Problem 31. Derive Hamilton s equations for a one-dimensional harmonic oscillator. Seminar 8. HAMILTON S EQUATIONS Solution: The Lagrangian L = T V = 1 m q 1 kq (1) yields and hence the Hamiltonian is Note that p = L q = m q q = p m, () H = qp L = ( p ) 1 p ( m m q 1 kq ) = p m + 1 kq. (3) H = T + V. (4) The Hamiltonian (3) creates Hamilton s equations as follows: { q = H = p p m ṗ = H = kq (5) q These two equations immediately yield the equation of motion for a harmonic oscillator, q = k q, (6) m or where q + ω 0q = 0, (7) ω 0 = k m. (8) Problem 3 Derive Hamilton s equations for a mathematical pendulum.
Solution: Letting the potential energy be zero at θ = 0, we may write the Lagrangian as L = 1 ml θ mgl(1 cos θ), (9) which gives and p θ = L θ = ml θ θ = p θ ml (10) H = p θ θ L = pθ θ ( 1 ml θ mgl(1 cos θ) ) = p θ ( pθ ml ) 1 ml( p θ ml ) + mgl(1 cos θ) = 1 ml p θ + mgl(1 cos θ). (11) Hence the canonical equations of Hamilton take the form { θ = H p θ ṗ θ = H = p θ ml θ = mgl sin θ (1) These equations immediately results in the celebrated equation of motion for a mathematical pendulum, θ + g l sin θ = 0. (13) Problem 33 Derive Hamilton s equations for an linear array of coupled harmonic oscillators. As a model, consider a light elastic string of the elastic constant k, clamped at both ends and loaded with n particles, each of mass m, equally spaced along the length of the string. Solution: Let us label the displacements of the various particles from their equilibrium positions by the coordinates q 1,..., q n. Although in the actual physical situation the particles can move both in the longitudinal and transverse directions. However, for simplicity we shall assume that the motion is either purely longitudinal or purely transverse, as shown in igure (will be shown during the seminar). In both cases the
kinetic energy of the system is given by T = m ( q 1 +... + q n). (14) The potential energy requires a special consideration in each case. Namely, in the case of longitudinal motion the stretch of the section of string between any particle j and its neighbour particle j + 1 is q j+1 q j, (15) and hence the potential energy of this section of the string is 1 k(q j+1 q j ). (16) or the case of transverse motion, the distance between particles j and j + 1 is [d + (q j+1 q j ) ] 1/ = d + 1 d (q j+1 q j ) +..., (17) in which d is the equilibrium distance between two adjacent particles. The stretch of the section of string connected the two particles is then approximately l = 1 d (q j+1 q j ). (18) Thus, if is the force of tension in the string, the potential energy of the section under consideration is given by l = d (q j+1 q j ). (19) It follows that the total potential energy of the system in either the longitudinal or the transverse type of motion is expressible as a quadratic function of the form V = K [q 1 +... + (q j+1 q j ) +... + q n], (0) in which { k, for longitudinal motion K =, for transverse motion (1) d The Lagrangian for either case is, therefore, has the form L = m ( q 1 +... + q j +... + q n) K [q 1 +... + (q j+1 q j ) +... + q n]. () This yields the congugate momenta p j = L q j = m q j. (3)
Combining these equations, we obtain the Hamiltonian for the system at hand as n n H = q j p j L = [m q j m qj + K ] (q j+1 q j ) = 1 n [ p j m +K(q j+1 q j ) ]. (4) j=0 j=0 or this Hamiltonian, the canonical equations of Hamilton gives j=0 { qk = H p k = p k m ṗ k = m q k ṗ k = H q k = K(q k q k 1 ) + K(q k+1 q k ) Substituting ṗ k from the second of these equations into the first one, we derive the set of the equations of motion (5) m q k = K(q k q k 1 ) + K(q k+1 q k ), (6) where k = 1,,..., n. or the trial solution of the harmonic form, q k = a k cos ωt, (7) the differential equations (6) for the generalized coordinates q k turn to the algebraic equations for the amplitude of oscillations of the kth particles, To close this system of equations, we set mω a k = K(a k 1 a k + a k+1 ). (8) a 0 = a n+1 = 0. (9) These equations express the boundary conditions at the eindpoints of the string which are motionless, by definition. We can proceed with the solution of this set of n equations in different ways. One possibility is to use the secular equation in order to find n values of ω through finding the n roots of this equation. However, in the case of n particle this method requires a little bit lengthy algebra. To avoid this, it is instructive to show how these roots can be found directly from the system (8). or this, let notice that Eq. (8) can be treated as the recursion formula for the amplitudes of oscillations of the adjacent particles, and introduce a quantity φ related to the amplitudes as a k = A sin k φ. (30) Direct substitution of this expression into the recursion formula yields mω A sin k φ = KA[sin(kφ φ) sin kφ + sin(kφ + φ)], (31)
