preguntas resueltas.pdf

8/10/2019 PREGUNTAS RESUELTAS.pdf

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HW

Problem set 1, Due Jan 15. Total 80 points. The problem marked by * will not

be graded.1 . Find an equation describing the shortest path between the ( x,y,z ) two points on the

conical surface z = 1 x2 + y2, and its general solutionSolution:Cylindrical coordinates: x = r sin ; y = r cos which gives us z = 1 r

L = ds = dr 2 + r2d2 + dz 2 =

2 + r2

2r dr (0.1)

dL/d = 0.

Euler equation:

ddr

r 2

2 + r2 2= 0

r 2 = C 2 + r2 2 =

2C r

r2

C 2

= 2C drr r 2 C 2 + C 1 =21/ 2 arccos |

C r |+ C 1 (0.2)

2 . Find the curve y(x) that passes through the endpoints (0 , 0) and (1, 1) and minimizes

the functional

I [y] = 1

0 [(dy(x)

dx )2 y2(x)]dx

Solution:

f = ( y )2 y2 (0.3)Euler equation:

2y(x) + 2 y = 0

y = C 1 sin t + C 2 cost

C 2 = 0, C 1 = 1/ sin 1 (0.4)


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3 . Consider a spherical pendulum of a mass m and a length l. Write expressions for

a. the Lagrangian in terms of generalized coordinates, and the Lagrange equations

L = 1

2ml 2(

d

dt)2 +

1

2ml 2 sin2 (

d

dt)2 + mgl cos

12

ml 2d2d2t ml

2 cos (ddt

)2 + mgl sin = 0 (0.5)

12

ml 2 ddt

(sin2 ddt

) = 0 (0.6)

b. the generalized momenta,

p = ml 2ddt

p = ml2 sin2 ddt

(0.7)

c. generalized forces

F = 0

F =

mgl sin + ml2 2 sin cos (0.8)

4 . A double pendulum consists of two simple pendule, with one pendulum suspended

from the the blob of the other. The two pendule have equal lengths and bobs have equal

masses. They are conned to move in the same plane. Find Lagranges equations of motion

for the system. Do not assume small angles.Solution:

The Lagrangian has a form:

L = ml 2 21 + m2 l2 22 + ml2 1 2 cos(1 2) + 2 mgl cos 1 + mgl cos 2

L1

= 2mgl sin 1 ml 2 1 2 sin(1 2),L2

= mgl sin 2 + ml2 1 2 sin(1 2),L 1

= 2 ml 2 1 + ml2 2 cos(1 2)L 2

= ml2 2 + ml2 1 cos(1 2) (0.9)

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Lagrange equations are:

ddt

[2ml 2 1 + ml2 2 cos(1 2)] + 2mgl sin 1 + ml2 1 2 sin(1 2) = 0d

dt[ml 2 2 + ml2 1 cos(1

2)] + mgl sin 2

ml 2 1 2 sin(1

2) = 0

(0.10)

.

5 . A pendulum consists of a mass m suspended by massless spring with unextended

length b and spring constant k.

.

a. Find Lagranges equations of motion.

Solution:

x = l sin ; y = l cos

x = l sin + l cos ; y = l cos l sin ;L =

m2

(l2 + l2 2) + mgl cos k2

(l b)2 (0.11)Lagrange equations:

ml ml 2 mg cos + k(l b) = 0ml 2 + 2ml l + mgl sin = 0 (0.12)

b. Suppose that the point of support of the pendulum is moving in vertical direction

according the low y = a sin ct Fig.1). Find the Lagrange equations.Solution:

y = l cos + a sin ct = y + a sin ct, y = l cos l sin + ac cos ct = y + ac cos ctL =

m2

(l2 + l2 2) + m yac cos ct + mgl cos k(l b)2(0.13)The term which depends only on time was omitted from the Lagrangian. It is convenient

to use the identity:ddt

(y cos ct) = dydt

cosct cy sin ct (0.14)

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FIG. 1:

Omitting the total derivative from the Lagrangian we have

L = m2

(l2 + l2 2) + mac2y sin ct + mgl cos k(l b)2 =m2

(l2 + l2 2) + mac2l cos sin ct + mgl cos k(l b)2 (0.15)(You do not have to use this trick. Lagrange Eqs. will be the same)

Lagrange Eqs:

ml ml 2 mg cos + k(l b) + mac2 cos sin ct = 0ml 2 + 2ml l + mgl sin mac 2l sin sin ct = 0 (0.16)

6 . A spherical pendulum is rotating about the vertical axis with frequency (See Fig.2).

Write a Lagrange equation.Solution:

L = 12

ml 2(ddt

)2 + 12

ml 22 sin2 + mgl cos

ddt

=

d2d2t

2 sin cos + gl sin = 0 (0.17)

7 . Two identical pendule of mass m and length l are attached to a bar of a mass M , which

can slide without friction in horizontal direction. (see Fig. 3). Write Lagrange equations.

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FIG. 2:

FIG. 3:

Solution:

x1,2 = x + l sin 1,2, y = l cos

L = M + 2m

2 x2

+ ml2

2 21 +

ml 2

2 22 + mlx cos 1 1 + mlx cos 2 2 + mgl cos 1 + mgl cos 2

(M + 2m)x + ml ddt

(cos 1 1 + cos 2 2) = 0

ml 1 + ddt

(x cos 1) + g sin 1 = 0

2 + ddt

(x cos 2) + g sin 2 = 0(0.18)

8 . A point particle with a mass m moves along a circle of radius l in a vertical plane

under the inuence of the gravity eld (mathematical pendulum). Estimate the period of

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FIG. 4:

the pendulum if a. E 2mgl . Here E is the energy of the pendulum.Solution:

T = lg (0.19)b. 0 < 2mgl E 2mgl .Solution:

T = lg ln 2mgl2mgl E (0.20)9* . A particle is oscillating on a curve y(x) with a frequency , which is independent of

the amplitude. Find the curve.Solution:

Let s be a length of the curve. (See g.4)

dsdt = mg sin = mg

dyds

dsdt

= 2s2s = g

dyds

s = ( 2/ 2g)s2

s = g2

sin (0.21)

Thus the curve has maximum hight Y = g/ 22.

