Waves & Normal Modes. Matt Jarvis

Transcription

1 Waves & Noral Modes Matt Jarvis January 19, 016

2 Contents 1 Oscillations Siple Haronic Motion - revision Noral Modes 5.1 The coupled pendulu The Decoupling Method The Matrix Method Initial conditions and exaples Energy of a coupled pendulu Unequal Coupled Pendula The Horizontal Spring-Mass syste Decoupling ethod The Matrix Method Energy of the horizontal spring-ass syste Initial Condition Vertical spring-ass syste The atrix ethod Interlude: Solving inhoogeneous nd order differential equations Horizontal spring-ass syste with a driving ter The Forced Coupled Pendulu with a Daping Factor Noral odes II - towards the continuous liit N-coupled oscillators Special cases General case N very large Longitudinal Oscillations Waves I The wave equation The Stretched String Solutions to the wave equation d Alabert s solution

3 Chapter 1 Oscillations Before we go into the ain body of the course on waves and noral odes, it is useful to have a sall recap on what we know about siple systes where we only have a single ass on a pendulu for exaple. This would all coe under the reit of siple haronic otion, which fors the basis of soe of the probles that we will encounter in this course Siple Haronic Motion - revision First, consider Hooke s Law, F = kx, 1.1) where F is the force, x is the displaceent with respect to the equilibriu position, and k is the constant of proportionality relating the two. The usual ai is to solve for x as a function of tie t in the oscillation of the spring or pendulu for exaple. We know that F = a, therefore F = kx = d x dt. 1.) This equation tells us that we need to find a solution for which the second derivative is proportional to the negative of itself. We know that functions that obey this are the sine, cosine and exponentials. So we can try a fairly general solutions of the for, xt) = A cosωt + φ) or xt) = Ae iωt 1.3)

4 3 The phase φ just provides a linear shift on the tie axis, the scale factor ω expands or contracts the curve on the tie axis and the constant A gives the aplitude of the curve. To check that this all works we can substitute Eq. 1.3 into Eq. 1., obtaining k [A cosωt + φ)] = [ ω A cosωt + φ) ] 1.4) = k + ω ) [A cosωt + φ)] = 0 1.5) Since this equation ust hold at all ties, t, we ust therefore have, k ω = 0 = ω = k. 1.6) You can also do this slightly ore rigourously by writing the differential equation as kx =. dv dt 1.7) but this is awkward as it contains three variables, x, vv and t. So you can t use the standard strategy of separation of variables on the two sides of the equation and then integrate. But we can write F = a =. dv dt =.dx dt. dv dv = v. dx dx. 1.8) Which then leads to, F = a = kx = v dv ) dx = kx.dx = v.dv 1.9) Integrating, we find 1 kx = 1 v + E, 1.10) where the constant of integration, E happens to be the energy. It follows that v = ± E 1 kx, 1.11) which can be written as, dx E 1 kx E = ± dt. 1.1)

5 4 A trig substitution turns the LHS into an arcsin or arccos function, and the result is xt) = A cosωt + φ) where ω = k 1.13) which is the sae result given in Eq General solutions to Hooke s law can obviously also encopass cobinations of trig functions and/or exponentials, for exaple, xt) = A sinωt + φ) 1.14) xt) = A sinωt + φ) = A cos φ sin ωt + A sin φ cos ωt 1.15) therefore, xt) = A 1 sin ωt + A cos ωt 1.16) is also a solution. Finally, for the coplex exponential solution xt) = Ce iβt 1.17) F = kx = Ce iβt = d x dt = β Ce iβt 1.18) = β = k = ω 1.19) = xt) = A e iωt + B e iωt 1.0) Using Euler s forula, e iθ = cos θ + i sin θ 1.1) xt) = A cos ωt + A i sin ωt + B cos ωt B i sin ωt 1.) If A = ia B ) and B = A + B ) then xt) = A sin ωt + B cos ωt 1.3)

6 Chapter Noral Modes Many physical systes require ore than one variable to quantify their configuration; for exaple a circuit ay have two connected current loops, so one needs to know what current is flowing in each loop at each oent. Another exaple is a set of N coupled pendula each of which is a one-diensional oscillator. A set of differential equations one for each variable will deterine the dynaics of such a syste. For a syste of N coupled 1-D oscillators there exist N noral odes in which all oscillators ove with the sae frequency and thus have fixed aplitude ratios if each oscillator is allowed to ove in x diensions, then xn noral odes exist). The noral ode is for whole syste. Even though uncoupled angular frequencies of the oscillators are not the sae, the effect of coupling is that all bodies can ove with the sae frequency. If the initial state of the syste corresponds to otion in a noral ode then the oscillations continue in the noral ode. However in general the otion is described by a linear cobination of all the noral odes; since the differential equations are linear such a linear cobination is also a solution to the coupled linear equations. A noral ode of an oscillating syste is the otion in which all parts of the syste ove sinusoidally with the sae frequency and with a fixed phase relation. The best way to illustrate the existence and nature of noral odes is to work through soe exaples, and to see what kind of otion is produced. 5

7 6.1 The coupled pendulu Figure.1: The Coupled Pendulu Rather than a single pendulu, now let us consider two pendula which are coupled together by a spring which is connected to the asses at the end of two thin strings. The spring has a spring constant of k and the length, l of each string is the sae, as shown in Fig..1 Unlike the siple pendulu with a single string and a single ass, we now have to define the equation of otion of the whole syste together. However, we do this in exactly the sae way as we would in any siple pendulu. We first deterine the forces acting on the first ass left hand side). Like in the siple pendulu case, we assue that the displaceents for the equilibriu positions are sall enough that the restoriing force due to gravity is approxiately given by g tan θ and acts along the line of asses. This force related to gravity produces the oscillatory otions if the pendulu is offset fro the equilibriu position, i.e. Likewise, for the second ass d x dt = ẍ = g sin θ x = g x l..1) d y dt = ÿ = g sin θ y = g y l..) However, in addition to this gravitational force we also have the force due to the

