Chapter 7 Lagrangian and Hamiltonian Mechanics

Home , Atwood machine, Double pendulum, Hamiltonian mechanics, Harmonic oscillator, Lagrangian mechanics

Abstract Chapter 7 is devoted to problems solved by Lagrangian and Hamiltonian mechanics.

7.1 Basic Concepts and Formulae

Newtonian mechanics deals with force which is a vector quantity and therefore dif- ﬁcult to handle. On the other hand, Lagrangian mechanics deals with kinetic and potential energies which are scalar quantities while Hamilton’s equations involve generalized momenta, both are easy to handle. While Lagrangian mechanics con- tainsn differential equations corresponding ton generalized coordinates, Hamil- tonian mechanics contains 2n equation, that is, double the number. However, the equations for Hamiltonian mechanics are linear. The symbolq is a generalized coordinate used to represent an arbitrary coordi- natex,θ,ϕ, etc. IfT is the kinetic energy,V the potential energy then the LagrangianL is given by

L T V (7.1) = − Lagrangian Equation: d dL ∂L 0(K 1,2...) (7.2) dt d q − ∂q = = ˙ K K ∂v where it is assumed thatV is not a function of the velocities, i.e. 0. Eqn (2) ∂ qK = is applicable to all the conservative systems. ˙ Whenn independent coordinates are required to specify the positions of the masses of a system, the system is ofn degrees of freedom.

r HamiltonH s 1 ps qs L (7.3) = = ˙ − wherep s is the generalized momentum and qK is the generalized velocity. ˙

287 288 7 Lagrangian and Hamiltonian Mechanics

Hamiltonian’s Canonical Equations

∂H ∂H qr , pr (7.4) ∂pr =˙ ∂qr =−˙

7.2 Problems

7.1 Consider a particle of massm moving in a plane under the attractive force μm/r 2 directed to the origin of polar coordinatesr,θ. Determine the equations of motion. 7.2 (a) Write down the Lagrangian for a simple pendulum constrained to move in a single vertical plane. Find from it the equation of motion and show that for small displacements from equilibrium the pendulum performs simple harmonic motion. (b) Consider a particle of massm moving in one dimension under a force with the potentialU(x) k(2x 3 5x 2 4x), where the constantk> 0. Show that the pointx =1 corresponds− to+ a stable equilibrium position of the particle. Find the= frequency of a small amplitude oscillation of the particle about this equilibrium position. [University of Manchester 2007] 7.3 Determine the equations of motion of the masses of Atwood machine by the Lagrangian method. 7.4 Determine the equations of motion of Double Atwood machine which consists of one of the pulleys replaced by an Atwood machine. Neglect the masses of pulleys. 7.5 A particular mechanical system depending on two coordinatesu andv has kinetic energyT v 2 u2 2 v2, and potential energyV u 2 v 2. Write down the Lagrangian= for˙ the+ system˙ and deduce its equations= of motion− (do not attempt to solve them). [University of Manchester 2008] 7.6 Write down the Lagrangian for a simple harmonic oscillator and obtain the expression for the time period. 7.7 A particle of massm slides on a smooth incline at an angleα. The incline is not permitted to move. Determine the acceleration of the block. 7.8 A block of massm and negligible size slides on a frictionless inclined plane of massM at an angleθ with the horizontal. The plane itself rests on a smooth horizontal table. Determine the acceleration of the block and the inclined plane. 7.9 A bead of massm is free to slide on a smooth straight wire of negligible mass which is constrained to rotate in a vertical plane with constant angular speedω about afixed point. Determine the equation of motion andfind the distancex from thefixed point at timet. Assume that att 0 the wire is horizontal. = 7.2 Problems 289

7.10 Consider a pendulum consisting of a small massm attached to one end of an inextensible cord of lengthl rotating about the other end which isﬁxed. The pendulum moves on a spherical surface. Hence the name spherical pendulum. The inclination angleϕ in the xy-plane can change independently. (a) Obtain the equations of motion for the spherical pendulum. (b) Discuss the conditions for which the motion of a spherical pendulum is converted into that of (i) simple pendulum and (ii) conical pendulum. 7.11 Two blocks of massm andM connected by a massless spring of spring constantk are placed on a smooth horizontal table. Determine the equations of motion using Lagrangian mechanics.

7.12 A double pendulum consists of two simple pendulums of lengthsl 1 andl 2 and massesm 1 andm 2, with the cord of one pendulum attached to the bob of another pendulum whose cord isﬁxed to a pivot, Fig. 7.1. Determine the equations of motion for small angle oscillations using Lagrange’s equations.

Fig. 7.1

7.13 Use Hamilton’s equations to obtain the equations of motion of a uniform heavy rod of massM and length 2a turning about one end which isﬁxed. 1 2 1 2 2 7.14 A one-dimensional harmonic oscillator has HamiltonianH 2 p 2 ω q . Write down Hamiltonian’s equation andﬁnd the general solution.= + 7.15 Determine the equations for planetary motion using Hamilton’s equations.

7.16 Two blocks of massm 1 andm 2 coupled by a spring of force constantk are placed on a smooth horizontal surface, Fig. 7.2. Determine the natural frequencies of the system.

Fig. 7.2 290 7 Lagrangian and Hamiltonian Mechanics

7.17 A simple pendulum of lengthl and massm is pivoted to the block of massM which slides on a smooth horizontal plane, Fig. 7.3. Obtain the equations of motion of the system using Lagrange’s equations.

Fig. 7.3

7.18 Determine the equations of motion of an insect of massm crawling at a uniform speedv on a uniform heavy rod of massM and length 2a which is turning about aﬁxed end. Assume that att 0 the insect is at the middle point of the rod and it is crawling downwards.= 7.19 A uniform rod of massM and length 2a is attached at one end by a cord of lengthl to aﬁxed point. Calculate the inclination of the string and the rod when the string plus rod system revolves about the vertical through the pivot with constant angular velocityω.

7.20 A particle moves in a horizontal plane in a central force potentialU(r). Derive the Lagrangian in terms of the polar coordinates (r,θ). Find the corresponding momentap r andp θ and the Hamiltonian. Hence show that the energy and angular momentum of the particle are conserved. [University of Manchester 2007] 7.21 Consider the system consisting of two identical masses that can move hori- zontally, joined with springs as shown in Fig. 7.4. Letx,y be the horizontal displacements of the two masses from their equilibrium positions. (a) Find the kinetic and potential energies of the system and deduce the Lagrangian. (b) Show that Lagrange’s equation gives the coupled linear differential equations m x 4kx 3ky m¨y=3−kx +4ky ¨ = −

Fig. 7.4 7.2 Problems 291

7.22 Two identical beads of massm each can move without friction along a horizontal wire and are connected to aﬁxed wall with two identical springs of spring constantk as shown in Fig. 7.5.

Fig. 7.5

(a) Find the Lagrangian for this system and derive from it the equations of motion. (b) Find the eigenfrequencies of small amplitude oscillations. (c) For each normal mode, sketch the system when it is at maximum displacement.

Note: Your sketch should indicate the relative sizes as well as the directions of the displacements. [University of Manchester 2007]

7.23 Two beads of mass 2m andm can move without friction along a horizontal wire. They are connected to aﬁxed wall with two springs of spring constants 2k andk as shown in Fig. 7.6:

(a) Find the Lagrangian for this system and derive from it the equations of motion for the beads. (b) Find the eigenfrequencies of small amplitude oscillations. (c) For each normal mode, sketch the system when it is at the maximum displacement.

Fig. 7.6 292 7 Lagrangian and Hamiltonian Mechanics

7.24 Three identical particles of massm, M andm with M in the middle are connected by two identical massless springs with a spring constantk. Find the normal modes of oscillation and the associated frequencies. 7.25 (a) A bead of massm is constrained to move under gravity along a planar rigid wire that has a parabolic shapey x 2/l, wherex andy are, respectively, the horizontal and the vertical= coordinates. Show that the Lagrangian for the system is m( x)2 4x 2 mgx2 L ˙ 1 = 2 + l2 − l

(b) Derive the Hamiltonian for a single particle of massm moving in one dimension subject to a conservative force with a potentialU(x). [University of Manchester 2006]

7.26 A pendulum of lengthl and massm is mounted on a block of massM. The block can move freely without friction on a horizontal surface as shown in Fig. 7.7. (a) Show that the Lagrangian for the system is M m m L + ( x)2 ml cosθ x θ l2(θ) 2 mgl cosθ = 2 ˙ + ˙ ˙+ 2 ˙ +

(b) Show that the approximate form for this Lagrangian, which is applicable for a small amplitude swinging of the pendulum, is M m m θ 2 L + ( x)2 ml xθ l2(θ) 2 mgl 1 = 2 ˙ + ˙ ˙+ 2 ˙ + − 2

(c) Find the equations of motion that follow from the simpliﬁed Lagrangian obtained in part (b), (d) Find the frequency of a small amplitude oscillation of the system. [University of Manchester 2006]

Fig. 7.7 7.2 Problems 293

7.27 (a) A particle of massm slides down a smooth spherical bowl, as in Fig. 7.8. The particle remains in a vertical plane (the xz-plane). First, assume that the bowl does not move. Write down the Lagrangian, taking the angleϑ with respect to the vertical direction as the generalized coordinate. Hence, derive the equation of motion for the particle.

Fig. 7.8

(b) Assume now that the bowl rests on a smooth horizontal table and has a massM, the bowl can slide freely along thex-direction.