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Problem set 2 , 3 0 points. Due. Jan. 22. The problem, marked by *, will not be

graded.

1. Find the deection angle of fast particles ( E V ) moving in a potential

U (r ) = V exp(r2

R2 )

as a function of the impact parameter .Solution:

p = U (| + vt|)dt =

2V v

xex2

where x = /R

= vRE

e2 /R 2 (0.22)

2. For what values of the angular momentum M it is possible to have nite orbits in the

potential ? The particle have a mass m.

U = exp(r ) (0.23).Solution:

U eff = U (r ) + M 2

2m2r 2 (0.24)

To have nite orbits the effective potential should have minimum. Eq. U eff = 0 can

be reduced to the form f (x) = M 3/m , where f = x(x + 1)ex . This equation has realroots only if M 2/m is less than the max. of f (x). (x > 0). This maximum has a value

(2 + 51/ 2)exp[1/ 2(1 + 51/ 2)].3. Two masses hang from a support by string of equal length l. The masses are coupled

by a spring of spring constant k, and upstretched length l. (See Fig.5.)a. Write Lagrange equations.Solution:

L = 12

ml 2 21 + 12

ml 2 22 12

kl2(2 1)2 mgl21 mgl22ml 21 + kl2(1 2) + mgl1 = 0ml 22 + kl2(2

1) + mgl2 = 0 (0.25)

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FIG. 5:

b. What are frequencies of small oscillations of the system?Solution:

We a looking for a solution in the form 1,2 = a1,2eit .

ml 22a1 + kl2(a1 a2) + mgla1 = 0ml 22a2 + kl2(a2 a1) + mgla2 = 0

(ml 22 kl2 mgl)2 = k221 = (2kl2 + mgl)/ml 2, 22 = g/l (0.26)

4. Find the frequency of the small oscillations for particles moving in the following 1D

potential:

U (x) = V cosx Fx.Solution:

U = V sin x 0 F = 0sin x 0 = F/V

U = V 2 cos x 02 =

V 2

m 1 ( F V )2(0.27)

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5. Find frequencies of small oscillations of the double pendulum considered in HW1,

problem 4.

Solution:

Linearized Lagrange Eqs:

2mgl1 ddt

[2ml 2 1 + ml2 2] = 0

mgl2 ddt

[ml 2 2 + ml2 1] = 0

(0.28)

21,2 = 2

2 1gl (0.29)

6. Consider the system shown in Fig.3. What are frequencies of small oscillations?

Lianearizing Lagrange Eqs. we have

(M + 2m)x + m1 + 2) = 0

ml 21 + mx + mg1 = 0

ml 22 + mx + mg2 = 0 (0.30)

21 = gl

, 22 = gl

M + 2mM

(0.31)

7.* Find the inaccessible region of space for a beam of particles with a mass m ying

parallel to the z-axis with a velocity v and being scattered by a potential U (r ) = /r .

.

Solution:

The solution of the Kepler problem for the orbit is

P r

= e cos( 0) 1where p = M 2/m ; e = 1 + 2EM 2/m 2, while 0 = 1/e is determined by the conditionthat = 0 at r = . The inaccesable region is bounded be the envelope of the family of orbits. To nd it we differentiate the equation for the orbit

M 2

mr + 1 cos

M

2E m

sin = 0

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with respect to M 2M mr 2E m sin = 0

Eliminating M from the two Eqs. for r, , M we get the equation of caustics:

2Er

= 1 + cos

and the inaccessible region is

r < 2

E (1 + cos )is bounded by a paraboloid of revolution.

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Problem set 3 Due Jan.29. 10 points each. Problems marked by * will not be

graded.

1. A two-dimensional oscillator has kinetic and potential energies

T = 12m(x2 + y2)

U = 12

k(x2 + y2) + xy (0.32)

a. Show by a coordinate transformation that this oscillator is equivalent to an anisotropic

oscillator with lagrangian

L = 12

m(2 + 2) A2 2 +

B2

2 (0.33)

Solution:x = 1 cos 2 sin y = 1 sin + 2 cos

L = 12

m(2 + 2) + k2

(21 + 22) + [(

21 21)sin cos + 12(cos2 sin2 )] (0.34)

If cos2 = 0 we have

L = L = 12

m(2 + 2) + 21k +

2 + 22

k 2

(0.35)

b. Find eigenfrequencies of the oscillatorSolution:

21,2 = k

m (0.36)

2. The Lagrangian of a system has a form

L = 1

2(m1 x2 + m2 y2) + xy

1

2(x2 + y2)

a Find the eigenfrequencies of the system.Solution:

Lagrange equations:

m1x + y + x = 0

m2y + x + y = 0

(0.37)

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(m12 1)(m22 1) 24 = 0

2 = (m1 + m2) (m1 + m2)2 + 4 22( 2

m1m2)

b Find the normal coordinates of the system.

.

x = Q1 cos Q2 sin y = Q1 sin + Q2 cos

(0.38)

This represent a rotation of the system of coordinates. The potential energy does not

change its form under the rotation, while the coefficient of Q1 Q2 is cos2+( m2m1)sin2.Thus to diagonalise the problem we determine the parameter from the condition cot 2 = 0.

3.