8 7 spring that is connected to the asses. This spring introduces additional forces on the two asses, with the force acting in the opposite direction to the direction of the displaceent, if we assue that the spring obeys Hooke s law. Therefore, the equations of otion becoe odified: ẍ = g x l kx + ky = g x l + ky x),.3) and ÿ = g y ky + kx l = g y.4) l ky x). We now have two equations and two unknowns. How can we solve these?.1.1 The Decoupling Method The first ethod is quick and easy, but can only be used in relatively syetric systes, e.g. where l and are the sae, as in this case. The underlying strategy of this ethod is to cobine the equations of otion given in Eqs..3 and.4 in ways so that x and y only appear in unique cobinations. In this proble we can siply try adding the equations of otion, which is one of the two useful cobinations that we will coe across. ẍ + ÿ) = g x l + ky x) g y l ky x) = g x + y) l.5) This type of equation should look failiar, where the left-hand side is a second derivative of the displaceent with respect to tie, and the right-hand side is a constant ultiplied by a displaceent in x and y. It becoes ore clear if we define q 1 = x + y and q 1 = ẍ+ = ÿ..6) Eq..5 then becoes, q 1 = g q 1 = q 1 = g l l q 1.7) This is now easily recognisable as the the usual SHM equation, with q 1 = ω q 1, where

9 8 Figure.: The centre of ass otion of the coupled pendulu as described by q 1 = x + y. ω 1 = g l..8) We can then iediately write a solution for the syste as q 1 = A 1 cosω 1 t + φ 1 ),.9) where A 1 and φ 1 are arbitrary constants set by the initial or boundary conditions. The otion described by q 1 = x + y tells us about the coupled otion of the two pendula in ters of how they oscillate together around a centre of ass. There is no dependence on the spring whatsoever, as ω 1 does not have a ter which involves k. This otion can be visualised as shown in Fig Given what we have learnt, the obvious other cobination of x and y is to subtract Eq..4 fro Eq..3: ẍ ÿ) = g x l + ky x) + g y + ky x) l g = l + k ) x y).10) Now let us define q = x y and q = ẍ ÿ..11) This gives us the following, g q = l + k ) q..1) We have SHM again, but this tie with the q coordinate, i.e. q = ω q, where in this case

10 9 Figure.3: The relative otion of the coupled pendulu as described by q = x y. ω = g l + k..13) Again, we can write a solution iediately as q = A cosω t + φ ),.14) with A and φ arbitrary constants defined by the initial and boundary conditions. In this case the q represents the relative otion of the coupled pendulu. As should be clear fro the dependence of ω on the spring constant, this otion ust describe how the otion of the syste depends on the copression and expansion of the spring. The variables q 1 and q are the odes or noral coordinates of the syste. In any noral ode, only one of these coordinates is active at any one tie. It is actually ore coon to define the noral coordinates with a noralising factor of 1/, such that q 1 = 1 x + y) and q = 1 x y)..15) The factor of 1/ is chosen to give standard for for the kinetic enery in ters of the noral odes. We will coe to this later. We now have the two noral odes that describe the syste, and the general solution of the coupled pendulu is just the su of these two noral odes. q 1 + q = x + y + x y = x

11 10 which leads to, x = A 1 cosω 1 t + φ 1 ) + A cosω t + φ ).16) and q 1 q = x + y x + y = y which leads to, y = A 1 cosω 1 t + φ 1 ) A cosω t + φ ).17) The constant are then just set by the initial conditions..1. The Matrix Method This ethod is a bit ore involved but in principle it can be used to solve any set up. But first of all we will use the exaple of the coupled pendulu shown in Fig..1. One strategy that we can use is to look for siple kinds of otion where the asses all ove with the sae frequency, building up to a general solution by cobinig these siple kinds of otion. So in the atrix ethod we start by having a guess at the solutions to the syste. By rearranging Eqs..3 and.4, we find ẍ + g + kx ky = 0 l g = ẍ + x l + k ) k.18) y = 0 and siilarly, g ÿ + y l + k ) k x = 0..19) We can write this as a atrix in the following way, [ d + g dt l + k ) ] [x k ] k d + g dt l + k ) = y [ ] 0..0) 0

12 11 Given this equation, we could ultiply both sides of the equation by the inverse of the atrix, which would lead to x, y) = 0, 0). This is obviously a solution and would ean that the pendula just hang there and do not ove. However, we want to find a general solution which would involve describing how the asses ove, not just when they are stationary. We are expecting oscillatory solutions, so let s just try one, such that [ ] [ ] x X = R e iωt,.1) y Y where X and Y are coplex constants. Substituting this trial solution into Eq..0 we find, [ ω + g l + k k k ω + g l + k ] [ ] X = Y This is the Eigenvector Equation with ω being the eigenvalues. [ ] 0..) 0 Therefore we need to find the non-trivial solutions and the only way to escape the trivial solution is to ensure that both X and Y ust be zero when the inverse of the atrix does not exist. To find the inverse of a atrix involves finding cofactors and deterinants, which can be a bit essy. However, the key thing to reeber is that A 1 = 1 A CT,.3) where A is the deterinant of the atrix A and C T is the transposed atrix of the cofactors. So deterining the inverse of a atrix always involves dividing by the deterinant of that atrix. Therefore, if the deterinant is zero then the inverse does not exist. This is what we want in order to solve Eqn... So setting the deterinant to zero, we find the solution to be ω + g l + k k k ω + g l + k = 0..4) ω + g l ) + k ) k = 0.5)

13 1 therefore, ω + g l ) + k ) k = 0 = ω + g l + k = ± k.6) So we get two solutions for ω, ω 1 = g l and ω = g l + k.7) equation. The ± abiguity in ω is ignored as you get sinusoidal solutions. To coplete the solution we substitute the values for ω back into the eigenvector For ω 1 = g/l, [ k k k k ] [ ] X1 = Y 1 [ ] 0 0 which leads to X 1 = Y 1. So X, Y ) is proportional to the vector 1, 1). For ω = g/l + k/, [ k k k k ] [ ] X = Y [ ] 0 0 which leads to X = Y. So X, Y ) is proportional to the vector 1, 1)..8).9) We can now write the general solution as the su of the two solutions that we have found, [ ] [ ] [ ] xt) 1 = X yt) 1 e iω1t 1 + X 1 e iω t 1.30) X and Y are coplex constants, lets define the as A 1 e iφ 1 when X = Y, and A e iφ when X = Y. Substituting back into Eq..30, we find x = A 1 e iφ 1 e iω 1t + +A e iφ e iω t.31) y = A 1 e iφ 1 e iω 1t A e iφ e iω t.3) Using Euler s forula Eq. 1.1) and just considering the real coponents and reoving the iaginary parts,