(i) Write down the Lagrangian in terms of the angleθ and thex- coordinate of the bowl,x. (ii) Starting from the corresponding Lagrange’s equations, obtain an equation giving x in terms ofθ, θ and θ and an equation giving θ ˙ ¨ ¨ in terms of x and¨ θ. (iii) Hence, and¨ assuming thatM >>m, show that for small displacements about equilibrium the period of oscillation of the particle is smaller by a factor [M/(M m) 1/2 as compared to the case where the bowl isﬁxed. [You may+ neglect] the terms inθ 2θ orθ θ 2 compared ¨ ˙ to terms in θ orθ.] ¨ [University of Durham 2004]

7.28 A system is described by the single (generalized) coordinateq and the LagrangianL(q, q). Deﬁne the generalized momentum associated withq and the corresponding˙ Hamiltonian,H(q,p). Derive Hamilton’s equations from Lagrange’s equations of the system. For the remainder of the question, consider the system whose Lagrangian,L(q, q). Find the corresponding Hamil- tonian and write down Hamilton’s equations.˙

7.29 Brieﬂy explain what is the “generalized (or canonical) momentum conjugate to a generalized coordinate”. What characteristic feature should the Lagrangian function have for a generalized momentum to be a constant of motion? A particle P can slide on a frictionless horizontal table with a small opening at O. It is attached, by a string of lengthl passing through the opening, to a particle Q hanging vertically under the table (see Fig. 7.9). The two particles have equal mass,m. Letτ denote the distance of P to the opening,θ the angle between OP and someﬁxed line through O andg the acceleration of 294 7 Lagrangian and Hamiltonian Mechanics

gravity. Initially,r a, Q does not move, and P is given an initial velocity of magnitude(ag) 1/=2 at right angles to OP.

Fig. 7.9

(a) Write the Lagrangian in terms of the coordinatesr andθ and derive the corresponding equations of motion. (b) Using these equations of motion and the initial conditions, show that r a3g/r 3 g. ¨ = (c) Hence, (i)− show that the trajectory of P is the circler a, (ii) show that P describes small oscillations about this circle if it= is slightly displaced from it and (iii) calculate the period of these oscillations:

2 2 2 2 v r r θ , wherev p is the speed of P [ p =˙ + ˙ ]

7.30 A particle of massm is constrained to move on an ellipseE in a vertical plane, parametrized byx a cosθ,y b sinθ, wherea,b> 0 anda b and the positivey-direction= is the upward= vertical. The particle = is connected to the origin by a spring, as shown in the diagram, and is subject to gravity. The 1 2 potential energy in the spring is 2 kr wherer is the distance of the point mass from the origin (Fig. 7.10).

Fig. 7.10 7.2 Problems 295

(i) Usingθ as a coordinate,ﬁnd the kinetic and potential energies of the particle when moving on the ellipse. Write down the Lagrangian and show that Lagrange’s equation becomesm(a 2 sin2 θ b 2 cos2 θ)θ ¨ (a2 b 2)(k m θ 2) sinθ cosθ mgb cosθ. + = ˙ (ii) Show− thatθ − π/2 are two equilibrium− points andﬁnd any other equilibrium points,=± giving carefully the conditions under which they exist. You may either use Lagrange’s equation or proceed directly from the potential energy. (iii) Determine the stabilities of each of the two equilibrium pointsθ π/2 (it may help to consider the casesa>b anda

7.31 In prob. (7.12) on double pendulum ifm 1 m 2 m andl 1 l 2 l, obtain the frequencies of oscillation. = = = = 7.32 Use Lagrange’s equations to obtain the natural frequencies of oscillation of a coupled pendulum described in prob. (6.46). 7.33 A bead of massm slides freely on a smooth circular wire of radiusr which rotates with constant angular velocityω. On a horizontal plane about a point ﬁxed on its circumference, show that the bead performs simple harmonic motion about the diameter passing through theﬁxed point as a pendulum of lengthr g/ω 2. = [with permission from Robert A. Becker, Introduction to theoretical mechanics, McGraw-Hill Book Co., Inc., 1954] 7.34 A block of massm is attached to a wedge of massM by a spring with spring constantk. The inclined frictionless surface of the wedge makes an angleα to the horizontal. The wedge is free to slide on a horizontal frictionless surface as shown in Fig. 7.11.

Fig. 7.11

(a) Show that the Lagrangian of the system is

(M m) 1 k L + x2 m s2 m x s cosα (s l) 2 mg(h s sinα), = 2 ˙ + 2 ˙ + ˙ ˙ − 2 − − − 296 7 Lagrangian and Hamiltonian Mechanics

wherel is the natural length of the spring,x is the coordinate of the wedge ands is the length of the spring. (b) By using the Lagrangian derived in (a), show that the equations of motion are as follows:

(m M) x m s cosα 0, + ¨ + ¨ = m x cosα m s k(s s 0) 0, ¨ + ¨ + − = wheres 0 l (mg sinα)/k. = + (c) By using the equations of motion in (b), derive the frequency for a small amplitude oscillation of this system. [University of Manchester 2008] 7.35 A uniform spherical ball of massm rolls without slipping down a wedge of massM and angleα, which itself can slide without friction on a horizontal table. The system moves in the plane shown in Fig. 7.12. Hereg denotes the gravitational acceleration.

Fig. 7.12

(a) Find the Lagrangian and the equations of motion for this system. (b) For the special case ofM m andα π/4ﬁnd = = (i) the acceleration of the wedge and (ii) the acceleration of the ball relative to the wedge. [Useful information: Moment of inertia of a uniform sphere of massm 2 and radiusR isI m R2.] = 5 [University of Manchester 2007]

7.3 Solutions

7.1 This is obviously a two degree of freedom dynamical system. The square of the particle velocity can be written as

v2 r2 (r θ) 2 (1) =˙ + ˙ Formula (1) can be derived from Cartesian coordinates

x r cosθ,y r sinθ = = x r cosθ r θ sinθ, y r sinθ r θ cosθ ˙ =˙ − ˙ ˙ =˙ + ˙ 7.3 Solutions 297

We thus obtain

x2 y2 r2 r 2θ 2 ˙ +˙ =˙ + ˙ The kinetic energy, the potential energy and the Lagrangian are as follows:

1 1 T mv 2 m( r2 r 2θ 2) (2) = 2 = 2 ˙ + ˙ μm V (3) =− r 1 μm L T V m( r2 r 2θ 2) (4) = − = 2 ˙ + ˙ + r

We taker,θ as the generalized coordinatesq 1,q 2. Since the potential energy V is independent of qi , Lagrangian equations take the form ˙ d ∂L ∂L 0(i 1,2) (5) dt ∂ qi − ∂qi = = ˙ ∂L ∂L μm Now m r, mr θ 2 (6) ∂ r = ˙ ∂r = ˙ − r 2 ∂L ˙ ∂L mr 2θ, 0 (7) ∂θ = ˙ ∂θ = ˙ Equation (5) can be explicitly written as d ∂L ∂L 0 (8) dt ∂ r − ∂r = ˙ d ∂L ∂L 0 (9) dt ∂θ − ∂θ = ˙ Using (6) in (8) and (7) in (9), we get

μm m r mr θ 2 0 (10) ¨ − ˙ + r 2 = d(r2θ) m ˙ 0 (11) dt =

Equations (10) and (11) are identical with those obtained for Kepler’s problem by Newtonian mechanics. In particular (11) signiﬁes the constancy of areal velocity or equivalently angular momentum (Kepler’s second law of planetary motion). The solution of (10) leads to theﬁrst law which asserts that the path of a planet describes an ellipse. This example shows the simplicity and power of Lagrangian method which involves energy, a scalar quantity, rather than force, a vector quantity in New- ton’s mechanics. 298 7 Lagrangian and Hamiltonian Mechanics

7.2 The position of the pendulum is determined by a single coordinateθ and so we takeq θ. Then (Fig. 7.13) =

Fig. 7.13

1 1 1 T mv 2 mω2l2 ml2θ 2 (1) = 2 = 2 = 2 ˙ V mgl(1 cosθ) (2) = − 1 L T V ml2θ 2 mgl(1 cosθ) (3) = − = 2 ˙ − − ∂T ∂T ml 2θ, 0 (4) ∂θ = ˙ ∂θ = ˙ ∂V ∂ 0, mgl sinθ (5) ∂θ = ∂θ = ˙ d ∂L ∂L 0 (6) dt ∂θ − ∂θ = ˙ d ∂ ∂ (T V) (T V) 0 dt ∂θ − − ∂θ − = ˙ d (ml2θ) mgl sinθ 0 dt ˙ + = orl θ g sinθ 0 (equation of motion) ¨+ = For small oscillation angles sinθ θ → gθ θ (equation for angular SHM) ¨=− l g 2π l ω 2 or time periodT 2π ∴ = l = ω = g 7.3 Solutions 299

7.3 In this system there is only one degree of freedom. The instantaneous conﬁgu- ration is speciﬁed byq x. Assuming that the cord does not slip, the angular velocity of the pulley is=x/R, Fig. 7.14. ˙ Fig. 7.14

The kinetic energy of the system is given by

2 1 2 1 2 1 x T m1 x m2 x I ˙ (1) = 2 ˙ + 2 ˙ + 2 R2

The potential energy of the system is

V m 1gx m 2g(l x) (2) =− − − And the Lagrangian is 1 I 2 L m1 m 2 x (m 1 m 2)gx m 2gl (3) = 2 + + R2 ˙ + − +

The equation of motion d ∂L ∂L 0 (4) dt ∂ x − ∂x = ˙ 300 7 Lagrangian and Hamiltonian Mechanics

yields I m1 m 2 x g(m 1 m 2) 0 + + R2 ¨ − − = (m1 m 2)g or x − I (5) ¨ = m1 m 2 + + R2 which is identical with the one obtained by Newton’s mechanics.