A particle moves in a central potential of the form U (r ) = k/ (n 1)r (n1) , where k andn are constants.

a. Find a criterium of existence of a circular orbit.Solution:

U eff = k

(n 1)r (n1) +

M 2

2mr 2

dU eff dr |r = R =

kRn

M 2

mR 3 = 0

R (n3) = mkM 2

(0.39)

b.Find a criterium of stability of the circular orbit.

Solution:

d2U ef f dr 2 |r = R =

nkR (n1)

M 2

mR 3 > 0

(3 n)M 2

m > 0 (0.40)

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4. Consider a mechanical system described by a Lagrangian L(q, q ) (for simplicity we

assume that L does not have explicit time dependence). Consider another Lagrangian,

L(q, q ) = L(q, q ) + ddt

f (q ) = L(q, q ) + f (q )q

where f (q ) is some function of the generalized coordinate q .

a. Write the Euler-Lagrange equation for the new Lagrangian L, and show that it coin-

cides with the Euler-Lagrange equation for the old Lagrangian L.

ddt

d(f (q )q )dq

f (q )q dq

= f q f q = 0 (0.41)b. Explain this coincidence using Hamiltons action principle. (What is the difference

between the action for L and L?).See a comment in LL, p4 after Eq. 2.8.

5. Let the central force be F (r ) = (b/r 2 c/r 4)n r , where b > 0, c > 0, and n r is theunit vector in the radial direction.

a. Find a radius r of a circular orbit.

Solution:

dU eff

dr =

d

dr[U +

M 2

2mr2 ] =

(b/r 2 c/r 4)r M 2

mr 3 = 0

M 2rm br

2 + c = 0

r = R = M 2

2bm +

12b

[(M 2

m )2 + 4 bc]1/ 2 (0.42)

b. Check stability of the orbit.

Solution:

d2U eff dr 2 |r = R =

3M mR 4

2bR3

+ 4cR5

= K/ 2 =

bR2 = M 2R/m + c

K/ 2 = 3M 3R/m 2bR2 + 4 c = M 2R/m + 2c > 0 (0.43)c. What is the angular frequency of the motion as a function of M, b and c, where M is

the angular momentum?

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Solution:

M = R2 (0.44)

d. Find eigenfrequency of small oscillations of r(t).Solution:

2 = k/m (0.45)

6.

a. Find the integrals of motion (other than the total energy E ) for the potential U =12 r

2 = 12 (x2 + y2).

Solution:

E x = mx2 + 12

x 2, E y = my2 + 12

y2 (0.46)

b. Is M z is an integral which is independent of E, E x , E y ?Solution: No

7. Find the frequency of the small oscillations for particles moving in the following 1D

potential:

U (x) = V cosx

Fx

.Solution:

U = V sin x 0 F = 0sin x 0 = F/V

U =

V 2 cos x 0

2 = V 2

m 1 ( F V )2(0.47)

8. Find frequencies of small oscillations of the double pendulum considered in HW1,

problem 4.Solution:

Linearized Lagrange Eqs:

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2mgl1 ddt

[2ml 2 1 + ml2 2] = 0

mgl2

d

dt[ml 2 2 + ml2 1] = 0

(0.48)

21,2 = 2

2 1gl

(0.49)

9. Consider the system shown in Fig.3. What are frequencies of small oscillations?Solution:

Linearizing Lagrange Eqs. we have

(M + 2m)x + m1 + 2) = 0

ml 21 + mx + mg1 = 0

ml 22 + mx + mg2 = 0 (0.50)

21 = gl

, 22 = gl

M + 2mM

(0.51)

10*.A particle of mass m is moving in the presence of a force F = K r 2 er/a . Determine

conditions for a such that the circular motion is stable.Solution:

F = dU/dr, dU eff /dr = 0, dU eff /dr > 0dU eff /dr =

F + d/dr (M 2/ 2mr 2) =

K

r2 e

r/a

M 2

mr3 = 0

kRe R/a = M 2/m

d2U eff /dr 2|r = R = 1R4

(2M 2

m 2KRe R/a

KR 2

a > 0

M 2

mR 4(1

Ra

) > 0 (0.52)

11*.

A comet is moving along a parabolic trajectory. (See g. 6). U = GmM/r , whereM is the mass of sun, and m is the mass of the particle. Find the time the comet spends

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within the orbit of the Earth E. Assume that the Earths orbit is circular. (The value of M

is given).Solution:

In the case of parabolic trajectory

E = mr 2

2 +

M 2

2mr 2 + U = 0 (0.53)

If r = rmin = p, then r = 0, U + M 2/ 2mr 2 = 0, and M = (2 mp )1/ 2

drdt

= 2m ( r M 22mr 2 )t = 2

rmax

r min

dr

2m (

r M

2

2mr 2 )=

= 23/ 2

3 ma3

(1 + 2 p

a ) 1 parmin = p, rmax = a

a

p

rdr r p

= 23 ma 3 (1 + 2 pa ) 1 p/a (0.54)

12*.

Calculate the rate of precession of the planetary perihelion, to lowest order in , if U =

GMmr + r 2 = U 0 + U .Solution:

At = 0 the trajectory of the planet is an ellipse and = 0.

= 2 rmax

r min

Mdrr 2 2m[E u]M 2/r 22 M

rmax

r mindr 2m[E u]M 2/r 2 =

2m M

rmax

r min

2mUdr

2m[E U 0]M 2/r 2

(0.55)

Since r = (dr/d ) = M mr 2 (dr/d ), andM 2

mr 2drd

= 1m 2m[E U 0]M 2/r 2, (0.56)

we have

= M

[2mM

0r 2Ud] =

M

[2mM

0d] =

2 mM 2

(0.57)

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FIG. 6:

Problem set 4 Due Feb. 5. 10 point each. Problems marked by * will not be graded.