which easily reduces to or mω = K( cos φ) = 4K sin φ, (3) K ω = m sin φ = ω 0 sin φ. (33) We see that to find the possible frequencies ω we need to know only the quantity φ. This quantity is not determined as yet, but it can be easily done by using the boundary conditions (9). The first og them, a 0 = 0, is satisfied automatically by the choice of the amplitude in the form given by Eq. (30). The other condition, namely, a n+1 = 0, will be met if we set in which N is an integer, because we then have as requires. ()φ = Nπ, (34) a n+1 = A sin ()φ = A sin Nπ = 0, (35) With this φ, the frequencies given by Eq. (33), become Nπ ω = ω 0 sin. (36) n + Moreover, from Eqs. (30) and (34) we see that the amplitude of the oscillation a k is Nπk a k = A sin. (37) We recall that the value of k denotes a particular particle in the array, while the value of N refers to the normal mode in which the system as a whole can oscillate. By plotting the amplitudes as given by Eq. (37), we can illustrate the different normal modes graphically. It will be shown on igure during the seminar the three first normal modes of oscillations in an array, consisting of three particles. The actual motion of the system, when it is oscillating in an N pure mode, obeys the equation q k = a k cos ω N t = A sin Nπk cos ω N t. (38) inally, the general type of motion is a linear combination of all normal modes, that is n Nπk q k = A N sin cos (ω N t δ N ). (39) N=1
Here we introduced the amplitude A N and phase shift δ N for the N mode whose values are determined from the initial conditions. It is instructive to cast a special look at the case where the number of masses on the string is very large (something like a very large number of very closely spaced atoms). Let the number of particles n increase but the spacing d between neighboring particles decrease, such that the length of the string L = ()d is held constant. Then for the modes N, such that N << n, the argument of the sine in the r.h.s. of Eq. (36) defining the normal frequencies is small, and we can write approximately Nπ ω N ω 0. (40) Recalling that for the transverse oscillations, we may rewrite Eq. (40) as ω N ω 0 = 1 1 K =, (41) m md 1 1 Nπ Nπ =. (4) md m/d ()d But ()d = L, the total length of the string, and m/d is, by definition, its mass per unit length, or the linear mass density, We thus have approximately ω N = N π L µ m d. (43) 1 µ (N = 1,,...) (44) We see that the normal frequencies are integral multiples of the form where ω N = Nω 1, (45) ω 1 = π L 1 µ is the lowest,or fundamental, frequency. Of course, this is only approximation, but for N << n it is very nice approximation! (46) Then, let examine the actual displacements of the particles under these conditions. irst, let denote the kth particle by its distance down the string from the fixed end x, so that x = kd. (47)
In terms of this new variable we have knπ () = Nπ(kd) ()d = Nπx L. (48) Replacing the argument of the sine term in the expression for q k given by Eq. (38), we can rewrite this equation as Nπx πx q N (x, t) = A sin cos ω N t A sin cos πf N t. (49) L By this, we have defined the wavelength λ N and the frequency f N by Physically, Eq. λ N = L N, λ N f N = ω N π. (50) (49) represents the displacement of any point along a continuous string when it is vibrating in its Nth mode with the frequency f N wave of wavelength λ N. as a standing Each vibrational mode consists of an integral number of half-wavelength units constrained to fit within the length L such that the eindpoints do not vibrates, that is, they are said be the nodes. In conclusion, it is worth to notice that an analysis of the dynamics of a line of interconnected masses was first attemted by Newton, himself. Two of his successors, the remarkable Bernoullis (John and his son Daniel), have ultimate success with the problem. They first demonstrated that a system of n masses has exactly n independent modes. urthermore, in 1753, Daniel has shown that the general motion of this system can be described as a superposition of its normal modes. This investigation by the Bernoullis, - Leon Brillouin writes, - may be said to form the beginning of theoretical physics as distinct from mechanics, in the sense that it is the first attempt to formulate laws for the motion of a system of particles rather than for that of a single particle..