14 13 x = A 1 cosω 1 t + φ 1 ) + A cosω t + φ ).33) y = A 1 cosω 1 t + φ 1 ) A cosω t + φ ).34) which are the sae noral odes and frequencies that we found fro the decoupling ethod. These are the general solutions of a coupled pendulu. The advantage of using the coplex exponential is only evident if there is a ixture of single and double derivatives as in the case of a daped pendulu discussed later. In the undaped case just discussed it would be equally siple to start with a noral ode trial solution proportional to cosωt + φ)..1.3 Initial conditions and exaples In this section we will go through an exaples using initial conditions, and the otions that these produce in the coupled pendulu. Further exaple will be given in the lectures. Let s consider the case where, xt = 0) = a, yt = 0) = 0 and the initial velocity of the asses ẋ = y) = 0. Before we go into the aths, what does this look like? We know fro SHM of a single pendulu that the velocity of the pendulu is zero at its axiu displaceent fro equilibriu. But in this case the second pendulu that is displaced in the y direction starts at it s equilibriu position. This ust ean that the spring connecting the two is either stretched or copressed, depending on the definition of the direction of x and y. It is useful to sketch these initial conditions so that you know roughly what to expect in ters of the excitation of the noral odes. Using our solutions to the coupled pendulu fro Eqs..16 and.17, for t = 0 we get x0) = A 1 cosφ 1 ) + A cosφ ) = a y0) = A 1 cosφ 1 ) A cosφ ) = 0.35)

15 14 Adding the Eqs..35 together we find A 1 cos φ 1 = a, subtracting we find A cos φ = a. So A 1 and A ust have non-zero solutions. We also have inforation about the initial velocity of the two asses. This therefore leads us to looking at the first differential of Eqs..16 and.17 with respect to tie. ẋ0) = A 1 ω 1 sinφ 1 ) A ω sinφ ) = 0 ẏ0) = A 1 ω 1 sinφ 1 ) + A ω sinφ ) = 0.36) Again adding and subtracting Eqs..36, we find A 1 sin φ 1 = 0 and A sin φ = 0. As both A 1 and A have non-zero values, then φ 1 = φ = 0. Therefore, A 1 cos φ 1 = a = A 1 and A cos φ = a = A, thus A 1 = A = a/. Substituting these back into Eqs..16 and.17, we find x = a cos ω 1t + cos ω t).37) y = a cos ω 1t cos ω t).38) In this case both noral odes are excited as q 1 = x + y) 0 and q = x y) 0. As you will see in lectures, we also have cases where only one noral ode is active. But let us continue with this syste and try and see what it is actually doing. We can rewrite Eqs..37 and.38 using soe trig identities as, ) ) ω1 + ω ω1 ω x = a cos t cos t ) ) ω1 + ω ω1 ω y = a sin t sin t.39).40) This shows that we have two distinct frequencies that the syste is oscillating in. A high-frequency ode where the frequency is ω 1 + ω )/ and a low-frequency ode with ω 1 ω )/. This eans that if we were to visualise the oscillations in x and y as a function of tie then we would see a broad envelope defined by the low-frequency ter, but within this enevelope we see uch higher frequency oscillations Fig..1.3).

16 15 Figure.4: Oscillatory pattern for a syste where both noral odes are excited. We can also calculate the period of the oscillations in the usual way. For exaple the period od the envelope is T env = π ω = 4π ω 1 ω..41) Here we have a beat frequency, where one coplete period of the envelope is equal to beats. Beats is when energy is transferred between pendula..1.4 Energy of a coupled pendulu So we have now found the solutions to the coupled pendulu in ters of how the two pendula oscillate with tie. We can also deterine how the energy in the syste varies between different types of energy, the two ain fors of energy being kinetic and potential energy. We know that for a perfect syste with no friction, then the total energy of the syste is given by the su of the potential and kinetic energies, U = KE + P E = T + V. The total kinetic energy is given by T = 1 ẋ + 1 ẏ.4)

17 16 and the potential energy coes fro two sources in the coupled pendulu, naely the spring and gravity. V spring = 1 ky x).43) V gravity = gh = gl l cos θ) = lg1 cos θ).44) Using sin θ = 1 cos θ and sin θ = x/l we find for the x-pendulu gh = gl l x.45) and for the y-pendulu so that gh = gl l y.46) V gravity = g l kx y )..47) Therefore the total potential energy is just the su of the gravitational and spring coponents, V total = 1 g l + k ) x + y ) kxy..48) This is not the only way to find the total energy of the syste. We can also use our knowledge of how the energy is related to the force, and therefore the equations of otion. We know, F X = V x x and F y = V y y,.49) therefore Integrating these equations we find, F x = V x x = ẍ = g x + ky x) l F y = V y y = ÿ = g y.50) l ky x). V x, y) = g x l + 1 kx kxy + fy) + C V x, y) = g y l + 1 ky kxy + fx) + C.51)

18 17 Neglecting the arbitrary constant C and this just relates to how we define the zero potential, then we find that the total potential energy is V total = 1 g l + k ) x + y ) kxy..5) as before. So we have expressions for the kinetic energy and the potential energy, so we can now write the total energy of the syste as U = T + V = 1 ẋ + ẏ ) + 1 g l + k ) x + y ) kxy.53) but this doesn t really convey uch about what is going on in ters of the energy associated with each noral ode. So why don t we go back and rewrite this in ters of the noral coordinates q 1 and q, where we defined q 1 = 1 x + y) and q = 1 x y)..54) and found that ω 1 = g g l. and ω = l + k..55) Substituting these into Eq..53, and with a bit of rearranging, we obtain U = 1 q ) 1 ω 1q1 + q + 1 ) ω q..56) The first ter here is then the energy associated with the first noral ode and the second ter is the energy associted with the second noral ode. Therefore the total energy of the syste is siply the su of the energies in each ode. So now we have gone through ost of the atheatics that we will use when looking at coupled systes, although we have started with the siplest exaple. Let us now have a look ata slightly ore coplicated syste, where the two pendula that are coupled by a spring are no longer the sae.

