Homework 3 Solutions 1. - blogs.umass.edu · Homework 3 Solutions 84 1. CHAPTER 3 3-5. Diffe...

6
1 Homework 3 Solutions 1. 3-5. Differentiating the equation of motion for a simple harmonic oscillator, 0 sin x A t ω = (1) we obtain 0 0 cos x A t t ω ω ' = ' (2) But from (1) 0 sin x t A ω = (3) Therefore, ( ) 2 0 cos 1 t x ω = A (4) and substitution into (2) yields 2 2 0 x t A x ω ' '= (5) Then, the fraction of a complete period that a simple harmonic oscillator spends within a small interval 'x at position x is given by 2 2 2 2 0 2 t x x A x A W ωW π ' ' ' = = x (6) A 1 A 2 A 3 A 3 A 2 A 1 t / τ x This result implies that the harmonic oscillator spends most of its time near x = ±A, which is obviously true. On the other hand, we obtain a singularity for t W ' at x = ±A. This occurs because at these points x = 0, and (2) is not valid.

Transcript of Homework 3 Solutions 1. - blogs.umass.edu · Homework 3 Solutions 84 1. CHAPTER 3 3-5. Diffe...

Page 1: Homework 3 Solutions 1. - blogs.umass.edu · Homework 3 Solutions 84 1. CHAPTER 3 3-5. Diffe rentiating th e equation of motion for a simple har m onic oscill ator, xA sin Z 0 t (1)

1

Homework 3 Solutions

1.84 CHAPTER 3

3-5. Differentiating the equation of motion for a simple harmonic oscillator,

0sinx A tω= (1)

we obtain

0 0cosx A t tω ω= (2)

But from (1)

0sinx

tA

ω = (3)

Therefore,

( )20cos 1t xω = − A (4)

and substitution into (2) yields

2 2

0

xt

A xω=

− (5)

Then, the fraction of a complete period that a simple harmonic oscillator spends within a small interval x at position x is given by

2 2 2 2

0 2

t x x

A x Aω π= =

− − x (6)

–A1–A2–A3 A3A2A1

∆t ⁄τ

x

This result implies that the harmonic oscillator spends most of its time near x = ±A, which is obviously true. On the other hand, we obtain a singularity for t at x = ±A. This occurs because at these points x = 0, and (2) is not valid.

3-6.

x1 x2

k

x

m1 m2

Suppose the coordinates of m and are and x and the length of the spring at equilibrium is . Then the equations of motion for m and are

1 2m 1x 2

1A 2m

( )1 1 1 2m x k x x= − − +�� A (1)

( )2 2 2 1m x k x x= − − +�� A (2)

Page 2: Homework 3 Solutions 1. - blogs.umass.edu · Homework 3 Solutions 84 1. CHAPTER 3 3-5. Diffe rentiating th e equation of motion for a simple har m onic oscill ator, xA sin Z 0 t (1)

2

90 CHAPTER 3

Substitution of (12) into (6) yields

( ) ( )

( ) ( )

( )

0 0

0 0

0

1 12

2

cos cos

i t i t i t i t

i t t i t ti t i t

Fe e e e

k

Fe e e e

k

Ft t t

k

ω ω ω ω

ω ωω ω

ω ω

− −+

− − − −

= − + −

= − + −

= − − (13)

Thus,

( )0 0cos cos ; F

t t t t tk

ω ω− = − − 0x x (14)

3-10. The amplitude of a damped oscillator is expressed by

( ) ( )1costx t Ae tβ ω δ−= + (1)

Since the amplitude decreases to 1 after n periods, we have e

1

21nT n

πβ β

ω= = (2)

Substituting this relation into the equation connecting 1ω and 0ω (the frequency of undamped oscillations), 2 2

1 02ω ω β= − , we have

2

2 2 210 1 1 2 2

11

2 4n nω

ω ω ωπ π

= + = + (3)

Therefore,

1 2

12 2

0

11

4 nωω π

= + (4)

so that

12 2

2

11

8 nωω π

3-11. The total energy of a damped oscillator is

( ) ( ) ( )21 12 2

E t mx t kx t= +� 2 (1)

where

( ) ( )1costx t Ae tβ ω δ−= − (2)

( ) ( ) ( )1 1 1cos sintAe t tβx t β ω δ ω ω δ−= − − − −� (3)

2.

Page 3: Homework 3 Solutions 1. - blogs.umass.edu · Homework 3 Solutions 84 1. CHAPTER 3 3-5. Diffe rentiating th e equation of motion for a simple har m onic oscill ator, xA sin Z 0 t (1)

3

88 CHAPTER 3

2 sin2

4 cos2

ds dv a

dt dt

da

dt

φ φ

φ

= =

= − (8)

from which

2

24 cos2

dv a

dtφ

= −� (9)

Letting cos2

be the new variable, and substituting (7) and (9) into (1), we have

4maz mgz− =�� (10)

or,

04g

z za

+ =�� (11)

which is the standard equation for simple harmonic motion,

20 0z zω+ =�� (12)

If we identify

0g

ω =A

(13)

where we have used the fact that . 4a=A

Thus, the motion is exactly isochronous, independent of the amplitude of the oscillations. This fact was discovered by Christian Huygene (1673).