1.

Consider a pendulum of a length l and mass m. Find a correction (2) to the frequency

of small oscillations 0 = g/l , which is quadratic in the amplitude of oscillations 0.Solution: = 20 sin 20 ( +

16

3 + ...) (0.58)

Since we are interested in a correction to 0 proportional to 20 we do not need higher powers

on is the expansion of sin. We can rewrite this equation as

202

= 20( + 16

3) + (202 1) (0.59)

and look for a solution in a form = (1) + (2) + .... , and = 0 + (1) + (2) +

....Here superscripts (1), (2), ... indicate that corresponding terms are proportional to

0, 20, .... respectively.

In the rst order in 0 we have

(1) = 20(1)(1) = 0 cos0t (0.60)

In the second order in 0 we have

(2) + 20(2) =

(1)

01 = (1) 00 cos0t (0.61)

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FIG. 7:

The requirement of the absence of resonance terms in rhs of the equation gives as (1) = 0.

This is a natural result: the frequency should be an even function of the amplitude of

oscillations. Then (1) = 0.

In the third order we have

(3) + 20(3) =

206

((1) )3 + (1)

0(1) =

cos 0t[3020

8 0(2) 0] + 20

2024

cos 30t (0.62)

(cos3 = 34 cos + 14 cos 3). The requirement of the absence of resonance terms in rhs of

the equation gives as

(2) = 20

8 0 (0.63)

.

2.

Consider a system shown in Fig.7. The point of support of the oscillator oscillates ac-

cording to the low y = a sin ct. Assume that c g/l ..a. Write down equations of motion averaged over the period of the oscillations 2 /c .

.Solution:

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y = a cos ct l cos , X = x + l sin y =

ac sin ct + l sin , X = x + l cos

L = 12

ml 2 2 + m + M 2

x2 + mlx cos acl sin ct sin + mgl cos (0.64)Using the fact that (cos sin ct) = sin sin ct + c cos cos ct we get, and neglecting atotal derivative we get

L = 12

ml 2 2 + M + m

2 x2 + mlx cos mlac 2 cos ct cos + mgl cos (0.65)

Thus the expression for the oscillating in time generalized force is

f = mlac2 sin cos ct (0.66)

(mgl 2 plays the role of the effective mass.) Averaging over the period 2 /c (See LL. Eq.

30.8) we get an expression for an effective potential U eff = mgl[cos + (ac )2

4gl sin2 ] and

for a Lagrangian

Leff = ml2

22 +

M + m2

x2 + mlx cos U eff (0.67)and Lagrange equations

ml 2 + ml ddt

(x cos ) = dU eff /d + mlx sin (m + M )x + ml

ddt

( cos ) = 0 (0.68)

b. How many stationary solutions do these equations have?

Solution:

In a stationary state all time derivatives are zero. So we have mgl[sin + (ac )2

4gl sin2] =

mgl sin (1 + (ac )24gl cos) = 0, which gives us sin = 0. Thus there are two solutions = 0,

a. = 0 is always stable solution. (( d2U eff /d 2)|=0 > 0)b. (d2U eff /d 2)|= > 0, and = is stable provided ( ac)2 > 2gl.c. What are frequencies of oscillations near the stationary solutions?

4.

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FIG. 8:

A ladder of mass m and length 2l stands against a frictionless wall with its feet on africtionless oor. (See Fig.8). The initial angle is 0. It is let go

.

a. Write Lagrange equations.Solution:

The kinetic energy ism2

(x2c + y2c ) +

12

I 2 = 23

ml 2 2 (0.69)

Here xc = l cos and yc = l sin are coordinates of the center of mass, and I = ml2/ 3 is

the moment of inertia relative to the center of mass.

L = 23

ml 2 2 mgl sin 43

l = g cos (0.70).

b. What will be the angle when the ladder loses contact with the wall?

Solution:

The energy conservation low

23

ml 2 2 = mgl sin 0 mgl sin xc = l 2 l sin = N/m (0.71)

Here N is the the force acting from the vertical wall to the lader. It the pint when the ladder

loses contact with the wall N = 0, which gives us =

2 cot , and using the lagrange

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FIG. 9:

Eq. we get 4l 2/ 3 = g sin . than from the energy conservation low we get

sin = 23

sin 0 (0.72)

5.

A homogeneous cylinder of radius r and mass m is rolling inside a cylindrical surface of

radius R. (There is no sliding.) (See Fig.9)

a. Write Lagrange equation.

Solution:

V c = (R r ), = V /rL =

m2

(R r )2 2 + 12

I (R r )2 2/r 2 + mg(R r )cos = 3m

4 (R a)2 2 + mg(R r )cos (0.73)

Here I = mR 2/ 2 is the moment of inertia of the cylinder.

.

b. Find a frequency of small oscillations about the stable equilibrium position.

. 6.

A uniform thin cylinder rod of length l and mass m is supported at its ends by two

massless springs with identical spring constants k. (See Fig.10). Find eigenfrequencies of

small oscillations. (Springs can move only vertically.)

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FIG. 10:

Solution:

Equations for the displacement of the center of mass and for small amplitude rotation

about the center of mass

12

(y1 + y2) = k(y1 + y2)I =

12

lk((y1 + y2) (0.74)

with I = ml 2/ 12.

Symmetric mode: ys = y1 + y2, 2 = 2 k/m .

Asymmetric mode: ya = y1 y2, 2 = 6 k/m ..

7* .

A particle of mass m is attached to a rigid support by a spring with force constant k.

(See Fig.11) To the mass of this oscillator an identical oscillator is attached. The whole

system is in the gravitation eld of the Earth. Find the normal frequencies and the normal

modes of the system. Consider vertical and horizontal motions separately.

