Assignment 6. Using the result for the Lagrangian for a double pendulum in Problem 1.22, we get

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1 Assignment 6 Goldstein 6.4 Obtain the normal modes of vibration for the double pendulum shown in Figure.4, assuming equal lengths, but not equal masses. Show that when the lower mass is small compared to the upper one, the two resonant frequencies are almost equal. If the pendula are set in motion by pulling the upper mass slighlty away from the vertical and then releasing it, show that subsequent motion is such that at regular intervals one pendulum is at rest while the other has its maximum amplitude. This is the familiar phenomenon of beats. Using the result for the Lagrangian for a double pendulum in Problem., we get L = (m + m ) l θ + m l θ m l θ θ cos(θ + θ ) + (m + m ) gl cos θ + m gl cos θ where we have set the lengths to l and kept the masses distinct. Note too that we have followed Goldstein s convention in the figure and taken θ as a counterclockwise measured angle and θ as a clockwise measured angle. As we are working within the assumption of small oscillations, we can make further simplifications to the Lagrangian. Working to second order in the potential, we can let cos θ θ / and similarly for cos θ. However, we only work to zeroth order in the kinetic energy and so cos(θ + θ ). While this seems contradictory, it is the basic assumption of small oscillations. The Lagrangian becomes L = (m + m ) l θ + m l θ m l θ θ (m + m ) gl θ m gl θ where we have dropped some constant terms from the potential energy piece as they do not contribute to the equations of motion. If we define θ η = θ the Lagrangian is L = η T T η ηt V η where the kinetic and potential energy matrices are T = l m + m m m m m + m V = gl m With these in hand, we can calculate the characteristic frequences from the secular equation: = det ( V ω T ) (m = + m )(gl l ω ) l m ω l m ω m (gl l ω ) = l 4 m (ω ω ) ω 4 m where, for convenience, we have defined ω = g/l and = m + m. The eigenfrequencies are ω = ω ± m m

2 The normal modes come from solving V η ± = ω±t η ± for η ±. equation we get the relevant equations Putting the eigenvalues into this matrix which, for the eigenfrequencies, reduce to and provide the (unnormalized) eigenvectors ( ω ω ±) θ + m ω ± θ = m ω ± θ + m ( ω ω ±) θ = m η + = m θ = ± θ η = m Note that if m << m, then we can write the frequencies as ω ± ω ( ± ɛ) where ɛ = m /m and the eigenfrequencies are very nearly the same. The normal modes are just the combination of these eigenvectors in the form ζ = A η where A is the matrix comprised of the above eigenvectors m m A = The time dependance is included in a simple way since the normal modes are just solutions of the harmonic oscillator equation for each eigenmode with the corresponding eigenfrequency: Putting all of this together, we get ζ ± + ω ± ζ ± = ζ ± (t) = C ± e iω±t where the constants C ± are complex and the normal modes are ζ ± (t) = θ (t) ± m θ (t) = C ± e iω± t where we know to take the real part. The initial conditions to consider are θ () = θ together with the vanishing of θ, θ and θ at t =. This yields the conditions θ = Re (C ± ) We thus have θ (t) ± = Re ( iω ± C ± ) Combining these so as to get θ and θ alone, we have θ (t) = θ θ (t) = θ m θ (t) = θ cos(ω ± t) cos(ω + t) + cos(ω t) cos(ω + t) cos(ω t) m

3 In order to analyze beat phenomena, it is useful to consider two different frequencies, the average and difference of the eigenfrequencies: In terms of these, we have and we get ω = (ω + + ω ) ω = (ω + ω ) ω ± = ω ± ω θ (t) = θ cos( ω t) cos( ωt) θ (t) = θ sin( ω t) sin( ωt). m In this form we see the overall oscillation of both pendula with frequency ω > ω together with a modulating oscillation in the amplitude of frequency ω. This modulating amplitude is 9 out of phase between the two pendula. Said another way, when the modulated amplitude for θ is a maximum (say ωt n = nπ), the amplitude of θ is zero and vice versa. This is the phenomena of beats. 3