19 18 Figure.5: The Unequal Coupled Pendulu. Unequal Coupled Pendula In Fig.. we show a syste where there are still two pendula coupled by a spring, with spring constant k, but this tie the length of the pendula are different, with the left-hand pendulu having length l 1 and the right-hand pendulu having a length of l. This alters the equations of otion and therefore also the solutions to the syste. So the equations of otions of this syste are siilar to Eqs..3 and.4, and are: ẍ = g x l 1 + ky x),.57) and ÿ = g y l ky x)..58) Unlike in the previous coupled pendulu, this syste is not syetric. Therefore let us go straight to the atrix ethod in order to solve for x and y. As before, we equate Eqs..57 and.58 to zero and write the in a atrix forat. [ d dt + g l 1 + k k ] [x k ] d + g dt l + k = y [ ] )

20 19 As before, we define x and y in ters of copex constants and assue an oscillatory solution of the for e iωt. and substituting to for the Eigenfunction equation, [ ] [X ω + g l 1 + k k ] k ω + g l + k = Y [ ] [ ] x X = R e iωt,.60) y Y [ ] 0..61) 0 Again we have the trivial soluton for x = y = 0, but for the non-trivial solutions we adopt the sae process and set the deterinant of the atrix to zero, i.e. ω + g l 1 + k k k ω + g l + k = 0..6) Calculating the deterinant, we find ω + g + k ) ω + g + k ) ) k = 0.63) l 1 l Expanding this and the solving for ω gives the following, where β 1, = g l 1,. ω1, = 1 β 1 + β ) + β k ± 1 β ) + ) k.64) We can check whether this agrees with the previous solution for the equal coupled pendulu fro Section.1, by setting l 1 = l = l and checking that we get the sae values for ω 1 and ω that are given in Eq..7. So now that we have the equations for ω 1 and ω, we can find xt) and yt) in the usual way, i.e. by substituting Eq..64 into the atrix given in.61. Doing this, we find Y1 X 1 X1 Y 1 ) ) 1 = ω 1 + β 1 + k/) k/ = ω 1 + β + k/) k/.65).66)

21 0 Y X ) 1 = ω + β 1 + k/) k/.67) X Y ) = ω + β + k/) k/.68) Substituting in for ω 1, we find, X1 Y 1 ) 1 = k [ ] 1 β β1) + β1 β ) + k/).69) X Y ) 1 = k [ ] 1 β β1) β1 β ) + k/).70) Y1 X 1 ) = k [ ] 1 β1 β) + β1 β ) + k/).71) Y and therefore X ) = k [ ] 1 β1 β) β1 β ) + k/).7) siilarly, X1 ) Y 1 X1 ) Y 1 1 = 1 = 1 X ) Y 1 X ) Y.73).74) and we can define this ratio of the aplitudes, r = Y1 X 1 ) = k therefore rx 1 = Y 1 and X = ry. [ ] 1 β1 β) + β1 β ) + k/).75) We can therefore right the general solution as a function of this aplitude ratio, [ ] x = y [ ] 1 A r 1 cosω 1 t + φ 1 ) + [ ] r A 1 cosω t + φ ).76)

22 1 Figure.6: Oscillatory pattern for the unequal coupled pendulu. Both noral odes are excited again and a Beats solution is apparent but in this case with r < 1 there is an incoplete transfer of energy between the pendula. So again we can set up soe initial conditions and solve for these. As we did in Sec..1 let us consider x = a; y = 0; ẋ = ẏ = 0, to see what difference the unequal pendulu length akes. Without going through all of the aths which you should try), we find that A 1 = a/1 + r ); A = ra/1 + r ); φ 1 = φ = 0. Hence, substituting these in to the general solution we find xt) = acos ω 1 t + r cos ω t)/1 + r ).77) yt) = arcos ω 1 t cos ω t)/1 + r ).78) which can be written in ters of the two active frequencies ω 1 ω )/ and ω 1 +ω )/ as before. Doing this we get ) ) ω1 + ω ω1 ω xt) = a cos t cos t a 1 r 1 + r ) ) ) ω1 + ω ω1 ω sin t sin t.79)

23 ) ) ) r ω1 + ω ω1 ω yt) = a 1 + r sin t sin t.80).3 The Horizontal Spring-Mass syste We now ore away fro pendula and consider other systes which produce noral odes. Here we look at the horizontal spring-ass syste Fig..3) where three spring are connecting two asses of ass to two fixed points at either end. Figure.7: The horizontal spring-ass syste. As usual we first set up the equations of otion: ü 1 = αku 1 ku 1 u ).81) ü = αku + ku 1 u ).8) We have what looks like a syetrical syste, so we can probably use the decoupling ethod. So let s use this first and then check that it all looks fine with the atrix ethod..3.1 Decoupling ethod In the decoupling ethod we define new coordinates which describe the coupled otion of the two asses, with the usual coordinates being u 1 + u and u 1 u, along with a noralising factor of 1/ that ensures that the energies all work out as expected. So as before, we have q 1 = 1 u 1 + u ) and q = 1 u 1 u ) q 1 = ü 1 + ü ) and q = ü 1 ü )..83)

24 3 First, we add Eqs..81 and.8, ü 1 + ü ) = q 1 = αku 1 + u ) ku 1 + ku + ku 1 ku = αku 1 + u ).84) Then subtract Eq..8 fro.81 = q 1 = αk q 1.85) ü 1 ü ) = q = αku 1 u ) ku 1 + ku ku 1 + ku = α + )kq ).86) α + )k = q = q.87) We can see that by using the decoupling ethod we have arrived iediately that we have two equations that have the usual SHM structure, as such we can easily obtain the values for oega: ω 1 = αk and ω = α + k.88).3. The Matrix Method As stated earlier, the atrix ethod can be used for all systes, not only syetric systes so we can check that what we obtain using the decoupling ethod is in agreeent with an independent ethod. So we follow exactly the sae process as we did for the coupled pendula, and 1) write out the equations of otion with a hoogeneous atrix equation: [ d dt + αk + k k ) Substitute in the trial solution: ] [u1 k ] d + αk dt + k = u [ ] 0..89) 0 [ u1 u ] [ ] X = R e iωt..90) Y

25 4 [ ω + αk + k k k ω + αk + k ] [ ] X = Y [ ] 0 0 3) Deand that the resulting operator atrix is singular, i.e. A = 0: 4) Calculate the deterinant: ω + αk + k k k ω + αk + k.91) = 0.9) 5) Equate to find ω. In this case we can factorise, ω + αk + k ) ) k = 0,.93) ω αk ) ω ) α + )k = 0.94) So we have the sae solutions as we found using the decoupling ethod. ω 1 = αk and ω = α + k.95) 6) To coplete the solution we substitute the values for ω back into the eigenvector equation Eq..91). For ω 1 = αk/: For ω = α + )k/: [ k k k k ] [ ] X = Y [ ] ) [ k k k k ] [ ] X = Y [ ] ) These are exactly the sae as we found for the coupled pendulu, with X = Y and X = Y, we therefore have the sae for for the general solution:

26 5 x = A 1 cosω 1 t + φ 1 ) + A cosω t + φ ).98) y = A 1 cosω 1 t + φ 1 ) A cosω t + φ ).99).3.3 Energy of the horizontal spring-ass syste Unlike in the case of the coupled pendulu, we do not have any gravitational coponent to the potential energy, and all energy in this syste is contained within the springs. So let us now just follow through the sae ethod as we used for the coupled pendulu and deterine the total energy of the syste. Again using q 1 = 1 u 1 + u ) and q = 1 u 1 u ). The total kinetic energy of the syste is siply: K = 1 u 1 + u ) = 1 q 1 + q ).100) For the potential energy, we have the coponent for the two springs at either end which are fixed to the wall, and a relative displaceent ter for the iddle spring. V = 1 αku ku u 1 ) + 1 αku.101) Substituting in q 1 and q, we find V = 1 kq kq α + ).10) Using the values for ω 1, that we found in the last section ω 1 = αk/ and ω = α + )k/: V = 1 ω 1q ω q.103)

27 6 energy Finally, cobining the Kinetic Energy and the Potential Energy, we find the total U = 1 q ) 1 ω 1q1 + q + 1 ) ω q.104) Again, the su of the energies in each noral ode..3.4 Initial Condition As with the coupled pendulu is is possible to use the general solution to find the otion of the syste given a set of initial conditions. If one were to use the sae set of initial conditions as stated in Sec..1.3 then we also find a siilar solution here, i.e. ) ) ω1 + ω ω1 ω u 1 = a cos t cos t,.105) ) ) ω1 + ω ω1 ω u = a sin t sin t,.106) where both noral odes are excited and we have a Beats solution, defined by the lowfrequency envelope ω 1 ω )/..4 Vertical spring-ass syste The final ideal oscillating syste that we are going to look at is the vertical spring-ass syste Fig..4). As usual the first thing that we need to do is to write down the equations of otion. ẍ = kx kx y) = ky x).107) ÿ = ky + kx = kx y).108) One thing to notice here is that we have unequal asses, this probably eans that the decoupling ethod is not going to work try it for yourself). So let us skip straight to using the atrix ethod.

28 7 Figure.8: The vertical spring-ass syste.4.1 The atrix ethod Write the equations of otion as a hoogeneous atrix equation: [ d dt + k k Substitute in the trial solution k d + k dt ] [x ] = y [ ] ) 0 [ u1 u ] [ ] X = R e iωt..110) Y [ ω + k k k ω + k ] [ ] X = Y [ ] 0 0 Deans that the resulting operator atrix is singular, i.e..111) A = 0 and get the eigenvalue equation: ω + k ) 1 ) k = 0.11) Fro this we can easily show, using the equation to solve a quadratic, that the noral frequencies ω 1 and ω are ω 1, = k 1 ± 1 ).113)

29 8 We can now obtain the noral odes of the vertical spring-ass syste by substituting the values for ω 1, in to the eigenvector equation. ω1, + k ) ) k X Y = 0,.114) For noral ode 1, ω 1 ) yields, X/Y = 1/, and for noral ode ω ) yields X/Y = 1/. We can easily visualise the otion that these two noral odes represent, as they are again analogous to the couple pendulu, with one representing a centre of ass otion, where both X and Y are oving in the sae direction, or relative otion where they are oving in opposite directions Fig..4.1).. Figure.9: The vertical spring-ass syste noral odes. On the left-hand side is the centre of ass otion described by X/Y = 1/ and on the right-hand side is the relative otion described by X/Y = 1/..5 Interlude: Solving inhoogeneous nd order differential equations In the next section we will tackle a proble which involves dealing with a inhoogeneous nd order differential equation, where the siple solution obtained by setting the deterinant of the atrix is zero only akes up a part of the general solution. As a precursor to this, in this section we will go through solving a siple nd order differential equation that is no equal to zero. Let us consider the proble:

30 9 dx dt + dy dt + y = t dy dt + 3x + 7y = et 1 We first want to find the copleentary function CF). To get the CF we can write these equations in atrix forat and set the RHS to zero: d dt + 1 ] [ ] x 3 7 d = y dt [ dx dt [ ] 0 0 We then try a solution to this, as usual let s try x = Xe iωt and y = Y e iωt. substitute this into the atrix and find the deterinant and set it equal to zero. Fro this we find the eigenvalues ω = 1 and ω = 3.115) We ω ω ω = 0.116) In the usual way, we then substitute these values of ω back into the eigenvector equation to find the relation between X and Y. Afer a bit of siple aths we find that for ω = 1 X = Y and for ω = 3 X = 4Y/3. Hence the CF is given by [ ] [ ] x 1 = X y a e t, for ω = 1.117) 1/ and [ ] [ ] x 1 = X y b e 3t, for ω = 3.118) 3/4 But the RHS is not actually equal to zero, and the CF only akes up a part of the general solution. In this case we need to find the particular integral PI) as well. Let s deal with the polynoial part first and try, [ dx dt and try the following as solutions, d dt + 1 ] [ ] x 3 7 d = y dt [ ] t 1.119) [ ] x = y [ ] X0 + X 1 t..10) Y 0 + Y 1 t

31 30 With this trial solution dx dt Eq..119, we obtain, = X 1 and dy dt = Y 1, therefore substituting these into X 1 + Y 1 + Y 0 + Y 1 t = t it therefore follows that Y 1 = 1 and X 1 + Y 1 + Y 0 = 0 = X 1 + Y 0 = 1.11) We also have, 3X 0 + X 1 t) + 7Y 0 + Y 1 t) Y 1 = 1 it therefore follows that 3X 0 + 7Y 0 = 0 and 3X 1 + 7Y 1 = 0 = X 1 = 7 3.1) Therefore, we have Y 0 = = 4 3 and X 0 = 7 3 Y 0 = 8 9. We therefore get, [ ] x = y [ t t Finally, we now have to look at the exponential ter. Let s try, so dx dt = Xet and dy dt = Y et. Fro this we find the following, [ ] x = y ].13) [ ] X e t.14) Y [ ] [ ] + 1 X = 3 7 Y [ ] ) Therefore, X + 3Y = 0 X = 3 Y 3X + 5Y = )Y = 1. So we find that Y = and X = 3.. To find the general solution we just bring all of these together, i.e. CF + PI 1 + PI. [ ] [ ] [ ] x 1 = X y a e t 1 + X 1/ b e 3 t + 3/4 [ ] 3 e t + [ t t ].16) where the first two ters coe fro the CF and the second two ters coes fro the two PIs.