7.4 By problem the masses of pulleys are negligible. The double machine is an Atwood machine in which one of the weights is replaced by a second Atwood machine, Fig. 7.15. The system now has two degrees of freedom, and its instantaneous conﬁguration is speciﬁed by two coordinatesx andx .l andl denote the length of the vertical parts of the two strings. Massm 1 is at depthx below the centre of pulley A,m 2 at depthl x x andm 3 at depthl l x x . The kinetic energy of the system is− given+ by + − −

1 2 1 2 1 2 T m1 x m2( x x) m3( x x) (1) = 2 ˙ + 2 − ˙ +˙ + 2 − ˙ −˙

while the potential energy is given by

V m 1gx m 2g(l x x ) m 3g(l x l x ) (2) =− − − + − − + −

Fig. 7.15 7.3 Solutions 301

The Lagrangian of the system takes the form

1 2 1 2 1 2 L T V m1 x m2( x x) m3( x x) = − = 2 ˙ + 2 − ˙ +˙ + 2 − ˙ −˙

(m 1 m 2 m 3)gx (m 2 m 3)gx m 2gl m 3g(l l ) (3) + − − + − + + + The equations of motion are then d ∂L ∂L 0 (4) dt ∂ x − ∂x = ˙ d ∂L ∂L 0 (5) dt ∂ x − ∂x = ˙ which yield

(m1 m 2 m 3) x (m 3 m 2) x (m 1 m 2 m 3)g (6) + + ¨ + − ¨ = − − (m3 m 2) x (m 2 m 3) x (m 2 m 3)g (7) − ¨ + + ¨ = − Solving (6) and (7) we obtain the equations of motion.

7.5 T v 2 u2 2 v2 (1) = ˙ + ˙ V u 2 v 2 (2) = − L T V v 2 u2 2 v2 u 2 v 2 (3) = − = ˙ + ˙ − + ∂L ∂L 2v 2 u, 2u (4) ∂ u = ˙ ∂u =− ˙ ∂L ∂L 4 v, 2v( u2 1) (5) ∂ v = ˙ ∂v = ˙ + ˙ The equations of motion d ∂L ∂L 0 (6) dt ∂ u − ∂u = ˙ d ∂L ∂L 0 (7) dt ∂ v − ∂v = ˙ yield

d 2 (v2 u) 2u 0 dt ˙ + = orv 2 u 2 u v 2u 0 (8) ¨ + ˙ ˙ + = 2 v v( u2 1) 0 (9) ¨ + ˙ + = 302 7 Lagrangian and Hamiltonian Mechanics

7.6 Here we need a single coordinateq x: = 1 1 T m x2,V kx2 (1) = 2 ˙ = 2 1 1 L m x2 kx2 (2) = 2 ˙ − 2 ∂L ∂L m x, kx (3) ∂ x = ˙ ∂x =− ˙ d ∂L ∂L 0 (4) dt ∂ x − ∂x = ˙ m x kx 0 (5) ¨ + = Letx A sinωt (6) = x Aω 2 sinωt (7) ¨ =− Inserting (6) and (7) and simplifying

mω 2 k 0 − + = k 2π m ω orT 0 2π = m = ω = k

whereT 0 is the time period. 7.7 Only one coordinateq x (distance on the surface of the incline) is adequate to describe the motion:= 1 1 T m x2,V mgx sinα,L m x2 mgx sinα = 2 ˙ =− = 2 ˙ + ∂L ∂L m x, mg sinα ∂ x = ˙ ∂x = ˙ Equation of motion d ∂L ∂L 0 dt ∂ q − ∂q = ˙ yields

d ∂ (m x) (mgx sinα) 0 dt ˙ − ∂x = or x g sinα ¨ = 7.8 This is a two degree of freedom system because both massm andM are moving. The coordinate on the horizontal axis is described byx for the inclined plane andx for the block of massm on the incline. The origin of the coordinate 7.3 Solutions 303

system isﬁxed on the smooth table, Fig. 7.16. Thex- andy-components of the velocity of the block are given by

Fig. 7.16

vx x x cosθ (1) =˙ +˙ vy x sinθ (2) =−˙ 2 2 2 2 2 v v v x 2 x x cosθ x (3) ∴ = x + y =˙ + ˙ ˙ +˙ Hence, the kinetic energy of the system will be

1 2 1 2 2 T M x m( x 2 x x cosθ x ) (4) = 2 ˙ + 2 ˙ + ˙ ˙ +˙ while the potential energy takes the form

V mgx sinθ (5) =− and the Lagrangian is given by

1 2 1 2 2 L M x m( x 2 x x cosθ x ) mgx sinθ (6) = 2 ˙ + 2 ˙ + ˙ ˙ +˙ + The equations of motion d ∂L ∂L 0 (7) dt ∂ x − ∂x = ˙ d ∂L ∂L 0 (8) dt ∂ x − ∂x = ˙ yield

m( x cosθ x) mg sinθ 0 (9) ¨ +¨ − = M x m( x x cosθ) 0 (10) ¨ + ¨ + ¨ = 304 7 Lagrangian and Hamiltonian Mechanics

Solving (9) and (10)

g sinθ (M m)g sinθ x + (11) ¨ = 1 m cos 2 θ/(M m) = M m sin 2 θ − + + g sinθ cosθ mg sinθ cosθ x (12) ¨ =− (M m)/m cos 2 θ =− M m sin 2 θ + − + which are in agreement with the equations of prob. (2.14) derived from Newto- nian mechanics.

1 7.9 T m( x2 ω 2x2) (1) = 2 ˙ + because the velocity of the bead on the wire is at right angle to the linear velocity of the wire:

V mgx sinωt (2) = becauseωt θ, whereθ is the angle made by the wire with the horizontal at timet, andx= sinθ is the height above the horizontal position:

1 L m( x2 ω 2x2) mgx sinωt (3) = 2 ˙ + − ∂L ∂L m x, mω 2x mg sinωt (4) ∂ x = ˙ ∂x = − ˙ Lagrange’s equation d ∂L ∂L (5) dt ∂ x − ∂x ˙ then becomes

x ω2x g sinωt 0 (equation of motion) (6) ¨ − + = which has the solution

ωt ωt g x Ae Be − sinωt (7) = + + 2ω2 whereA andB are constants of integration which are determined from initial conditions.

Att 0,x 0 and x 0 (8) = = ˙ = ωt ωt g Further x ω(Ae Be − ) cosωt (9) ˙ = − + 2ω 7.3 Solutions 305

Using (8) in (7) and (9) we obtain

0 A B (10) = + g 0 ω(A B) (11) = − + 2ω g g Solving (10) and (11) we getA ,B (12) =− 4ω2 = 4ω2 The complete solution forx is

g ωt ωt x (e− e 2 sinωt) = 4ω2 − + 7.10 Let the length of the cord bel. The Cartesian coordinates can be expressed in terms of spherical polar coordinates (Fig. 7.17)

Fig. 7.17

x l sinθ cosφ = y l sinθ sinφ = z l cosθ =− V mgz mgl cosθ (1) = =− v2 x2 y2 z2 l 2(θ 2 sin2 θφ2) =˙ +˙ +˙ = ˙ + ˙ 1 T ml2(θ 2 sin2 θφ2) (2) = 2 ˙ + ˙ 1 L ml2(θ 2 sin2 θφ2) mgl cosθ (3) = 2 ˙ + ˙ + ∂L ∂L ml 2θ, ml 2 sinθ cosθ φ2 mgl sinθ (4) ∂θ = ˙ ∂θ = ˙ − ˙ ∂L ∂L ml 2 sin2 θφ, 0 (5) ∂φ = ˙ ∂φ = ˙ 306 7 Lagrangian and Hamiltonian Mechanics

The Lagrangian equations of motion d ∂L ∂L d ∂L ∂L 0 and 0 (6) dt ∂θ − ∂θ = dt ∂φ − ∂φ = ˙ ˙ g give θ sinθ cosθ φ2 sinθ 0 (7) ¨− ˙ + l = d ml2 (sin2 θφ) 0 (8) dt ˙ = Hence sin2 θφ constant C (9) ˙= = and eliminating φ in (7) with the use of (9) we get a differential equation inθ ˙ only.

g cosθ θ sinθ C 2 0 (10) ¨+ l − sin3 θ =

∂L 2 2 The quantityP φ ml sin θφ (5) = ∂φ = ˙ ˙ is a constant of motion and is recognized as the angular momentum of the system about thez-axis. It is conserved because torque is not produced either by gravity or the tension of the cord about thez-axis. Thus conservation of angular momentum is reﬂected in (5). Two interesting cases arise. Supposeφ constant. Then φ 0 andC 0. ˙ In this case (10) reduces to = = = g θ sinθ 0 (11) ¨+ l =

which is appropriate for simple pendulum in which the bob oscillates in the vertical plane. Supposeθ constant, then from (9) φ constant. Putting θ 0 in (7) ˙ ¨ we get = = = g g ω φ (12) = ˙= l cosθ = H

and time period 2π H T (13) = ω = g

appropriate for the conical pendulum in which the bob rotates on horizontal plane with uniform angular velocity with the cord inclined at constant angleθ with the vertical axis. 7.3 Solutions 307

7.11 The two generalized coordinatesx andy are indicated in Fig. 7.18. The kinetic energy of the system comes from the motion of the blocks and potential energy from the coupling spring:

Fig. 7.18

1 1 T m x2 M y2 (1) = 2 ˙ + 2 ˙ 1 V k(x y) 2 (2) = 2 − 1 1 1 L m x2 M y2 k(x y) 2 (3) = 2 ˙ + 2 ˙ − 2 − ∂L ∂L m x, k(x y) (4) ∂ x = ˙ ∂x =− − ˙ ∂L ∂L M y, k(x y) (5) ∂ y = ˙ ∂y = − ˙ Lagrange’s equations are written as d ∂L ∂L d ∂L ∂L 0, 0 (6) dt ∂ x − ∂x = dt ∂ y − ∂y = ˙ ˙ Using (4) and (5) in (6) we obtain the equations of motion

m x k(x y) 0 (7) ¨ + − = m y k(y x) 0 (8) ¨ + − =

7.12 This problem involves two degrees of freedom. The coordinates areθ 1 andθ 2 (Fig. 7.19)

1 2 1 2 T m1v m2v (1) = 2 1 + 2 2 2 2 v (l 1θ1) (2) 1 = ˙ 2 2 2 v (l 1θ1) (l 2θ2) 2l 1l2θ1θ2 cos(θ2 θ 1) (by parallelogram law) (3) 2 = ˙ + ˙ + ˙ ˙ −