4 Goldstein 6.8 The equilibrium configuration of a molecule is represented by three atoms of equal mass at the vertices of a 45 right triangle connected by springs of equal force constant. Obtain the secular determinant for the modes of vibration in the plane and show by rearrangement of the columns that the secular equation has a triple root ω =. Reduce the determinant to one of third rank and obtain the nonvanishing frequencies of free vibration. Perhaps the hardest part of this problem is setting things up in a clear way. We begin by orienting an x-y coordinate system so that the atom at the right angle of the 45 triangle is at the origin and the hypotenuse is in the first quadrant. With this orientation, the equilibrium positions (no motion) of the particles are (, ), (, a) and (a, ). For clarity, we will further label our particles clockwise from the origin so that particle in equilibrium is at (, ), particle is at (, a) and particle 3 is at (a, ). The absolute coordinate locations of these particles are then (x, y ), (x, y ) and (x 3, y 3 ). Each of these coordinates is, of course, a function of time. We also give the displacements of these particles from their equilibrium positions, namely (η, ξ ) = (x, y ) (η, ξ ) = (x a, y ) (η 3, ξ 3 ) = (x 3, y 3 a) It is these displacements that we will use as our generalized coordinates. The kinetic energy is obvious. Indeed, the kinetic energy matrix will be the identity matrix multiplied by the mass, m. The potential energy is a bit harder to come by. The potential energy in the spring between particles and is given by displacement of the spring away from its equilibrium length. Said another way, we want the difference between the length of the spring at a moment of time, t, and the equilibrium length a. This is (x x ) + (y y ) a Thus the potential energy associated with that spring is V = k (η + a η ) + (ξ ξ ) a The other pieces of the potential energy will look similar, namely V 3 = k (η3 η ) + (ξ 3 + a ξ ) a V 3 = k (η3 η a) + (ξ 3 + a ξ ) a The total potential energy is the sum of these three pieces. It is a mess, to say the least. It is also not a very convenient form for the methods we want to use. But we need to realize that we are considering small oscillations and we must expand the potential about the equilibrium positions. The practical way of saying and doing this is to note that the equilibrium lengths a are much larger than either types of displacements, η, ξ. So we expand the square roots in powers of the (small) displacements: V = k = k a + a (η η ) + a (η η ) + a (ξ ξ ) a ( a + ) a (η η ) + a k (η η ) where we have thrown away all terms inside the original square brackets second order and higher in the displacements. Similar analyses with the other potential terms yields V 3 k (ξ 3 ξ ) V 3 k 4 (η 3 η ) (ξ 3 ξ ). 4

5 The total potential energy is thus V = k (η η ) + k (ξ 3 ξ ) + k (η 3 η ) + k (ξ 3 ξ ) k (η 3 η )(ξ 3 ξ ) The potential energy matrix for η, ξ, η, ξ, η 3, ξ 3 is thus and the characteristic equation becomes V ω T = k 3/ / / / V = k / / / / / / / / / / / 3/ mω /k mω /k 3/ mω /k / / / / / mω /k / / / / / mω /k / / / / 3/ mω /k The best way to evaluate this is to use some computer algebra program. Doing so yields V ω T = k 5 ( mω ) 3( mω k )( mω k )( mω 3k ) which is a sixth order polynomial in ω. One of the solutions is a third order zero (ω = ). These zero eigenfrequencies correspond to the three possible rigid body motions (modes) of the triangular molecule, two independent translations in the x-y plane and a single rotation in the plane (about an axis perpendicular to the plane a z axis if you will). The remaining frequencies are the frequencies of free vibration, namely ω = k m, k m, 3k m 5