32 31 Figure.10: The driven horizontal spring..6 Horizontal spring-ass syste with a driving ter This brings us on to the penultiate syste that we are going to consider in this part of the course. Consider two asses oving without friction connected to two springs, of spring constant k and k, with the spring of k connected to a wall, which is driven by an external k force to have a tie-dependent displaceent, given by xt) = A sin ). t The equations of otion are therefore, ü 1 = k[xt) u 1 ] ku 1 u ) ü = ku 1 u ).17) So as in Sec..5 we first find the copleentary function, and consider the hoogeneous case with xt) = 0. In this case the equations of otion for the following, [ ] d [u1 + 3k dt k ] [ ] 0 = 0 k d dt + k u.18) Let us try the usual for of a solution, [ u1 u ] [ ] X = R e iωt.19) Y and the equate the deterinant of this atrix to zero, and find the eigenvalues. ω + 3k ) ω + k ) k = 0

33 3 ω = k ± 1 16 ) k ) k 8 ω = k ± ) Substituting these values for ω back into the eigenvector equation ω + 3k/)X ky = 0 we find; for ω = + )k/ X3k + )k) = ky therefore Y = 1 + )X.130) for ω = )k/ X3k )k) = ky therefore Y = 1 )X.131) So the ratio of the aplitudes in the CF are ) Y = 1 + and X 1 ) Y = 1 X Now we have to go back to the original equations of otion and find a solution for the particular integral. Writing the equations of otion in atrix for, we have Try the ansatz, [ d dt + 3k k k d + k dt [ u1 u ] = R ] [u1 ] u [ P Q = Ak ] e i k [ ] ] R [e i k t 0 t ).13).133) fro which we find that d P/dt = k exp[i k/t]/, therefore we have [ k + 3k k k 0 ] [ ] P = ka Q [ ] 0.134) This equation is of the for MU = V, rearranging to find U = M 1 V. So we need to calculate the inverse of the atrix M. So to calculate the inverse, we are required to find the deterinant and the transpose of the cofactor atrix see Eq..3). The deterinant is siply M = k ), and the cofactor atrix is

34 33 [ ] 0 k adj = k k and therefore M 1 = k ) [ ] 0 k ) [ ] 0 1 = k 1 k k.135) Substituing back into Eq..134 we obtain Therefore, [ ] P = ka Q [ ] 0 k [ ] [ ] P 0 = Q A ) [ 0 ] ).137) So finally we get to the general solution which is the su of the CF and PI: [ u1 u ] [ ] { [ 1 = A 1 1 k cos + ) [ ] { [ 1 +A 1 + k cos ) + [ ] 0 cos A ] 1 t + φ1 } ] 1 t + φ } k t.138).7 The Forced Coupled Pendulu with a Daping Factor Figure.11: The Forced Coupled Pendulu. The coupled pendulu in this case has both a driving ters given by F cos αt and a retarding force γ v, where v is the velocity of the asses.

35 34 The last exaple that we will work through in this section of the course, is one in which we not only have a driving force, but also a daping ter. This is therefore ore akin to a real-life syste where things are not frictionless. In fact such a syste fors of the basis for any echanical systes in the real world. So let us consider the syste which is shown in Fig..7. However, in addition to the gravitational force and the force due to the spring, we also apply a driving force to the x ass, as denoted by the F cos αt ter, which provides a cyclic puping otion. We also consider a daping force acting on both asses, which is related to the velocity of the asses by F ret = γv. As before we set up the equations of otion. ẍ = γẋ g x + ky x) + F cos αt.139) l ÿ = γẏ g y + ky x).140) l These are exactly the sae equations of otion as foind for the coupled pendulu as you would expect, but with the additional ters due to the daping force in both expressions) and the driving force in the expression for the x otion). This looks like a non-trivial proble, so let us go straight to using the atrix ethod. So the general atrix for the equations of otion is given by [ d dt + g l + k + γ k d dt d k dt + g l + k + γ d dt ] [x ] = F y [ ] 1 Re iαt ).141) 0 We use the fact that cos αt = Re iαt ) here, but you will obtain the sae solution if you use cos αt. The right-hand side is not equal to zero, so this is an inhoogeneous atrix as we could infer directly fro the equations of otion. Therefore, we need to find the solution to the hoogeneous equivalent the copleentary function; CF), and the particular integral. To find the CF, we write down the hoogeneous equations and solve as before;

36 35 so that [ d Try the following, dt + g l + k + γ k d dt dx dt = iωxeiωt dy = iωy eiωt and dt d k dt + g l + k + γ d dt ] [x ] = y [ ] ) [ ] [ ] x X = R e iωt.143) y Y and d x dt = ω Xe iωt d y dt = ω Y e iωt and find the eigenvalues fro the following deterinant, ω + g l + k + γ iω k k ω + g l + k + γ iω = 0,.144) which leads to, ω + iω γ + g l ) + k ) k = 0 = ω + iω γ + g l + k ) ).145) k = ± Solve using the equations for solving a quadratic for both ±k/). For +k/), For k/), ω 1 = iγ g γ ) ± l.146) g ω 1 = iγ ± l + k ) γ ).147) You should notice that these are siilar to the eigenvalues in the siple coupled pendulu case, where ω 1 = g/l and ω = g/l + k/ Eq..7). We have the sae ters for this syste, as you ight expect, but also the additional ters that include the daping factor. So we can rewrite these in ters of these original values for ω 1, as, ω 1, = iγ γ ). ± ω1,.148)

37 36 There is no physical difference between the ± variants here, so fro now on we will just use the positive solution. We substitute these eigenvalues into the eigenvector equation to find the ratio of the aplitudes in each noral ode, and thus find the CF. With a bit of siple aths, for ω 1, we find X = Y and for ω we find X = Y. Reebering Eq..143, x = RXe i ω 1,t ) and y = RY e i ω 1,t ) we find that, [ ] { [ ] [ x γt = e ) 1 γ ) ) 1 ] [ ] [ A y 1 cos ω1 1 γ ) ) 1 ]} t + φ1 + A 1 cos ω t + φ 1.149) where ω 1 = g/l and ω = g/l + k/. As we ight expect, we introduce an exponential decay factor, which describes the daping force acting on the two asses. As a note, you can solve this part with the decoupling ethod and then using a trial solution of the for q = Re iωt ) and you get the sae results. However, to obtain the general solution, which includes the driving force ter, we need to calculate the PI as well. Starting fro Eqn..141, we now try the ansatz, [ ] {[ ] } x P = R e iαt y Q.150) which requires us to solve a atrix of the for MU = V, as we did in Sec..6, so we rearrange to find U = M 1 V. Therefore we have, [ ] P = M 1 F Q [ ] ) and use Eq..3 to find M 1. So we first find the deterinant of M,