For small angles, cos(θ2 θ 1) 1 − 308 7 Lagrangian and Hamiltonian Mechanics

Fig. 7.19

1 2 2 1 2 2 2 2 T m1l θ m2 l θ l θ 2l 1l2θ1θ2 (4) 2 1 ˙1 + 2 1 ˙1 + 2 ˙2 + ˙ ˙ V m 1gl1(1 cosθ 1) m 2gl1(1 cosθ 1) m 2gl2(1 cosθ 2) = − + − + − 2 θ1 m2g 2 2 m 1gl1 l1θ l 2θ (5) 2 + 2 1 + 2 2 1 2 2 1 2 2 2 2 θ1 L m1l1 θ1 m2 l1 θ1 l 2 θ2 2l 1l2θ1θ2 m 1gl1 = 2 ˙ + 2 ˙ + ˙ + ˙ ˙ − 2 m2g 2 2 l1θ l 2θ (6) − 2 1 + 2 ∂L 2 2 m 1l θ1 m 2l θ1 m 2l1l2θ2 (7) ∂θ = 1 ˙ + 1 ˙ + ˙ ˙1 ∂L m 1gl1θ1 m 2gl1θ1 (m 1 m 2)gl1θ1 (8) ∂θ1 =− − =− + ∂L 2 m 2l θ2 m 2l1l2θ1 (9) ∂θ = 2 ˙ + ˙ ˙2 ∂L m 2gl2θ2 (10) ∂θ2 =−

Lagrange’s equations are d ∂L ∂L d ∂L ∂L 0, 0, (11) dt ∂θ − ∂θ = dt ∂θ − ∂θ = ˙1 1 ˙2 2

using (7, (8), (9) and 10) in (11) we obtain the equations of motion

(m1 m 2)l1θ1 m 2l2θ2 (m 1 m 2)gθ1 0 (12) + ¨ + ¨ + + = l2θ2 gθ 2 l 1θ1 0 (13) ¨ + + ¨ = 7.3 Solutions 309

7.13 Writingp θ andp φ for the generalized momenta, by (4) and (5) andV by (1) of prob. (7.10)

∂L 2 Pθ P θ ml θ, θ (1) ∂θ = = ˙ ˙= ml2 ˙ ∂L 2 2 pφ p φ ml sin θφ, φ (2) ∂φ = = ˙ ˙= ml2 sin2 θ ˙ ∂L ∂L H q i L orH L 2T q i = ˙ ∂ qi − + = = ˙ ∂ qi ˙ ˙ ∂L ∂L 1 2T θ φ p2 cosec2θp 2 (3) = ˙ ∂θ + ˙ ∂φ = ml2 θ + φ ˙ ˙ 1 H T V p2 cos ec2θp 2 mgl cosθ (4) ∴ = + = 2ml2 θ + φ −

The coordinateφ is ignorable, and thereforep φ is a constant of motion determined by the initial conditions. We are then left with only two canonical equations to be solved. The canonical equations are

∂H ∂H qj , p j ˙ = ∂p j ˙ =− ∂q j ∂H pθ θ 2 (5) ˙= ∂p θ = ml 2 ∂H pφ cosθ pθ mgl sinθ (6) ˙ =− ∂θ = ml2 sin3 θ −

wherep φ is a constant of motion. By eliminatingp θ we can immediately obtain a second-order differential equation inθ as in prob. ( 7.10).

1 1 7.14 H p2 ω2q2 (1) = 2 + 2 ∂H q p (2) ∂p =˙ = ∂H p ω 2q (3) ∂q =−˙ =

Differentiating (2)

q p ω 2q (4) ¨ =˙ =− Letq x, then (4) can be written as = x ω2x 0 (5) ¨ + = 310 7 Lagrangian and Hamiltonian Mechanics

This is the equation for one-dimensional harmonic oscillator. The general solution is

x A sin(ωt ε) B cos(ωt ε) (6) = + + + which can be veriﬁed by substituting (6) in (5). HereA,B andε are constants to be determined from initial conditions.

7.15 Letr,θ be the instantaneous polar coordinates of a planet of massm revolving around a parent body of massM:

1 T m( r2 r 2θ 2) (1) = 2 ˙ + ˙ 1 1 V GMm (2) = 2a − r

whereG is the gravitational constant and 2a is the major axis of the ellipse:

∂T pr pr m r, r (3) = ∂ r = ˙ ˙ = m ˙ ∂T 2 pθ pθ mr θ, θ (4) = ∂θ = ˙ ˙= mr 2 ˙ 1 p2 1 1 H p2 θ GMm (5) = 2m r + r 2 + 2a − r

and the Hamiltonian equations are

2 ∂H pr ∂H pθ G Mm r, 3 2 pr (6) ∂pr = m =˙ ∂r =− mr + r =−˙

∂H pθ ∂H 2 , 0 p θ (7) ∂p θ = mr ∂θ = =−˙

Two equations in (7) show that

2 pθ constant mr θ (8) = = ˙ meaning the constancy of angular momentum or equivalently the constancy of areal velocity of the planet (Kepler’s second law of planetary motion). Two equations in (6) yield

2 pr p G Mm G Mm r ˙ θ r θ 2 (9) ¨ = m = m2r 3 − r 2 = ˙ − r 2 7.3 Solutions 311

Equation (9) describes the orbit of the planet (Kepler’sﬁrst law of planetary motion)

1 2 1 2 7.16 T m1 x m2 x (1) = 2 ˙1 + 2 ˙2

1 2 V k(x 1 x 2) (2) = 2 −

1 2 1 2 1 2 L m1 x m2 x k(x 1 x 2) (3) = 2 ˙1 + 2 ˙2 − 2 − Equations of motion are d ∂L ∂L 0 (4) dt ∂ x1 − ∂x 1 = ˙ d ∂L ∂L 0 (5) dt ∂ x2 − ∂x 2 = ˙ Using (3) in (4) and (5)

m1 x1 k(x 1 x 2) 0 (6) ¨ + − = m2 x2 k(x 1 x 2) 0 (7) ¨ − − = It is assumed that the motion is periodic and can be considered as superposition of harmonic components of various amplitudes and frequencies. Let one of these harmonics be represented by

2 x1 A sinωt, x 1 ω A sinωt (8) = ¨ =− 2 x2 B sinωt, x 2 ω B sinωt (9) = ¨ =− Substituting (8) and (9) in (6) and (7) we obtain

2 (k m 1ω )A kB 0 − − = 2 kA (k m 2ω )B 0 − + − = The frequency equation is obtained by equating to zero the determinant formed by the coefﬁcients ofA andB: 2 (k m 1ω ) k − − 2 0 k(k m 2ω ) = − − Expansion of the determinant gives

4 2 m1m2ω k(m 1 m 2)ω 0 − + = 312 7 Lagrangian and Hamiltonian Mechanics

2 2 ω m1m2ω k(m 1 m 2) 0 [ − + ] = which yields the natural frequencies of the system: k(m 1 m 2) k ω1 0 andω 2 + = = m1m2 = μ

whereμ is the reduced mass. The frequencyω 1 0 implies that there is no genuine oscillation of the block but mere translatory= motion. The second fre- quencyω 2 is what one expects for a simple harmonic oscillator with a reduced massμ. 7.17 Letx(t) be the displacement of the block andθ(t) the angle through which the pendulum swings. The kinetic energy of the system comes from the motion of the block and the swing of the bob of the pendulum. The potential energy comes from the deformation of the spring and the position of the bob, Fig. 7.20.

Fig. 7.20

The velocityv of the bob is obtained by combining vectorially its linear velocity(l θ) with the velocity of the block( x). The height through which the ˙ bob is raised from the equilibrium position is˙l(1 cosθ), wherel is the length of the pendulum: −

v2 x2 l 2θ 2 2l xθ cosθ (1) =˙ + ˙ + ˙ ˙ 1 1 T M x2 m( x2 l 2θ 2 2l xθ) (2) = 2 ˙ + 2 ˙ + ˙ + ˙ ˙ ( forθ 0, cosθ 1) ∵ → → 1 V kx2 mgl(1 cosθ) = 2 + − 1 θ 2 kx2 mgl (3) = 2 + 2 7.3 Solutions 313

1 1 1 θ 2 L M x2 m( x2 l 2θ 2 2l xθ) kx2 mgl (4) = 2 ˙ + 2 ˙ + ˙ + ˙ ˙ − 2 − 2

Applying Lagrange’s equations d ∂L ∂L d ∂L ∂L 0, 0 (5) dt ∂ x − ∂x = dt ∂θ − ∂θ = ˙ ˙ we obtain

(M m) x ml θ kx 0 (6) + ¨ + ¨+ = lθ x gθ 0 (7) ¨+¨ + = 7.18 Considering that att 0 the insect was in the middle of the rod, the coordinates of the insectx,y=,z at timet are given by

x (a vt) sinθ cosφ = + y (a vt) sinθ sinφ = + z (a vt) cosθ = + and the square of its velocity is

x2 y2 z2 v 2 (a vt) 2(θ 2 φ2 sin2 θ) ˙ +˙ +˙ = + + ˙ + ˙ 2 m T Ma2(θ 2 φ2 sin2 θ) v2 (a vt) 2(θ 2 φ2 sin2 θ) ∴ = 3 ˙ + ˙ + 2 { + + ˙ + ˙ } V Mga cosθ mg(a vt) cosθ constant =− − + + L T V = − The application of the Lagrangian equations to the coordinatesθ andφ yields d 4 4 Ma2θ m(a vt) 2θ Ma2 m(a vt) 2 φ2 sinθ cosθ dt 3 ˙+ + ˙ − 3 + + ˙

Ma m(a vt) g sinθ (1) =−{ + + } d 4 and Ma2 m(a vt) 2 φ sin2 θ 0 (2) dt 5 + + ˙ =

Equation (2) can be integrated at once as it is free fromφ: 4 Ma2 m(a vt) 2 φ sin2 θ constant C (3) 5 + + ˙ = = 314 7 Lagrangian and Hamiltonian Mechanics

Equation (3) is the equation for the constancy of angular momentum about the vertical axis. When φ in (2) is eliminated with the aid of (3) we obtain a second-order ˙ differential equation inθ.