Solution:

Denote the unstretched length of the spring as l. The equilibrium length of the springs

will then be l1 = l + 2 mg/ for the upper spring, and l2 = l + mg/ for the lower springs.

There are many ways to introduce the generalized coordinates. The result should not

depend on the choice of the coordinates. We shall use the deviations of the mass point

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FIG. 11:

from the equilibrium position. Let the x axis be directed vertically, and y horizontally. Thecoordinates of the rst (upper) mass point is ( x1, y1), and of the second (lower) mass point

is (x2, y2), respectively.

The kinetic energy of the system is

T = m2

(x21 + y21 + x

22 + y

22)

The potential energy is

V = 2 (l1 + x1)2 + y21 l

2

+ 2 (l2 + x2 x1)2 + ( y2 y1)2 l

2

mg(x1 + x2)Expanding the potential energy in Taylor series over x1,2 and y1,2, keeping only terms up to

the second order inclusively, we nd

V = 2

[(l1 l)2 + ( l2 l)2]+ [(l1 l) mg]x1 + [(l2 l1) mg]x2+

2[x2

1 + ( x2

x1)2] +

2

l1 ll1

y21 +

l2 ll2

(y2

y1)2

The constant term (rst line) can be ignored. The terms linear in x1 and x2 (second line)

vanish due to the equilibrium condition. The last line is all that has to be taken into account.

Clearly, the vertical and the horizontal motions separate.

For the vertical motion ( x1, x2) the characteristic equation is

det2 m2

m2

= 0

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The two normal frequencies are

2 = 2m

(3 5)and the two normal modes are

11 52For the horizontal motion, let us introduce the notations

1 = l1 l

l1=

2mgl + 2 mg

2 = l2 l

l2=

mgl + mg

The characteristic equation is then

det

( 1 + 2)

m2 2

2 2 m2 = 0which gives two solutions

2 = 2m

1 + 2 2 21 + 4 22The two corresponding normal modes are

2

12 12 21 + 4 22

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FIG. 12:

Problem set 5 Due Feb. 12. 10 point each. Problems marked by * will not be graded.

1.

A torsion pendulum consists of a vertical wire attached to a mass which may rotate about

the vertical. Consider three torsion pendulums which consist of identical wires from which

identical homogeneous solid cubes are hung. Cube A is hung from a corner, cube B from

midway along an edge, and cube C from the middle of a face (See Fig.12). What are the

ratios of periods of the three pendulums? Briey explain or derive your answer.

Solution:

I 1 = I 2 = I 3. The ratios equal to one.

2.

A solid ball of radius r rolling with velocity v, collides inelastically with a step of height

h < r . (See Fig.13). Assume no slipping. What is the minimum velocity for which the ball

will trip up over the step?

Solution:

M = mv(r h) + I = 75

mvr mvhM = I = ( I + mr 2) =

75

mr 2 (0.75)

Here M and M are angular momenta about the point of impact before and after the

collision. As the center of mass of the ball is momenterely at rest after the collision, the

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FIG. 13:

angular momentum conservation low gives us

= v/r 5hv/ 7r 2 (0.76)The kinetic energy must be sufficient to provide for the increase in potential energy. I 2/ 2 =

mgh .

v = r 70gh7r 5h

(0.77)

3.

Consider a nonlinear pendulum described by an equation

x + 20x + x2 + x3 = f sin

02

t (0.78)

a. What is the lowest order in f in which a resonance takes place?Solution:

In the absence of nonlinear terms

x(0) = a sin(0t + ) + f 20 20 / 4 sin 0

2 t (0.79)

Nonlinear terms generate harmonics with frequency 0. So if a = 0, in lowest in f order

x(2) + 20x(2) = (x(1) )2 = (

4f 320

)2 sin2 20

2 t =

( 4f 320

)21 cos 0t

2 = F (1 cos 0t) (0.80)

There is a component of the force at the resonance frequency 0.

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b. Estimate the amplitude of the oscillations at which the resonance saturates. Neglect

numerical factors.

As the amplitude a of the harmonics with the frequency 0 grows, the frequency of the

oscillations changes. The a- dependence of the frequency is quadratic in a (See Eq.28.13 in

LL)

= [ 3 80

5 2

1230]a2 (0.81)

As the frequency changes, the system is moving out of the resonance and the amplitude of

oscillations with the frequency 0 saturates at a value (See Eq.22.4 in LL)

a = F 0

(0.82)

(0 ) Finallya[

F 3

80 52

1230

]1/ 3 (0.83)

4. Consider the problem 1 in $ 35 of LL (Fig. 48) and assume that the lowest point

of the symmetrical top oscillates vertically according to the low Z = a sin t . Write the

Lagrangian of the system.

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MechFig1MT.pdf

FIG. 14:

MT

1. 50 points.

A particle of mass m1 is constrained to move on a horizontal plane. A second particle

of mass m, is constrained to a vertical line. The two particles are connected by a massless

string which passes through a hole in the plane. (See Fig. 1.)a. Write a Lagrange equation.

Solution:

L = m1

2 (r 2 + r2 2) +

12

mr 2 mgrddt

(m1r 2 2) = 0

(m1 + m)r m1r

2

+ mg = 0 (0.84)

b. Find the dependence of the radius r = R(M ) as a function of the angular momentum

of the system M in a stationary state ( r = 0).

Solution:

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ddr

U eff = ddr

(mgr + M 2

2 m1r 2) = 0

r = R = ( M 2

gm1m)1/ 3

(0.85)

c. Find a frequency of oscillations.