38 37 where ω 1 = g/l and ω = g/l + k/. This is a bit unwieldy, so it is generally easier to write this in ters of polar coordinates, i.e. where [ M = α + iα γ g + l )] + k [ ] k = α + iα γ ) + ω 1 α + iα γ ).15) + ω B 1, = [ ω 1, α ) + M = B 1 e iθ 1.B e iθ,.153) αγ ) ] 1 and tan θ 1, = αγ ω 1, α ) So we now have the deterinant, and the next thing to find is the adjoint atrix, which is just the transpose of the cofactor atrix. wich we can write as [ α adjm = + iα γ + g k l + k ) k α + iα γ + g l + K ] ).154) α + iα γ g + l + k ) = 1 Siilarly, we can also write k = 1 α + iα γ + g ) + 1 α + iα γ l + g l + k ) α + iα γ + g l + k = 1 ) B 1 e iθ 1 + B e iθ. ) 1 α + iα γ + g ) l = 1 ) B e iθ B 1 e iθ ).156) Substiting these back to find the adjoint, adjm = 1 [ B1 e iθ 1 + B e iθ ) B e iθ B 1 e iθ ] 1) B e iθ B 1 e iθ ) 1 B1 e iθ 1 + B e iθ )..157) Therefore, M 1 = 1 [ eiθ1+θ) B1 e adjm = iθ 1 + B e iθ ) B e iθ B 1 e iθ ] 1) M B 1 B B e iθ B 1 e iθ ) 1 B1 e iθ 1 + B e iθ ) [ 1 B1 e = iθ + B e iθ ) 1 B e iθ 1 B 1 e iθ ].158) ) B 1 B B e iθ 1 B 1 e iθ ) B1 e iθ + B e iθ 1)

39 38 Recall, [ ] x = F [ ] ) y R M 1 1 e iαt,.159) 0 therefore [ ] x = y F B 1 B [ ] B1 cosαt + θ ) + B cosαt + θ 1 ) B cosαt + θ 1 ) B 1 cosαt + θ ).160) with B 1, = [ ω 1, α ) + αγ ) ] 1 and tan θ 1, = αγ ω 1, α ) So finally, the general solution is the su of the CF and the PI; [ ] { [ ] [ x γt = e ) 1 γ A y 1 cos ω1 1 { [ ] [ γt +e ) 1 γ A cos ω 1 F + B 1 B ) ) 1 t + φ1 ]} ) ) 1 t + φ ]} [ B1 cosαt + θ ) + B cosαt + θ 1 ) B cosαt + θ 1 ) B 1 cosαt + θ ) ]..161) The copleentary function is the transient solution deterined by the external conditions, whereas the particular integral gives the steady state solution deterined by the driving force.

40 Chapter 3 Noral odes II - towards the continuous liit In the last few sections we have built up the foundations for the study of waves in general. For the reainder of the course we will be focusing on various types of wave, and the general applicability of the equations which govern waves across any parts of physics. We will look at two general fors of waves; first we will consider the non-dispersive syste, which eans that the speed at which the wav travels is not dependent on the wavelength and frequency. Later one we will look at dispersive waves, where the speed at which the wave travels does depend on the frequency, this leads to the introduction of group velocity. 3.1 N-coupled oscillators Fro our work looking at noral odes, we know that we can describe a syste of coupled oscillators by the linear superposition of N noral odes, where N in the coupled pendulu case was the nuber of asses on pendula for exaple. In this section we will look what happens when we extend this superposition to the continuous liit, i.e. for a single piece of wire or string. Rather than individual asses here, we would have a string that you could consider as being the suation of very sall bits of string of length dl and a ass that relates to these very sall parts of the string. Fig shows a diagra of a stretched string, where we split the string up into sall ass eleents. In this exaple, we assue that all of the angles of the stretched 39

41 40 Figure 3.1: A siple string that has been stretched by a very sall aount at its centre. string fro the horizontal are sall. As before, we write down the equations of otion for each eleent of the string: In the vertical direction we have, F py = ÿ p = T sin α p 1 + T sin α p 3.1) For the horizontal direction we have F px = T cos α p 1 + T cos α p. 3.) As we assue α is sall, then sin α i tan α i α i. We can also assue, cos α i 1 α i, using the trig identities cos θ + sin θ = 1 and cos θ = cos θ sin θ. Therefore, in the horizontal direction we have zero net force as α p 1 α p, as we would expect if we only stretch the string in the vertical direction at its centre. For the vertical direction, Eq. 3.1 becoes F p = ÿ p = T l y p y p 1 ) + T l y p+1 y p ). 3.3) If we define ω0 = T/l then this becoes ÿ p + ω0y p ω0y p+1 y p 1 ) = ) Special cases Before we ove on to the general solution to this equation, let us first look at a few specific, special cases. Naely, the ost siple ones with N = 1 and N =.

42 41 Fig So if we have N = 1 then the stretch string just looks like the syste shown in Figure 3.: A stretched string with N = 1. Eq.3.6 becoes ÿ 1 + ω0y 1 = 0, 3.5) as y p+1 = y p 1 in this case, and the equation is the usual for for SHM, i.e. acceleration proportional to the displaceent. Therefore, we can iediately show that the angular frequency of the oscillation ω = ω 0 with ω 0 = T l for N = 1 Now let us look at the N = case. By just drawing what the possibilties are in this case Fig ), we know that we should have two solutions. Figure 3.3: Two possibilities for the starting points for a stretched string with N =. Obviously you can have the inverse of these but that is just the sae as inverting the direction of y. Writing 3.6 for N = we get, ÿ 1 + ω 0y 1 ω 0y = 0 ÿ + ω 0y ω 0y 1 = 0. Let us define the noral coordinates, as we did in Sec..1, i.e. 3.6) q 1 = 1 y 1 + y ) and q = 1 y 1 y )