7.19 Letρ be the linear density of the rod, i.e. mass per unit length. Consider an inﬁnitesimal element of length of the rod

Fig. 7.21

dT ρω 2(l sinθ x sinφ) 2dx = + 2a T dT ρω 2 (l2 sin2 θ 2lx sinθ sinφ x 2 sin2 φ)dx = = 0 + + 4 ω 2 Ml2 sin2 θ 2Mla sinθ sinφ Ma2 sin2 φ (1) = + + 3

where we have substitutedρ M/2a: = V Mg(l cosθ a cosφ) (2) =− + 4 L ω 2 Ml2 sin2 θ 2Mla sinθ sinφ Ma2 sin2 φ = + + 3 Mg(l cosθ a cosφ) (3) + + ∂L ∂L 0, ω 2(2Ml 2 sinθ cosθ 2Mla cosθ sinφ) Mgl sinθ (4) ∂θ = ∂θ = + − ˙ ∂L ∂L 8 0, ω 2 2Mla sinθ cosφ Ma2 sinφ cosφ Mga sinφ (5) ∂φ = ∂φ = + 3 − ˙ The Lagrange’s equations d ∂L ∂L d ∂L ∂L 0, 0 (6) dt ∂θ − ∂θ = dt ∂φ − ∂φ = ˙ ˙ 7.3 Solutions 315

yield

2ω2(l sinθ a sinφ) g tanθ (7) + = 4 2ω2(l sinθ a sinφ) g tanφ (8) + 3 =

Equation (7) and (8) can be solved to obtainθ andφ.

7.20 Express the Cartesian coordinates in terms of plane polar coordinates (r,θ)

x r cosθ,y r sinθ (1) = = x r cosθ r θ sinθ (2) ˙ =˙ − ˙ y r sinθ r θ cosθ (3) ˙ =˙ + ˙ Square (2) and (3) and add

v2 x2 y2 r2 r 2θ 2 (4) =˙ +˙ =˙ + ˙ 1 1 T mv 2 m( r2 r 2θ 2) (5) = 2 = 2 ˙ + ˙ V U(r) (6) = 1 L T V m( r2 r 2θ 2) U(r)(Lagrangian) (7) ∴ = − = 2 ˙ + ˙ −

Generalized momenta:

∂L ∂L ∂L 2 pk ,p r m r,p θ mr θ (8) = ∂ qk = ∂ r = ˙ = ∂θ = ˙ ˙ ˙ ˙ Hamiltonian:

1 H T V m( r2 r 2θ 2) V(r) (9) = + = 2 ˙ + ˙ +

Conservation of Energy: In generalH may contain an explicit time dependence as in some forced systems. We shall therefore writeH H(q,p,t). ThenH varies with time for two reasons:ﬁrst, because of its explicit= dependence ont, second because the variableq andp are themselves functions of time. Then the total time derivative ofH is

dH ∂H n ∂H n ∂H qβ pβ (10) dt = ∂t + ∂qβ ˙ + ∂p β ˙ β 1 β 1 = = 316 7 Lagrangian and Hamiltonian Mechanics

Now Hamilton’s equations are

∂H ∂H qβ , pβ (11) ∂p β =˙ ∂qβ =−˙

Using (11) in (10), we obtain dH ∂H n ∂H ∂H ∂H ∂H (12) dt = ∂t + ∂qβ ∂p β − ∂p β ∂qβ β 1 = whence dH ∂H (13) dt = ∂t Equation (13) asserts thatH changes with time only by virtue of its explicit time dependence. The net change is induced by the fact that the variation ofq andp with time is zero. Now in a conservative system, neitherT norV contains any explicit dependence on time. ∂H Hence 0. It follows that ∂t = dH 0 (14) dt = which leads to the law of conservation of energy

H T V E constant (15) = + = = The Hamiltonian formalism is amenable forﬁnding various conservation laws. Conservation of angular momentum: The Hamiltonian can be written as

n H pβ qβ L (16) = ˙ − β 1 = Using the polar coordinates (r,θ) 1 2 1 2 2 H p r r p θ θ m r mr θ U(r) (17) = ˙ + ˙− 2 ˙ + 2 ˙ −

Using (7) and (8) in (17) 7.3 Solutions 317

p2 p2 H r θ U(r) (18) = 2m + 2mr 2 +

Now the second equation in (11) gives

∂H pθ 0( θ is absent in(18)) (19) − ˙ = ∂θ = ∵

This leads to the conservation of angular momentum

pθ J constant (20) = = 7.21 Let each mass bem. 1 1 1 (a) T m x2 m y2 m( x2 y2) (1) = 2 ˙ + 2 ˙ = 2 ˙ +˙ 1 1 1 V kx2 3k(x y) 2 ky2 k(2x 2 3xy 2y 2) (2) = 2 + 2 − + 2 = − + 1 L T V m( x2 y2) k(2x 2 3xy 2y 2) (3) = − = 2 ˙ +˙ − − + d ∂L ∂L (b) 0 (4) dt ∂ x − ∂x = ˙ d ∂L ∂L and 0 (5) dt ∂ y − ∂y = ˙ yield m x 4kx 3ky (6) ¨ =− + m y 3kx 4ky (7) ¨ = − (c) Letx A sinωt andy B sinωt (8) = = x Aω 2 sinωt and y Bω 2 sinωt (9) ¨ =− ¨ =− Substituting (8) and (9) in (6) and (7) and simplifying we obtain

(4k ω 2m)A 3k B 0 (10) − − = 3k A (4k ω 2m)B 0 (11) − + − = The frequency equation is obtained by equating to zero the determinant formed by the coefﬁcients ofA andB: (4k ω 2m) 3k − − 0 (12) 3k(4k ω 2m) = − − 318 7 Lagrangian and Hamiltonian Mechanics

Expanding the determinant

(4k ω 2m)2 9k 2 0 (13) − − = This gives the frequencies k 7k ω1 ,ω 2 (14) = m = m

Periods of oscillations are 2π m T1 2π (15) = ω1 = k 2π m T2 2π (16) = ω2 = 7k k If we putω ω 1 in (10) or (11) we getA B and if we put = = m = 7k ω ω 2 in (10) or (11), we getA B. Theﬁrst one corresponds to = = m =− symmetric mode of oscillation and the second one to asymmetric one. The normal coordinatesq 1 andq 2 are formed by the linear combination of x andy:

q1 x y,q 2 x y (17) = − = + q1 q 2 q2 q 1 x + ,y − (18) ∴ = 2 = 2 Substituting (18) in (6) and (7)

m 3k ( q1 q2) 2k(q 1 q 2) (q2 q 1) (19) 2 ¨ + ¨ =− + + 2 −

m 3k ( q2 q1) (q1 q 2) 2k(q 2 q 1) (20) 2 ¨ −¨ = 2 + − − Adding(19) and(20),m q 2 kq 2 (21) ¨ =− Subtracting(20) from(19),m q 1 7kq 1 (22) ¨ =−

Equation (21) is a linear equation inq 2 alone, with constant coefficients. Simi- larly (22) is a linear equation inq 1 with constant coefficients. Since the coefficients on the right sides are positive quantities, we note that both (21) and (22) are differential equations of simple harmonic motion having the frequencies given in (14). It is the characteristic of normal coordinates that when the equa- 7.3 Solutions 319

tions of motion are expressed in terms of normal coordinates they are linear with constant coefﬁcients, and each contains but one dependent variable. Another feature of normal coordinates is that both kinetic energy and potential energy will have quadratic terms and the cross-products will be absent. Thus in this example, when (18) is used in (1) and (2) we get the expressions

m k T ( q2 q2),V (7q2 q 2) = 4 ˙ 1 +˙ 2 = 4 1 + 2

The two normal modes are depicted in Fig. 7.22. k 7k (a) Symmetrical withω 1 and (b) asymmetrical withω 2 = m = m

Fig. 7.22

1 7.22 (a) See Fig. 7.23 T m( x2 x2) (1) : = 2 ˙ 1 +˙ 2 1 2 1 2 V kx1 k(x 2 x 1) = 2 + 2 − 2 1 2 k x1 x 1x2 x2 (2) = − + 2 1 2 2 2 1 2 L T V m( x1 x2 ) k x1 x 1x2 x2 (3) =− = 2 ˙ +˙ − − + 2 d ∂L ∂L d ∂L ∂L 0, 0 (4) dt ∂ x1 − ∂x 1 = dt ∂ x2 − ∂x 2 = ˙ ˙ m x1 k(2x 1 x 2) 0 (5) ¨ + − = m x2 k(x 1 x 1) 0 (6) ¨ + − = (5) and (6) are equations of motion

(b) Let the harmonic solutions bex 1 A sinωt,x 2 B sinωt (7) = = 2 2 Then x 1 Aω sinωt, x 2 Bω sinωt, (8) ¨ =− ¨ =− Using (7) and (8) in (5) and (6) we get

(2k mω 2)A kB 0 (9) − − = kA (k mω 2)B 0 (10) − + − =

Fig. 7.23 320 7 Lagrangian and Hamiltonian Mechanics

The eigenfrequency equation is obtained by equating to zero the determinant formed by the coefﬁcients ofA andB: (2k mω 2) k − − 0 (11) k(k mω 2) = − − Expanding the determinant, we obtain

m2ω4 3mkω 2 k 2 0 − + = 3 √5 k ω2 ± (12) = 2 m k k ω 1 1.618 ,ω 2 0.618 (13) ∴ = m = m

(c) Insertingω ω 1 in (10) weﬁndB 1.618A. This corresponds to a symmetric= mode as both the amplitudes= have the same sign. Insertingω ω 2 in (10), weﬁndB 0.618A. This corresponds to asymmetric= mode. These two modes=− of oscillation are depicted in Fig. 7.24 with relative sizes and directions of displacement. k (a) Symmetric modeω 1 1.618 = m k (b) Asymmetric modeω 2 0.618 = m

Fig. 7.24

7.23 (a) Letx 1 andx 2 be the displacements of the beads of mass 2m andm, respectively.