Solution:

2 = 1

m1 + md2U eff

dr 2 |r = R = 1

1 + m1/m(3g/R ) (0.86)

2. 50 points

A particle of mass m = 1 moves in a potential

V = 1r 2

where r is the distance to the center.

a. Write down the Lagrangian of the system, using the polar coordinates r and .

Solution:

L = 12

(r 2 + r2 2) + 1r 2

b. The particle moves from r = along an orbit with angular momentum M . Showthat if M < 2 then the particle falls into the center of force (capture), and if M > 2 itwill escape to innity.

Solution:

The radial motion is determined by the effective potential, which is equal to the sum of

the potential energy and the centrifugal potential,

V eff (r ) = 1r 2

+ M 2

2r 2 =

M 2 22r 2

When M 2 > 2, the effective potential increases to + as r 0. A particle moving fromr = will bounce back. When M 2 < 2 the effective potential is negative and goes to as r 0. A particle moving from r = will fall to the center r = 0.

c. Suppose the velocity of the particle at t = is v. Write the condition for capturein term of the impact parameter b.

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Solution:

The angular momentum is conserved, and is equal to the angular momentum at t = ,which is bv. Therefore capture occurs when bv 0.

a. Find a radius of a circular orbit as a function of A, m, n , M .

Solution:A circular motion corresponds to a minimum of the effective potential

d2U eff dr 2 |r = R =

d2

dr 2[Ar n /n +

M 2

2mr 2]|r = R

R = ( M 2/mA )1

( n +2) (0.87)

b. What is the angular frequency of this motion?

mR 2 = M (0.88)

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c. What is the frequency of small oscillations of r(t) near the circular orbit?

K/ 2 = d2U eff

dr 2 |r = RK = ( n + 2)

M 2

m [

M 2

mA]4/ (n +2)

2 = k/m (0.89)

d. At what value of n = non-circular orbits are closed?

Solution: n = 1, 2.3.

A point of support of a pendulum of a length l and a mass m is moving horizontally with

an acceleration a.

Write a lagrange equation.

Solution:

x = l sin + at 2/ 2, x = l cos + at

y = l cos . y = l sin L =

ml2

22 + mlat cos mgl(1 cos )

ml 2 mlat sin + mlat cos mlat sin + sin (0.90)

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Problem set 5. Due Feb.26. (70 points). Problems marked by * will not be graded.

1.

Obtain the Hamilton equations by calculating

dq idt = [H ; q i];

dpidt

= [H ; pi] (0.91)

where [..,..] are the Poisson bracket. The form of the Hamilton function H should not be

specied.

Solution:

Since q and p are independent variables q i /p j = 0 and pi /q j = 0 and q i /q j = ij

one gets[H ; q i] =

H p i

(0.92)

[H ; pi] = H q i

(0.93)

2.Find the canonical transformation generating by the generating function

F = ( r P ) + ( a [r P ])

Solution:

p = F

r = P + [a r ];R = r + [a r ] (0.94)

The transformation is a rotation of the system of coordinates over the angle a

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3.

Is the following transformation is canonical one?

Q = ln(1 + q 1/ 2 cos p)

P = 2(1 + q 1/ 2 cos p)q 1/ 2 sin p

(0.95)

Solution:

[Q, Q ] = 0, [P, P ] = 0, [Q, P ] =

= Qq

P p

P q

Qp

=

q 1/ 2 cos p1 + q 1/ 2 cos p

[q sin2 p + (1 + q 1/ 2 cos p)q 1/ 2 cos p] ++

q 1/ 2 sin2 p1 + q 1/ 2 cos p

[cos p + (1 + q 1/ 2 cos p)q 1/ 2] = 1 (0.96)

So the transformation is canonical.

4.

Consider a top with principle moments of inertia I 3 > I 2 > I 1. Initially it rotates with

frequency 1 bout the axis 1. A a small torque 2 = 0 cos t is applied about axis 2.

Use Eulers equations and calculate 1, 2, 3 in the rst order in 0.

Solution:

I 1d1dt

+ ( I 3 I 2)23 = 0I 2

d2dt

+ ( I 1 I 3)13 = I

d3dt

+ ( I 2 I 1)12 = 0

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Since we expect that 2, 3 1 we can leanearise the eqs.I 1

d1dt

= 0

I 2d2

dt + ( I 3

I 2)13 =

I 3d3dt

+ ( I 2 I 1)12 = 0According to the rst eq. 1 = const . Solving the eqs. in a form 1,2(t) =

(0)1,2eit we get

(0)2 = 0iI 3

1(I 3 I 1)(I 2 I 3) 2I 2I 3(0.97)

(To solution of the uniform eq. should be added to this eq.)

5.

The surface of a sphere is vibrating SLOWLY in such a way that the principal moments

of inertia are harmonic functions of time.

I zz = 2mr 2

5

(1 + cos t) (0.98)

I xx = I yy = 2mr 2

5 (1

2

cost) (0.99)

where 1. Find z,y,z in the rst order in .

Solution:

The orbital moment is conserved. However, the principe moments of inertia are dened

in the body frame. Using Eulers equations we get:

dI zz zdt = 0

dI xx xdt

+ 32

I 0yz cost = 0

dI yy ydt

32

I 0xz cost = 0 (0.100)

where I 0 = 2mr 2/ 5. Thus

z = z0

1 + cost (0.101)

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FIG. 15:

Since 1, in our approximation z = const . Then, in the rst in order, one can neglect

dI xx,yy /dt , and the last two eqs. are

I 0dxdt

+ 32

I yz cost = 0

I 0dydt

32

I xz cost = 0 (0.102)

If z one can regard in these eqs as a constant. Then a solution of these equationsis x = 0 sin 0t, x = 0 cos0t, where 0 = 3

2z .

6.