43 4 which lead to, q 1 + ω 0q 1 = 0 therefore ω = ω 0 q + 3ω 0q = 0 therefore ω = 3ω 0. Therefore, we have two possible noral odes, as you would expect for N =. This is siply analagous to the coupled pendula and N-spring systes that we looked at in previous lectures. So now that we have discussed these two special cases, let us ove on an try to find a general solution to the syste, where N can be anything General case Fro Eq. 3.6 we have ÿ p + ω0y p ω0y p+1 y p 1 ) = 0. This is a second order differential equation and we expect an oscillatory solution, so let us consider a solution of the for y p = A p cos ωt. Note that here we have not included a phase offset ter, which we represented by φ in previous lectures. By oitting this ter here we are just iposing the fact that all the asses start at rest. Obviously we could reintroduce this ters later. So substituting the trial solution into Eq. 3.6 gives N equations: ω + ω 0)A p ω 0A p+1 A p 1 ) = 0 with p = 1,, 3..., N = A p+1 + A p 1 A p = ω + ω 0 ω 0 3.7) But we know that since ω is the sae for all asses, i.e. it doesn t atter where we are on the string, then the right-hand side cannot depend on p. So if the RHS does not depend on p, neither can the left hand side. We also know that if the string is fixed at both ends, then for p = 0 and p = N + 1, A p = 0. So we need to look for fors of A p which satisfy these criteria. Let us try a description of A p of the for, A p = C sin pθ, so the left-hand side of Eq. 3.7 becoes,

44 43 A p+1 + A p 1 A p = C [sinp + 1)θ + sinp 1)θ] C sinpθ) = C sinpθ) cos θ C sinpθ) = cos θ, 3.8) which satisfies the criteria for the equation to be independent of p. Now for the second criteria of when p = 0orp = N + 1, then A p = 0, we just have to require that N + 1)θ = nπ. So cobining all of this we have ) pnπ A p = C sin N ) Substituting this back into Eq. 3.7, [ )] nπ ω = ω0 1 cos N + 1 ) 3.10) nπ = ω = ω 0 sin N + 1) So now we can write the general solution for the displaceent in y. Cobining Eq. 3.6 with Eq and reintroducing the phase offset such that y p A p cosωt + φ p ), we obtain ) ) pnπ nπ y pn t) = C n sin cosω n t + φ n ) with ω n = ω 0 sin N + 1 N + 1) 3.11) where ω 0 = T l Although the value of n can be greater than N, this would just generate duplocate solutions, and there are always N noral odes in total. Fig shows the 5 noral odes for the case of N = 5. One thing to note is that all the asses are displaced in such a way that they fall on an underlying sine curve, as you would expect given the solution that we have just found.

45 44 Figure 3.4: Noral odes for the case of N = 5 with snapshots taken at t = 0. As in the case of the coupled pendula, with N = 5 there are five noral odes and the subsequent otion of the string can be described by the superposition of these odes, with teh initial conditions dictating which noral odes are active in a syste. Also, note that the displaceent of the asses on each string filled red circles) all fall on a sine curve N very large Obviously the syste we considered in the last section does not really represent what happens to a continuous, unifor piece of string. We generally do not have asses along the string that represent the nuber of displaceent points. But we can use this as a starting point for considering a real string, i.e. we can assue that N is very large and describe the asses in ters of the linear density of the string and assue that the string has a unifor density per unit length.

46 45 So let us consider a string-ass syste which has a length L and a total ass M. We can consider this string as being ade up of a series of sall eleents of length l and ass, such that L = N + 1)l and M = N, and define the linear density of the string as ρ = /l. Eq describes how the frequency of the oscillation for each noral ode. If we just consider the ode nubers n which are sall in coparison to N, which would be the case of N is very large as we are assuing then we essentially reove the sine dependence and find, ) ) nπ T nπ ω n = ω 0 sin = N + 1) l sin N + 1) ) T nπ = ω n =. /l N + 1)l As we have defined the linear density above, this then becoes, 3.1) ω n = nπ L T ρ. 3.13) This eans that all of the noral frequencies are integer ultiples of the lowest frequency given when n = 1, i.e. ω 1 = π L T ρ. So we now know that the noral frequencies are just given y integer ultiples of the lowest frequency ode. We now consider the displaceent of the string for the sae liit of n sall copared to N. Starting fro Eq. 3.11, we have ) pnπ y pn t) = C n sin cosω n t + φ n ), N + 1 but as the eleents of the string, each of length l, becoe saller and saller, we approach a continuous variable along the x axis, which we define as x = pl, such that y n x, t) = C n sin xnπ ) cosω n t + φ n ), 3.14) L

47 46 resulting in a sinusoidal wave in both x and t, i.e. we have derived the equation for the coplete otion of the string, at least for when n << N. Fro this we can calculate the vertical y) displaceent of the string at and tie t and at any point along the axis of the string x. Let us now look what happens when we ove to n = N. Fro Eq we know that this ust give us the highest frequency ode of the oscillations, ) nπ ω N = ω 0 sin ω ) N + 1) If we now consider the ratio of the displaceents of successive eleents of the string for the n = N ode using Eq. 3.11, i.e. y p y p+1 = sin ) pnπ N+1 p+1)nπ N+1 sin ) sinpπ) ) sinpπ + π) So everything successive eleent is approxiately displaced equally but in the opposite direction to the previous eleent. If we lost the approxiations, and calculated a ore rigourous solution then what we would find is that we obtain adjacent positive and negative displaceents Fig ) where the aplitude of the strin is the axiu at the centre. Figure 3.5: Illustration of the adjecent displaceet of the string for the n = N ode, resulting in the highest frquency ode. therefore So referring back to Eq. 3.1, we now know that for n = N, y p 1 y p y p+1, F p = ÿ p = T l y p y p 1 ) + T l y p+1 y p ) = T l y p) T l y p) = 4y p T l 3.17)

48 47 We defined ω 0 = T/l, therefore ÿ 4yω0, fro which we find the result ω N ω Longitudinal Oscillations In the previous section we considered only transverse otion of a string. Obviously waves can also exist as longitudinal waves, e.g. sound waves and oscillations on a spring. Here we will quickly deonstrate that the sae equations that we found in the previous section also hold for longitudinal waves. So rather than the string with N-ass eleents, let us consider a horizontal spring syste with N asses between siilar springs with spring constant k Fig ). Figure 3.6: Horizontal spring-ass syste with N-asses. If u p is the displaceent fro equilibriu of ass p, then the equation of otions of each ass is given by, ü p = ku p+1 u p ) + ku p 1 u p ) = ü p + ω0u p ω0u p+1 + u p 1 ) = 0 with ω0 = k 3.18). This is exactly the sae for as the equation of otion as given in Eq. 3.6 for the transerve oscillations of a string ade up of N asses. Therefore, the solutions will be exactly the sae.