1 1 T (2m) x2 (m) x2 (1) = 2 ˙ 1 + 2 ˙ 2

1 2 1 2 V 2kx 1 k(x 2 x 1) (2) = 2 · + 2 − 2 1 2 3 2 1 2 L m x x k x x 1x2 x (3) = ˙1 + 2 ˙2 − 2 1 − + 2 2

Lagrange’s equations are d ∂L ∂L d ∂L ∂L 0, 0 (4) dt ∂ x1 − ∂x 1 = dt ∂ x2 − ∂x 2 = ˙ ˙ 7.3 Solutions 321

which yield equations of motion

2m x1 k(3x 1 x 2) 0 (5) ¨ + − = m x2 k(x 1 x 2) 0 (6) ¨ − − = (b) Let the harmonic solutions be

x1 A sinωt,x 2 B sinωt (7) = = 2 2 x1 Aω sinωt, x 2 Bω sinωt (8) ¨ =− ¨ =− Substituting (7) and (8) in (5) and (6) we obtain

(3k 2mω 2)A kB 0 (9) − − = k A (mω 2 k)B 0 (10) + − = The frequency equation is obtained by equating to zero the determinant formed by the coefﬁcients ofA andB: (3k 2mω 2) k − − 0 k mω2 k = − Expanding the determinant

2m2ω4 5kmω 2 2k 2 0 − + = 2k k ω1 ,ω 2 = m = 2m 2k (c) Putω ω 1 in (9) or (10). WeﬁndB A. = = m =− k Putω ω 2 in(9) or(10). WeﬁndB 2A. = = 2m =+

The two normal modes are sketched in Fig. 7.25.

Fig. 7.25

(a) Asymmetric mode 2k ω1 B A = m =− 322 7 Lagrangian and Hamiltonian Mechanics k (b) Symmetric modeω 2 B 2A = 2m =+

7.24 There are three coordinatesx 1,x 2 andx 3, Fig. 7.26:

Fig. 7.26

1 1 1 T m x2 Mx2 m x2 (1) = 2 ˙ 1 + 2 2 + 2 ˙ 3 1 2 2 V k (x2 x 1) (x 3 x 2) = 2 − + −

1 2 2 2 k x 2x 1x2 2x 2x 2x3 x (2) = 2 1 − + 2 − + 3

1 2 1 2 1 2 1 2 2 2 L m x M x m x k x 2x 1x2 2x 2x 2x3 x (3) = 2 ˙ 1 + 2 ˙ 2 + 2 ˙ 3 − 2 1 − + 2 − + 3 Lagrange’s equations d ∂L ∂L d ∂L ∂L d ∂L ∂L 0, 0, 0 (4) dt ∂ x1 − ∂x 1 = dt ∂ x2 − ∂x 2 = dt ∂ x3 − ∂x 3 = ˙ ˙ ˙ yield

m x1 k(x 1 x 2) 0 (5) ¨ + − = M x2 k( x 1 2x 2 x 3) 0 (6) ¨ + − + − = m x3 k( x 2 x 3) 0 (7) ¨ + − + = Let the harmonic solutions be

x1 A sinωt,x 2 B sinωt,x 3 C sinωt (8) = = = 2 2 2 x 1 Aω sinωt, x 2 Bω sinωt, x 3 Cω sinωt (9) ∴¨ =− ¨ =− ¨ =− Substituting (8) and (9) in (5), (6) and (7)

(k mω 2)A kB 0 (10) − − = kA (2k Mω 2)B kC 0 (11) − + − − = kB (k mω 2)C 0 (12) − + − = 7.3 Solutions 323

The frequency equation is obtained by equating to zero the determinant formed by the coefﬁcients ofA,B andC (k mω 2) k0 − − k(2k Mω 2) k 0 − − − = 0 k(k mω 2) − − Expanding the determinant we obtain

ω2(k mω 2)(ω2 Mm 2km Mk) 0 (13) − − − = The frequencies are k k(2m M) ω1 0,ω 2 ,ω 3 + (14) = = m = Mm

The frequencyω 1 0 simply means a translation of all the three particles = without vibration. Ratios of amplitudes of the three particles can be found out k by substitutingω 2 andω 3 in (10), (11) and (12). Thus whenω ω 2 is = m substituted in (10), weﬁnd the amplitude for the central atomB 0. When B 0 is used in (11) we obtainC A. This mode of oscillation= is depicted in= Fig. 7.27a. =−

Fig. 7.27

 k(2m M) Substitutingω ω 3 + in (10) and (12) yields = = Mm 2m 2m B A C =− M =− M

Thus in this mode particles of massm oscillate in phase with equal amplitude but out of phase with the central particle. This problem has a bearing on the vibrations of linear molecules such as CO2. The middle particle represents the C atom and the particles on either side represent O atoms. Here too there will be three modes of oscillations. One will have a zero frequency,ω 1 0, and will correspond to a simple translation = of the centre of mass. In Fig. 7.27a the mode withω 1 ω 2 is such that the carbon atom is stationary, the oxygen atoms oscillating back= and forth in opposite phase with equal amplitude. In the third mode which has frequencyω 3, the carbon atom undergoes motion with respect to the centre of mass and is in opposite phase from that of the two oxygen atoms. Of these two modes 324 7 Lagrangian and Hamiltonian Mechanics

onlyω 3 is observed optically. The frequencyω 2 is not observed because in this mode, the electrical centre of the system is always coincident with the centre of mass, and so there is no oscillating dipole momenter available. Hence dipole radiation is not emitted for this mode. On the other hand in the third mode characterized byω 3 such a moment is present and radiation is emitted. x2 y (1) 7.25 (a) = l 2x x y ·˙ (2) ˙ = l 4x 2 v2 x2 y2 x2 1 =˙ +˙ =˙ + l2 1 1 4x 2 T mv 2 m x2 1 = 2 = 2 ˙ + l2 mgx2 V mgy = = l 1 4x 2 mgx2 L T V m x2 1 = − = 2 ˙ + l2 − l

1 L m( r2 r 2θ 2) U(r) (b) = 2 ˙ + ˙ − ∂L pr pr m r, r = ∂ r = ˙ ˙ = m ˙ ∂L 2 pθ pθ mr θ, θ = ∂θ = ˙ ˙= mr 2 ˙ 1 p2 H p2 θ U(r) = 2m r + r 2 +

7.26 (a) At any instant the velocity of the block is x on the plane surface. The linear velocity of the pendulum with respect˙ to the block isl θ, Fig. 7.28. ˙ The velocityl θ must be combined vectorially with x toﬁnd the velocity ˙ v or the pendulum with reference to the plane: ˙

v2 x2 l 2θ 2 2 xlθ cosθ (1) =˙ + ˙ + ˙ ˙ The total kinetic energy of the system

1 1 T M x2 m( x2 l 2θ 2 2 xlθ cosθ) (2) = 2 ˙ + 2 ˙ + ˙ + ˙ ˙ Taking the zero level of the potential energy at the pivot of the pendulum, the potential energy of the system which comes only from the pendulum is 7.3 Solutions 325

Fig. 7.28

V mgl cosθ (3) =− 1 1 L T V (M m) x 2 ml cosθ x θ ml2θ 2 mgl cosθ ∴ = − = 2 + ˙ + ˙ ˙+ 2 ˙ + (4)

where we have used (2) and (3). θ 2 (b) For small angles cosθ 1 , in theﬁrst approximation, and cosθ 1, − 2 in the second approximation. Thus in this approximation (4) becomes 1 1 θ 2 L (M m) x 2 ml xθ ml2θ 2 mgl 1 (5) = 2 + ˙ + ˙ ˙+ 2 ˙ + − 2

(c) The Lagrange’s equations d ∂L ∂L d ∂L ∂L 0, 0 (6) dt ∂θ − ∂θ = dt ∂ x − ∂x = ˙ ˙ lead to the equations of motion

x lθ gθ 0 (7) ¨ + ¨+ = (M m) x ml θ 0 (8) + ¨ + ¨= (d) Eliminating x between (7) and (8) and simplifying ¨ (M m) g θ + θ 0 (9) ¨+ M l = This is the equation for angular simple harmonic motion whose frequency is given by (M m)g ω + (10) = Ml 326 7 Lagrangian and Hamiltonian Mechanics

7.27 (a) First, we assume that the bowl does not move. Both kinetic energy and potential energy arise from the particle alone. Taking the origin at O, the centre of the bowl, Fig. 7.29, the linear velocity of the particle isv r θ. ˙ There is only one degree of freedom: =

Fig. 7.29

1 1 T mv 2 mr 2θ 2 (1) = 2 = 2 ˙ V mgr cosθ (2) =− 1 L mr 2θ 2 mgr cosθ (3) = 2 ˙ +

Lagrange’s equation d ∂L ∂L 0 (4) dt ∂ q − ∂q = ˙ becomes d ∂L ∂L 0 (5) dt ∂θ − ∂θ = ˙ which yields the equation of motion

mr 2θ mgr sinθ 0 ¨+ = g or θ sinθ 0 (equation of motion) (6) ¨+ r =

For small angles, sinθ θ. Then (6) becomes g θ θ 0 (7) ¨+ r = 7.3 Solutions 327 g which is the equation for simple harmonic motion of frequencyω = r or time period