Consider a pendulum shown in Fig.15 Assume that the length of the pendulum l =

l0 + l1 cost changes in time, and l1 l0.a. Find the interval of frequencies where the resonance takes place.

Solution:The Lagrangian of the system is

L = m2

l2(t)( )2 + mgl(t)cos (0.103)

(The term proportional to l2 can be dropped from the Lagrangian. A Lagrange equation

has a form:

+ 2ll = 2(t)sin (0.104)

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The second term in the Eq. 0.89 makes this problem slightly different from that in LL

Ch.30. (If g(t) oscillates in time there is no second term in the Eq. and the problem would

be exactly the same as in LL. In both cases the parametric resonances exists.)

Expanding the frequency of oscillations in powers of l1 (keeping only lowest in h terms),

linearizing the Eq. and introducing = 20 + we get

80h sin(20 + )t = 2(t) (0.105)

20 = gl0

; 2(t) = 20[1h cos(20 + )t],h = l1/ 2l0, (0.106)

Let us look for a solution in the form = a sin(0 + 12 )t + bcos(0 + 12 )t, At the ends of

the interval of where the resonance takes place a and b are time independent, and and one

can drop their time derivatives. (In other words, one can differentiate over time only sin

and cos. The terms proportional to 20 cancel. Using the fact that

cos(0 + 12

)t cos(20 + )t = 12

[cos 3(0 + 12

)t + cos(0 + 12

)t]

sin(0 + 12

)t cos(20 + 12

)t = 12

[sin 3(0 + 12

)t + sin[(0 + 12

)]t]

sin(0 + 1

2)t sin(20 +

1

2)t =

1

2[cos(0 +

1

2)t

cos 3(0 +

1

2)t] (0.107)

and dropping the harmonics oscillating with the frequency close to 3 0 we get

b( + 92

h0)sin(0 + 12

)t +

a( + 92

h0)(cos(0 + 12

)t = 0 (0.108)

This Eq. , should valid at arbitrary t, and we get for the endpoints = 92 h0), and forthe interval:

92

h0 < < 92

h0 (0.109)

b. Assume that g/l and write a Lagrange equation averaged over the period 2 /

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Solution:

The Lagrange equation is different from that considered in Ch. 30 LL. by the second

term. However, the form of the potential energy is the same. (After the averaging the second

term does not contribute to the Lagrange Eq. ) Thus, in notations of LL

f = mgl1 sin cos t

U eff = mgl0 cos + (mgl1)2

4 sin2

ml 20 = dU eff

d (0.110)

c. Now consider the case g/l Write an expression for the adiabatic invariant. Howthe energy of the pendulum depends on time?

7.

Calculate the following Poisson brackets:

a.[M i , M j ];b.[M i ; p j ];c.[M i , x j ], d.[(a M ), (b M )].Solution:

[M i , M j ] = ijkM k ; [M i ; p j ] = ijkpk , [M i ; p j ] = ijkpk ; [(a M )(b M )] = ([a b ]M )(0.111)

8.

Consider an innite system of of pendulums in a gravitational eld, shown in Fig.16

(Only a spring between the rst and the second pendulums is shown.)

a.

Find a Lagrangian of the system.Solution:

L =n

[ml 2

22n +

k2

(n +1 n )2 + mgl cos n ] (0.112)b.

Write the Lagrangian in a continuous limit where |n +1 n | 1. For simplicity use thesystem of units where l ,m, g, a, k = 1. Here a and k are spacing between springs and the

spring constant.

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FIG. 16:

Solution:

L = 12 dx[( (x, t ))2 + ( x)2 + 2 cos ] (0.113)

c.

Write Lagrange equations both in both discrete and continues limits.

solution:

ml 2n + k(2n n +1 n1) + mgl sin n = 0 2 sin = 0 (0.114)

d.

Write the expressions for generalized momenta and for the Hamiltonians in the continues

limit.

Solution:

p = L

=

H ( p, ) = dxp L = dx p2 + ()

2 + 2(1 cos )2

(0.115)

8*.

Find a soliton (solitary wave) solutions of the Lagrange eq. in the previous problem in

the continues limit. Use the boundary conditions () = 0 , () = 2 This solution

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should have a form (y) = (x ut ) which does not change its shape in time and move witha velocity u. Substitute this form into the lagrange eq., and reduce it to the problem of a

single particle moving in an effective potential U (). So plays a role of coordinate, and

y plays a role of time. Estimate (up to a numerical factor) a relation between the solitons

width and its velocity u.

Solution:

(1 u2)(d2d2y

)2 + 2 cos = 2E (0.116)

the boundary conditions gives E = 1. One can introduce a variable z = y/ (1 u2)1/ 2 whicheliminates u from the eq. So the width of the soliton scales as (1u2)1/ 2. The eq. is identicalto an eq. describing a particle with a mass (1 u2) moving in a well potential . Here y playsthe role of time. If (1 u

2

) > 0 the particles motion is nite, and it oscillates.Solution of this eq. with a center at y = y0 ((y0) = 0 is

y y01 u2

= / 2

/ 2

dsin

= ln tan( / 4) (0.117)

BTW, Sin-Gordon eq. has innite number of conservation lows. It is totaly integrable

system, and all exact solutions can be obtained. The famous method of obtaining these

solutions is a inverse scattering method. please read about it.

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Problem set 6 Due March. 5 Problems marked by * will not be graded.

1 A fast particle ( v A/k 2m) of mass m is moving in the potential eldU (r ) = A(x2 y2)sin kz (0.118)

at a small angle to the z-axis.

Describe the motion of the particle in xy-plane.Solution:

In the rst approximation z = vt. In the xy-plane there is a fast oscillating force acting

on the particle

f x = 2Ax sin kvt; f y = 2Ay sin kvtthe corresponding effective potential (See Ch. 30 Landau) has a form

U ef f (x, y) = m2

2 (x2 + y2) (0.119)

where = A/mkv .

