 4. By definition, F ) = e t ft)dt. Differentiating both ide with repect to [NO t think of t a a contant in thi calculation, ince i doing the changing here] we get where the chain rule ha been ued. So a required. F ) = F ) = te t ft)dt, e t [tft)]dt = L[tft)] Now the Laplace tranform of int) i / + 2 ). Hence the Laplace tranform of t int) i minu the derivative of thi, ie L[tint)] = 2 + 2 ) 2. For the reonance equation ÿ + y = cot), take the Laplace tranform of both ide; with the given initial data, we get 2 Y + Y = + 2 ),
and o hu Y ) = + 2 ) 2. yt) = 2 tint), which i indeed exactly the olution we got in Chapter 2 for reonance in thi cae. [he formula in the note i x = F t 2mω inωt); here F = ω = m =.] Notice how the Laplace method automatically take care of the extra factor of t. Quetion 5. he original ODE decribing thi ituation in the abence of friction wa Mẍ = Mg ρ A d + x)g, where the downward direction i poitive. We need an extra term to account for the udden force exerted by the rogue wave. Since the force i exerted uddenly, thi ugget that we need a Dirac delta function, o the force will be proportional to δt ). Now F = ma, Newton law, can be written a F = [time derivative of mv], where v i the velocity, o the change in the momentum i equal to the time integral of the force. Recall that δt )dt =, o clearly in our cae F = P δt ), ince P i the given change in the momentum. [Integrate both ide to verify thi, and remember that xt) here i the DOWNWARD diplacement o the upward force of the wave [a tated in the problem] i negative, like the buoyancy force. Note that the unit here are correct ince the delta function ha unit of /time; thi i becaue the time integral of the delta function i a pure number.] So we have Mẍ = Mg ρ A d + x)g P δt ), which, a in utorial 4, implifie to ẍ = ρ A g M x P δt ). M aking the Laplace tranform of both ide, remembering that the hip i initially at ret, we have or 2 X = ρag M X P M e, X) = P M e 2 + ω 2 = P ωm ωe 2 + ω 2,
where ω i the natural frequency of ocillation of the hip, ρ A g/m. Uing the t- hifting theorem, we can find the invere Laplace tranform: xt) = P in[ωt )]ut ). ωm he graph i flat until t =, then you get the uual Simple Harmonic Motion with angular frequency ω. hat make ene the udden impule given by the wave hould trigger off SHM. he amplitude i P/ωM, o thi i the maximum ditance the hip goe down if it doen t ink. Quetion 6 he force uddenly witche on at t = and then witche off at t = + τ. It hould therefore be proportional to ut - ) - ut - + τ)), ince thi function behave in jut that way. Note that it integral from to i τ. So we hould et F = P τ [ut ) ut + τ))] again you can check thi by integrating both ide and remembering that the change in momentum hould be the integral of the force, and that up i negative in thi problem. So now the differential equation in quetion 5 become ẍ = ρ A g M x P [ut ) ut + τ))]. Mτ ake the Laplace tranform of both ide to get and o 2 X) = ρ A g M X) P Mτ [e e +τ) ], X) = P Mτ [e e +τ) ] 2 + ω 2, where ω i defined a before. You can obtain the olution uing the t-hifting theorem a before. Uing L Hopital rule, one ee that τ [e e +τ) ] e a τ. Hence our expreion for X) doe indeed tend to the ame expreion for X) a in Quetion 5 if you let τ tend to zero. hat a expected, becaue the impule
hould be like a delta function if τ i very hort. So you can alway think of a delta function a a horthand for a difference of two tep function in thi manner. Quetion 7 Fourier coefficient for any function ft) with period are given by a n = 2 /2 /2 ft)co 2πnt) dt, and the Fourier erie i then b n = 2 a 2 + [ /2 /2 a n co 2πnt ft)in 2πnt) dt, ) + bn in 2πnt) ]. Now in the definition of Dirac comb, only one of the term in the infinite um i nonzero in the interval /2 to /2, namely the k = term, o we only have to include that term when we work out the integral. We have a n = 2 /2 /2 δt)co 2πnt) 2 dt = co) = 2, b n = 2 /2 /2 δt)in 2πnt) 2 dt = in) =, uing the property of the delta function explained in the note. So the Fourier erie of the comb i jut + 2 co 2πnt). hi may look ridiculou mathematically, but if you ue graphmatica/matlab/whatever to graph, for example, y = /2+cox)+co2x)+co3x)+co4x)+co5x)+co6x)+co7x)+co8x)+co9x), you can convince yourelf that if you take enough term you will indeed end up with omething that look like Dirac comb. You are umming wave which interfere detructively everywhere except at regularly paced point, where they uddenly interfere contructively a bit like a rogue wave!