 2π r T 2π (8) = ω = g

(b) (i) The bowl can now slide freely along thex-direction with velocity x. The velocity of the particle with reference to the table is obtained˙ by addingl θ to x vectorially, Fig. 7.29. The total kinetic energy then comes ˙ from the motion˙ of both the particle and the bowl. The potential energy, however, is the same as in (a):

v2 r 2θ 2 x 2 2r θ x cos(180 θ) (9) = ˙ + − ˙˙ −

from the diagonal AC of the parallelogram ABCD

1 1 T M x2 m(r 2θ 2 x2 2r xθ cosθ) (10) = 2 ˙ + 2 ˙ +˙ − ˙ ˙

V mgr cosθ (11) =− L T V(Lagrangian) = − 1 1 M x2 m(r 2θ 2 x2 2r xθ cosθ) mgr cosθ (12) = 2 ˙ + 2 ˙ +˙ − ˙ ˙ +

(ii) and (iii). In the small angle approximation the cosθ in the kinetic energy can be neglected as cosθ 1 but can be retained in the potential energy in order to avoid higher order→ terms. Equation (12) then becomes

1 1 L (M m) x 2 mr xθ mr 2θ 2 mgr cosθ (13) = 2 + ˙ − ˙ ˙+ 2 ˙ +

We have now two degrees of freedom,x andθ, and the corresponding Lagrange’s equations are

 d ∂L ∂L d ∂L ∂L 0, 0 (14) dt ∂ x − ∂x = dt ∂θ − ∂θ = ˙ ˙ 328 7 Lagrangian and Hamiltonian Mechanics

which yield the equations of motion

(M m) x mr θ 0 (15) + ¨ − ¨= x r θ gθ 0 (16) ¨ − ¨− = Equations (15) and (16) constitute the equations of motion. Eliminating x we obtain ¨ M m g θ + θ 0 (17) ¨+ M r =

which is the equation for simple harmonic motion with frequencyω = (M m) g + and time period M r 2π M r T 2π (18) = ω = (M m) g + On comparing (18) with (8) it is observed that the period of oscillation is smaller by a factor M/(M m) 1/2 as compared to the case where the bowl isﬁxed. [ + ]

7.28 Take the differential of the Lagrangian

L(q1,...,q n, q1,..., qn,t) ˙ ˙ n ∂L ∂L ∂L dL dqr d qr dt (1) = ∂qr + ∂ qr ˙ + ∂t r 1 = ˙ Now the Lagrangian equations are d ∂L ∂L 0 (2) dt ∂ qr − ∂qr = ˙ and the generalized momenta are deﬁned by

∂L p r (3) ∂ qr = ˙ Using (3) in (2) we have d ∂L pr (4) dt ∂ qr =˙ ˙ 7.3 Solutions 329

Using (2), (3) and (4) in (1)

n ∂L dL ( pr dqr p r d qr ) dt (5) − ˙ + ˙ + ∂t r 1 = Equation (5) can be rearranged in the form n n ∂L d pr qr L ( qr dp r pr dqr ) dt (6) ˙ − = ˙ − ˙ − ∂t r 1 r 1 = = The Hamiltonian functionH is deﬁned by

n H pr qr L(q 1,...,q n, q1,..., qn,t) (7) = ˙ − ˙ ˙ r 1 = Equation (6) therefore my be written as

n ∂L dH ( qr dpr pr dqr ) dt (8) = ˙ − ˙ − ∂t r 1 =

While the Lagrangian functionL is an explicit function ofq 1,...,q n, q1,..., qn andt, it is usually possible to expressH as an explicit function only ˙ ˙ ofq 1,...,q n,p 1,...,p n,t, that is, to eliminate then generalized velocities from (7). Then equation of type (3) are employed for this purpose. Each provides one of thep’s in terms of the q s. Assuming that the elimination of the generalized velocities is possible, we˙ may write

H H(q 1,...,q n,p 1,...,p n,t) (9) = H now depends explicitly on the generalized coordinates and generalized momenta together with the time. Therefore, taking the differential dH, we obtain n ∂H ∂H ∂H dH dqr dp r dt (10) = ∂qr + ∂pr + ∂t r 1 = Comparing (8) and (10), we have the relations

∂H ∂H qr , pr (11) ∂pr =˙ ∂qr =−˙ ∂H ∂L (12) ∂t =− ∂t 330 7 Lagrangian and Hamiltonian Mechanics

Equations (11) are called Hamilton’s canonical equations. These are 2n is number. For a system withn degrees of freedom then Lagrangian equations (2) of the second order are replaced by 2n Hamiltonian equations of theﬁrst

order. We note from the second equation of (11) that if any coordinateq l is not contained explicitly in the Hamiltonian functionH, the conjugate momentum

pl is a constant of motion. Such coordinates are called ignorable coordinates.

7.29 The generalized momentump r conjugate to the generalized coordinateq r ∂L is deﬁned as p r . If the Lagrangian of a dynamical system does not ∂ qr = contain a certain˙ coordinate, sayq s, explicitly thenp s is a constant of motion. (a) The kinetic energy arises only from the motion of the particle P on the table as the particle Q is stationary. The potential energy arises from the particle Q alone. When P is at distancer from the opening, Q will be at a depthl–x from the opening:

1 1 T mv 2 m( r2 r 2θ 2) (1) = 2 p = 2 ˙ + ˙ V mg(l r) (2) =− − 1 L T V m( r2 r 2θ 2) mg(l r) (3) = − = 2 ˙ + ˙ + −

For the two coordinatesr andθ, Lagrange’s equations take the form d ∂L ∂L d ∂L ∂L 0, 0 (4) dt ∂ r − ∂r = dt ∂θ − ∂θ = ˙ ˙ Equations (4) yield

r r θ 2 g (5) ¨ = ˙ − d (mr 2θ) 0 dt ˙ = r 2θ C constant (6) ∴ ˙= = Equations (5) and (6) constitute the equations of motion. (b) Initial conditions: Atr a,r θ √ag = ˙= g θ (7) ∴ ˙= a

Using (7) in (6) withr a, we obtain = C2 a 3g (8) = 7.3 Solutions 331

Using (6) and (8) in (5)

rc2 a3g r g g (9) ¨ = r 4 − = r 3 −

(c) (i) Since Q does not move, P must be at constant distancer a from the opening. Therefore P describes a circle of constant radius= a. (ii) Let P be displaced by a small distancex from the stable circular orbit of radiusa, that is

r a x (10) = + r x (11) ∴¨ =¨

Using (10) and (11) in (9)

 a3 x 3 x g 1 g 1 − 1 ¨ = (a x) 3 − = + a − + 3gx or x ¨ − a 3gx or x 0 (12) ¨ + a =

which is the equation for simple harmonic motion. Thus the particle P when slightly displaced from the stable orbit of radiusa executes oscillations aroundr a. This aspect= of oscillations has a bearing on the so-called betatron oscillations of ions in circular machines which accelerate charged particles to high energies. If the amplitudes of the betatron oscillations are large then they may hit the wall of the doughnut and be lost, resulting in the loss of intensity of the accelerated particles.

7.30 (i) x a cosθ,y b sinθ,r 2 a 2 cos2 θ b 2 sin2 θ (1) = = = + x a θ sinθ, y b θ cosθ (2) ˙ =− ˙ ˙ = ˙ 1 1 T m( x2 y2) m(a 2 sin2 θ b 2 cos2 θ)θ 2 (3) = 2 ˙ +˙ = 2 + ˙ 1 1 V mgy kr 2 mgb sinθ k(a 2 cos2 θ b 2 sin2 θ) (4) = + 2 = + 2 + 1 L m(a 2 sin2 θ b 2 cos2 θ)θ 2 mgb sinθ = 2 + ˙ − 1 k(a 2 cos2 θ b 2 sin2 θ) (5) − 2 + 332 7 Lagrangian and Hamiltonian Mechanics

Lagrange’s equation d ∂L ∂L 0 dt ∂θ − ∂θ = ˙ yields

m(a 2 sin2 θ b 2 cos2 θ)θ m(a 2 sinθ cosθ b 2 sinθ cosθ) θ 2 + ¨+ − ˙ mgb cosθ k( a 2 sinθ cosθ b 2 sinθ cosθ) 0 + + − + = orm(a 2 sin2 θ b 2 cos2 θ)θ (a 2 b 2)(k m θ 2) sinθ cosθ + ¨− − − ˙ mgb cosθ 0 (6) + = (ii) Equilibrium point is located where the force is zero, or∂V/∂θ 0. Differentiating (4) =

∂V mgb cosθ k(b 2 a 2) sinθ cosθ (7) ∂θ = + − π Clearly the right-hand side of (7) is zero forθ =± 2 Writing (7) as

mgb k(b 2 a 2) sinθ cosθ (8) [ + − ] Another equilibrium point is obtained when

mgb sinθ (9) = k(a 2 b 2) − provideda>b. ∂2V (iii) An equilibrium point will be stable if > 0 and will be unstable if ∂θ 2 ∂2V < 0. Differentiating (8) again we have ∂θ 2

∂2V k(b 2 a 2)(cos2 θ sin 2 θ) mgb sinθ (10) ∂θ 2 = − − − π Forθ ,(10) reduces to = 2 k(a 2 b 2) mgb (11) − − mgb π Expression (11) will be positive ifa 2 >b 2 , andθ will be + k = 2 a stable point. 7.3 Solutions 333 π Forθ ,(10) reduces to =− 2 k(a 2 b 2) mgb (12) − + mgb π Expression (12) will be positive ifa 2 >b 2 , andθ will be − k =− 2 a stable point. A(θ) (iv)T 2π = V (θ)

π 2 2 V k(a b ) mgb, by(12) − 2 = − + 1 A(θ) is the coefﬁcient of θ 2 in(3) 2 ˙ A(θ) m(a 2 sin2 θ b 2 cos2 θ) ∴ = + π 2 ∴A ma − 2 = ma2 T 2π ∴ = k(a 2 b 2) mgb − + 7.31 In prob. (7.12) the following equations were obtained:

(m1 m 2)l 1θ1 m 2l2θ2 (m 1 m 2)gθ1 0 (1) + ¨ + ¨ + + = l2θ2 gθ 2 l 1θ1 0 (2) ¨ + + ¨ = Forl 1 l 2 l andm 1 m 2 m,(1) and(2) become = = = =

2lθ1 l θ2 2gθ 1 0 (3) ¨ + ¨ + = lθ2 l θ1 gθ 2 0 (4) ¨ + ¨ + = The harmonic solutions of (3) and (4) are written as

θ1 A sinωt,θ 2 B sinωt (5) = = 2 2 θ1 Aω sinωt, θ2 Bω sinωt (6) ¨ =− ¨ =− Substituting (5) and (6) in (3) and (4) and simplifying

2(lω2 g)A lω 2 B 0 (7) − + = lω2 A (lω 2 g)B 0 (8) + − = The frequency equation is obtained by equating to zero the determinant formed by the coefﬁcients ofA andB: 334 7 Lagrangian and Hamiltonian Mechanics 2(lω2 g)lω 2 − 0 lω2 (lω2 g) = − Expanding the determinant

l2ω4 4 lgω 2 2g 2 0

− +g = ω2 2 √2 = ± l g ∴ω 2 √2 = ± l g g ω 1 0.76 ,ω 2 1.85 ∴ = l = l

7.32 While the method employed in prob.(6.46) was based on forces or torques, that is, Newton’s method, the Lagrangian method is based on energy:

1 T m( x2 x2) (1) = 2 ˙ 1 +˙ 2

For small angles y1 and y2 are negligibly small ˙ ˙

1 2 V k(x 1 x 2) mgb(1 cosθ 1) mgb(1 cosθ 2) = 2 − + − + −

2 2 θ1 x1 For small angles 1 cosθ 1 . − = 2 = 2b2 2 x2 Similarly, 1 cosθ 2 − = 2b2

1 2 mg 2 2 V k(x 1 x 2) (x x ) (2) ∴ = 2 − + 2b 1 + 2 1 2 2 1 2 2 mg 2 2 L m( x x ) k(x 2x 1x2 x ) (x x ) (3) ∴ = 2 ˙ 1 +˙ 2 − 2 1 − + 2 − 2b 1 − 2

The Lagrange’s equations for the coordinatesx 1 andx 2 are d ∂L ∂L d ∂L ∂L 0, 0 (4) dt ∂ x1 − ∂x 1 = dt ∂ x2 − ∂x 2 = ˙ ˙ Using (3) in (4) we obtain

mg m x1 k x1 kx 2 0 (5) ¨ + + b − = mg m x2 kx 1 k x2 0 (6) ¨ − + + b = 7.3 Solutions 335

Assuming thatx 1 andx 2are periodic with the same frequency but different amplitudes, let

2 x1 A sinωt, x 1 Aω sinωt (7) = ¨ =− 2 x2 B sinωt, x 2 Bω sinωt (8) = ¨ =− Substituting (7) and (8) in (5) and (6) and simplifying

mg k mω 2 A kB 0 (9) + b − − = mg kA k mω 2 B 0 (10) − + + b − = The frequency equation is obtained by equating to zero the determinant formed by the coefﬁcients ofA andB: mg 2 k b mω k + − mg− 0 k k mω 2 = − + b − Expanding the determinant and solving gives g g 2k ω1 andω 2 , = b = b + m

In agreement with the results of prob.(6.46).

7.33 Let the origin be at thefixed point O and OB be the diameter passing through C the centre of the circular wire, Fig. 7.30. The position ofm is indicated by the angleθ subtended by the radius CP with the diameter OB. Only one general coordinateq θ is sufficient for this problem. Letφ ωt be the angle which the diameter= OB makes with thefixedx-axis. From= the geometry of the diagram (Fig. 7.30) the coordinates ofm are expressed as

Fig. 7.30 336 7 Lagrangian and Hamiltonian Mechanics

x r cosωt r cos(θ ωt) (1) = + + y r sinωt r sin(θ ωt) (2) = + + The velocity components are found as

x rω sinωt r( θ ω) sin(θ ωt) (3) ˙ =− − ˙+ + y rω cosωt r( θ ω) cos(θ ωt) (4) ˙ = − ˙+ + Squaring and adding and simplifying we obtain

x2 y2 r 2ω2 r 2(θ ω) 2 2r 2ω(θ ω) cosθ (5) ˙ +˙ = + ˙+ + ˙+ 1 T mr 2 ω2 ( θ ω) 2 2ω( θ ω) cosθ (6) ∴ = 2 [ + ˙+ + ˙+ ] HereV 0, and soL T . The Lagrange’s equation then simply reduces to = = d ∂T ∂T 0 (7) dt ∂θ − ∂θ = ˙ Cancelling the common factormr 2 (7) becomes

d (θ ω ω cosθ) ω( θ ω) sinθ 0 (8) dt ˙+ + + ˙+ = which reduces to

θ ω 2 sinθ 0 (9) ¨+ = which is the equation for simple pendulum. Thus the bead oscillates about the rotating line OB as a pendulum of lengthr a/ω 2. = 7.34 (a) The velocityv of massm relative to the horizontal surface is given by combining s with x. The components of the velocityv are ˙ ˙

vx x s cosα (1) =˙ +˙ vy s sinα (2) =−˙ v 2 v 2 v 2 x2 s2 2 x s cosα (3) ∴ = x + y =˙ +˙ + ˙ ˙ Kinetic energy of the system

1 1 T M x2 m( s2 x2 2 s x cosα) (4) = 2 ˙ + 2 ˙ +˙ + ˙ ˙ Potential energy comes exclusively from the massm (spring energy gravitational energy) + 7.3 Solutions 337

k V (s l) 2 mg(h s sinα) (5) = 2 − + − (M m) 1 k L T V + x2 m s2 m x s cosα (s l) 2 = − = 2 ˙ + 2 ˙ + ˙ ˙ − 2 − mg(h s sinα) (6) − −

(b) The generalized coordinates areq 1 x andq 2 s. The Lagrange’s equations are = = d ∂L ∂L d ∂L ∂L 0, 0 (7) dt ∂ x − ∂x = dt ∂ s − ∂s = ˙ ˙ Using (6) in (7), equations of motion become

(M m) x m s cosα 0 (8) + ¨ + ¨ = m x cosα m s k(s s 0) 0 (9) ¨ + ¨ + − = wheres 0 l (mg sinα)/k. = + Letx A sinωt ands s 0 B sinωt (10) = − = x ω 2 A sinωt, s Bω 2 sinωt (11) ¨ =− ¨ =− Substituting (10) and (11) in (8) and (9) we obtain

A(M m) B cosα 0 (12) + + = Amω2 cosα B(mω 2 k) 0 (13) + − = EliminatingA andB, weﬁnd k(M m) ω + (14) = m(M m sin 2 α) +

Components of the velocity of the ball as observed on the table are

7.35 vx x y cosα (1) =˙ +˙ vy y sinα (2) =˙ v2 v 2 v 2 x2 y2 2 x y cosα (3) = x + y =˙ +˙ + ˙ ˙ 1 1 1 1 2 T(ball) mv 2 Iω 2 mv 2 mr 2ω2 = 2 + 2 = 2 + 2 × 5 1 1 7 mv 2 mv 2 mv 2 (4) = 2 + 5 = 10 1 T(wedge) (M m) x 2 (5) = 2 + ˙ 338 7 Lagrangian and Hamiltonian Mechanics

7 1 T(system) m( x2 y2 2 x y cosα) (M m) x 2 (6) ∴ = 10 ˙ +˙ + ˙ ˙ + 2 + ˙ V(system) V(ball) mgy sinα (7) = =− 7 1 L m( x2 y2 2 x y cosα) (M m) x 2 mgy sinα (8) = 10 ˙ +˙ + ˙ ˙ +2 + ˙ + Lagrange’s equations d ∂L ∂L d ∂L ∂L 0, 0 (9) dt ∂ x − ∂x = dt ∂ y − ∂y = ˙ ˙ become 7m 7 x m y cosα (M m) x 0 (10) 5 ¨ + 5 ¨ + + ¨ = 7m 7 y m x cosα mg sinα 0 (11) 5 ¨ + 5 ¨ − = Solving (10) and (11) and simplifying

5mg sinα cosα x (for the wedge) ¨ =− 5M (5 7 sin 2 α)m + + 5(5M 12m)g sinα y + (for the ball) ¨ = 7(5M (5 7 sin 2 α)m) + + ForM m andα π/4 = = 5g x ¨ = 27 85√2 y ¨ = 189 Chapter 8 Waves

Abstract Chapter 8 deals with waves. The topics covered are wave equation, progressive and stationary waves, vibration of strings, wave velocity in solids, liquids and gases, capillary waves and gravity waves, the Doppler effect, shock wave, rever- beration in buildings, stationary waves in pipes and intensity level.

8.1 Basic Concepts and Formulae

The travelling wave: The simple harmonic progressive wave travelling in the posi- tivex-direction can be variously written as

2π y A sin (vt x) = λ − t x A sin 2π = T − λ A sin(ωt kx)

= − x A sin 2πf t (8.1) = − v whereA is the amplitude,f the frequency andv the wave velocity,λ the wave- length,ω 2πf the angular frequency andk 2π/λ, the wave number. Similarly,= the wave in the negativex-direction= can be written as

2π y A sin (vt x) (8.2) = λ + and so on. The superposition principle states that when two or more waves traverse the same region independently, the displacement of any particle at a given time is given by the vector addition of the displacement due to the individual waves. Interference of waves: Interference is the physical effect caused by the superposition of two or more wave trains crossing the same region simultaneously. The wave trains must have a constant phase difference.

339