Thus the particle performs harmonic oscillations in the xy plane with the frequency .

Note that kv and we can apply results of Ch. 30 to this problem.

2

Consider a conguration of the magnetic eld shown in Fig. 19. In the center there is a

source of particles which emits them isotropically with momentum p. Estimate a fraction of

particles which escape the magnetic trap.Solution:

The adiabatic invariant is

I = 12 P

t dr = 12 p

t dr + e2c Adr (0.120)Here p t is the component of momentum perpendicular to H . The modulus |p t | = const ,

and r = cpt /eH . Thus

I = rpt e2c

Hr 2 = cp2t2eH

= const (0.121)

Energy is also conserved ( p2t + p2)/ 2m = E . If at x = p = 0, then the particle do notleave the region. At this point

I = cE eH

(0.122)

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Inside the region

I = I = cE sin2 e(H H )

(0.123)

where sin = pt /p . Thus sin = [(H H )/H ]1/ 2.Thus the fraction of particles which escapes on innity is

2(1 cos )4

(0.124)

Solve the Hamiltonian-Jacoby equation for S (q,t , ) in the case of a single particle moving

under the Hamiltonian H = p2/ 2m.

S t

+ 12

(S q

)2 = 0 (0.125)

S = q 12

2t (0.126)

3

Find a solution solution u(x vt) of of Koteveg de Vries equationut

+ 3ux 3

+ uux

= 0 (0.127)

Express u(x, t ) in terms of an integral, and estimate the relation between the width and the

velocity of the soliton.Solution:

Let us look at a solution in the form u = u(x

ct). t u =

c xu, and introducing

y = x ct we get ddy

[d2ud2y cu + u

2/ 2 = 0 (0.128)

Integrating on time over y and putting the integration constant to zero we get an equation

describing a non-linear oscillator

d2ud2y

= dU dx

W =

cu2/ 2 + u3/ 6 (0.129)

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FIG. 17:

The potential W has a minimum at y = 2c width 3c. Thus the amplitude of the soliton

is of order c, and its width is of order 1/ c. The exact solution for the soliton isu =

3cch2(x ct)/ )

(0.130)

2 = 2 /c

4.

Consider a symmetrical top I 1 = I 2 = I 3 whose lower point is xed (See Fig. 17) in thepresence of gravity. The mass of the top is m, and the distance from the lower point to the

center of mass is l.

a.

Write an expression for its rotational kinetic energy about the center of mass using

Eulerian co-ordinates.Solution:

K =i

I ii

I 12

(2 + 2 sin2 ) + I 32

( + cos )2 (0.131)

b.

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Write the expression for a Lagrangian. (Do not forget that the top is rotating about the

xed point).

L =i

(I iiI 1 + ml2

2 (2 + 2 sin2 ) +

I 3

2 ( + cos )2 = mgl cos

5.

Write the adiabatic invariant I for the pendulum shown in Fig. 15 assuming that l(t)

changes slowly in time. (Assume that the amplitude of oscillations is small.) Given the

value of I , how the amplitude of the oscillations 0 changes in time ? How the frequency

changes in time?Solution:

2 = g/l (t).L =

12m

l2 2 + mgl cos

I =

pdq/2 = E/ =

l2(t)202

= l3/ 2g1/ 220/ 2 = const

0 = [2Il3/ 2(t)g1/ 2]1/ 2 (0.132)

6.

The Hamiltonian for the system has the form:

H = 1

2( 1

q 2 + p2q 4) (0.133)

a. Find equation of motion.

b. Find a canonical transformation that reduces the Hamiltonian to the form of harmonic

oscillator.Solution:

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P = pq 2, Q = 1q

H = 1

2(P 2 + Q2)

(Q, P ) p,q = 1q 2

(q 2) (0.134)So the transformation is canonical.

7*

Consider a particle moving on a ( xy) plane. A constant external magnetic eld B is

applied in the direction perpendicular to the plane (z direction). Chose Ax = By/ 2;Ay = Bx/ 2, show that the Lagrangian is invariant under innitesimal rotation:

x = y, y = x (0.135)where is a small number. What is the conserved quantity that can be extracted from this

invariance?Solution:

The Lagrangian is

L = m2 (x2 + y2) + ec A x = m2 (x2 + y2) + eB2c (xy yx)Under an innitesimal rotation x = y, y = x the change of the Lagrangian is

L = m(x x + y y) + eB2c

(xy + x y yx y x)= m (xy yx) +

eB2c

(yy xx + xx yy) = 0The corresponding conserved quantity is (the z -component of) the angular momentum

L = xpy ypx = m(xy yx) + eB2c (x2 + y2)

8*

Consider an innitely thin rectangular of mass m with sides a and b which rotates about

the diagonal. (See Fig.18) . Find a torque applied to the system.Solution:

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FIG. 18:

FIG. 19:

Eulers equations are:

I xdxdt

+ ( I z I y)yz = xI y

dy

dt + ( I x

I z)zx = y

I zdzdt

+ ( I y I x )yx = z (0.136)I x = 112 ma

2; I y = 112 mb2, x = b a 2 + b2 , y =

a a 2 + b2 .All time derivatives of the angular velocities are zero, z = 0. Therefore x,y = 0.

z = xy/ (I y I x ) (0.137)9. *.

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Also see LL Ch. 5 Problems 1-3. Ch.9 Problems 1,2, Ch.14. Problem1, Ch. 18 Problam

1, Ch 21; Problems 1,2, Ch.27 Problam 1, Ch 51 Problem 1.

preguntas resueltas.pdf

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