6 Goldstein 6. A uniform bar of length l and mass m is suspended by two equal springs of equilibrium length b and force constant k, as shown in a diagram in Goldstein. Find the normal modes of small oscillation in the plane. Referring to the figure, we will place a coordinate system in the following way. In the equilibrium configuration when the system is not moving, we place the origin of our coordinate system at the center of the bar. The right end of the bar is at (x, y) = (l/, ) while the left end of the bar is at (x, y) = ( l/, ). The upper right support for the right spring is at a coordinate location of (x, y) = (l/ + b sin θ, b cos θ ) while the left support has a coordinatization of (x, y) = ( l/ b sin θ, b cos θ ). The equilibrium length of both springs under the influence of gravity is b. It is important to note that this already takes into account the potential energy due to gravity. Because we will be considering small displacements (oscillations) away from equilibrium, any linear terms in the potential will already have been discarded. In other words, we are working at energies slightly above the minimum of the potential. There are three degrees of freedom in the problem. They include the two dimensional motion of the center of mass of the bar. These displacements we will call x and y, displacements from the origin of coordinates. There is also a rotational degree of freedom associated with the bar s motion about its center of mass. We will identify this with the angle φ and measure it from the horizontal (equilibrium) position and take it to be positive in the counterclockwise direction. Our kinetic energy becomes T = m (ẋ + ẏ ) + I φ where the moment of inertia of our rod about the center of mass is I = ml /. Because of how φ appears later, we will write the kinetic energy as T = m ẋ + ẏ ) + (l φ/) 3 Now we consider the potential terms. The potential energy in the right spring we will take to be related to the difference in the lengths of the stretched and unstretched springs, namely V = k { (x l + cos φ) ( l + b sin θ ) (y l + + sin φ) b cos θ b k { } (x ) ( b sin θ + y + lφ b cos θ ) b k x sin θ + ( y + lφ) cos θ where in the second line we have assumed small oscillations in the form that cos φ and sin φ φ. To get the third line, we have simply expanded and kept only terms from the square root that are linear in our small displacements, x, y and φ. Expanding the other potential energy we get something similar V = k { (x l cos φ) ( l b sin θ ) (y l } + sin φ) b cos θ b k x sin θ ( y lφ) cos θ } Combining these we get for the total potential (again, we drop the linear terms as we are working only with the small oscillations) V k ( lφ ) ( lφ ) x sin θ + y cos θ + cos θ + 4x sin θ cos θ 6

7 Constructing the kinetic and potential matrices, we have T = m m m/3 V = k cos θ sin θ sin θ cos θ sin θ cos θ cos θ The characteristic equation, det(v ω T) =, becomes = ( k cos θ mω ) ( k sin θ mω )( k cos θ 3 mω) 4k sin θ cos θ This has solutions for the eigenvalues of ω = ω = k m cos θ ω 3 = k m ( + cos θ ) The corresponding (unnormalized) eigenvectors can now be found as η = cos θ sin θ η = η 3 = sin θ 3 cos θ Note that the zero mode corresponds to a pure rigid body motion. It amounts to a horizontal motion of the center of mass combined with a rotation of the bar about the center of mass. No motion takes place in the vertical direction. The normal mode associated with η is a purely vertical oscillation of the center of mass. The third normal mode is the highest frequency mode and produces oscillations which combine a horizontal and rotational aspect. 7

8 Goldstein 6. Two particles move in one dimension at the junction of three springs, as shown in the figure in Goldstein. The springs all have unstretched lengths equal to a, and the force constants and masses are shown. Find the eigenfrequencies and normal modes of the system. After the other problems this is perhaps the easiest of the lot. For two particles moving in one dimension, we have only two degrees of freedom. The distance between the walls is 3a. The absolute coordinate location of the first particle is x and the second is x as measured from an origin at the left wall. The corresponding displacements from eqilibrium are η = x a The kinetic energy of the system is and the potential energy is η = x a T = m ( η + η ) V = k ( x a ) 3k ( + (x a) (x a) ) k ( + x a ) = k η + 3(η η ) + η Using the matrix approach, we have the kinetic and potential matrices given by m T = m 4k 3k V = 3k 4k The characteristic equation for this system thus becomes which gives us the eigenfrequencies of det ( V ω T ) = ( k mω )( 7k mω ) = ω = k m, 7k m We now want the normal modes of the system, i.e. the characteristic motions for η and η. To this end, we must solve V η = ω T η for η T = (η η ). In the case that ω + = k/m, we find the relevant equation to be η = η and our unnormalized eigenvector is η ω+ =. In the case that ω = 7k/m, we find the relevant equation to be η = η with an unnormalized eigenvetor of η ω+ =. Both, of course, can be normalized with an extra factor of /. The modes, then, correspond to symmetric motion in phase for the ω + mode and antisymmetric motion out of phase for the ω mode. 8