3-9. The equation of motion for 00 t t is

( ) ( )0mx k x x F kx F kx= − − + = − + +�� 0 (1)

while for , the equation is 0t t

( )0mx k x x kx kx= − − = − +�� 0 (2)

It is convenient to define

0x x= −

which transforms (1) and (2) into

m k F= − +�� ; 00 t t (3)

m k= −�� ; (4) 0t t

3.

OSCILLATIONS 89

The homogeneous solutions for both (3) and (4) are of familiar form ( ) i t i tt Ae Beω ω−= + , where

k mω = . A particular solution for (3) is F k= . Then the general solutions for (3) and (4) are

i t i tFAe Be

kω ω−

− = + + ; 00 t t (5)

i t i tCe Deω ω−+ = + ; t (6) 0t

To determine the constants, we use the initial conditions: ( ) 00x t x= = and x(t = 0) = 0. Thus,

( ) ( )0 0t t− − 0= = = =� (7)

The conditions give two equations for A and B:

( )

0

0

FA B

k

i A Bω

= + +

= −

(8)

Then

2F

A Bk

= = −

and, from (5), we have

( )0 1 cosF

x x tk

ω− = − = − ; 00 t t (9)

Since for any physical motion, x and must be continuous, the values of x� ( )0t t− = and

( )0t t− =� are the initial conditions for ( )t+ which are needed to determine C and D:

( ) ( )

( )

0 0

0 0

0 0

0 0

1 cos

sin

i t i t

i t i t

Ft t t Ce De

k

Ft t t i Ce De

k

ω ω

ω ω

ω

ω ω ω

−+

−+

= = − = +

= = = −� (10)

The equations in (10) can be rewritten as:

( )0 0

0 0

0

0

1 cos

sin

i t i t

i t i t

FCe De t

k

iFCe De t

k

ω ω

ω ω

ω

ω

+ = −

−− =

(11)

Then, by adding and subtracting one from the other, we obtain

( )

( )

0 0

0 0

12

12

i t i t

i t i t

FC e e

k

FD e e

k

ω ω

ω ω

= −

= −

(12)

Page 4: Homework 3 Solutions 1. - blogs.umass.edu · Homework 3 Solutions 84 1. CHAPTER 3 3-5. Diffe rentiating th e equation of motion for a simple har m onic oscill ator, xA sin Z 0 t (1)

4

OSCILLATIONS 89

The homogeneous solutions for both (3) and (4) are of familiar form ( ) i t i tt Ae Beω ω−= + , where

k mω = . A particular solution for (3) is F k= . Then the general solutions for (3) and (4) are

i t i tFAe Be

kω ω−

− = + + ; 00 t t (5)

i t i tCe Deω ω−+ = + ; t (6) 0t

To determine the constants, we use the initial conditions: ( ) 00x t x= = and x(t = 0) = 0. Thus,

( ) ( )0 0t t− − 0= = = =� (7)

The conditions give two equations for A and B:

( )

0

0

FA B

k

i A Bω

= + +

= −

(8)

Then

2F

A Bk

= = −

and, from (5), we have

( )0 1 cosF

x x tk

ω− = − = − ; 00 t t (9)

Since for any physical motion, x and must be continuous, the values of x� ( )0t t− = and

( )0t t− =� are the initial conditions for ( )t+ which are needed to determine C and D:

( ) ( )

( )

0 0

0 0

0 0

0 0

1 cos

sin

i t i t

i t i t

Ft t t Ce De

k

Ft t t i Ce De

k

ω ω

ω ω

ω

ω ω ω

−+

−+

= = − = +

= = = −� (10)

The equations in (10) can be rewritten as:

( )0 0

0 0

0

0

1 cos

sin

i t i t

i t i t

FCe De t

k

iFCe De t

k

ω ω

ω ω

ω

ω

+ = −

−− =

(11)

Then, by adding and subtracting one from the other, we obtain

( )

( )

0 0

0 0

12

12

i t i t

i t i t

FC e e

k

FD e e

k

ω ω

ω ω

= −

= −

(12)

90 CHAPTER 3

Substitution of (12) into (6) yields

( ) ( )

( ) ( )

( )

0 0

0 0

0

1 12

2

cos cos

i t i t i t i t

i t t i t ti t i t

Fe e e e

k

Fe e e e

k

Ft t t

k

ω ω ω ω

ω ωω ω

ω ω

− −+

− − − −

= − + −

= − + −

= − − (13)

Thus,

( )0 0cos cos ; F

t t t t tk

ω ω− = − − 0x x (14)

3-10. The amplitude of a damped oscillator is expressed by

( ) ( )1costx t Ae tβ ω δ−= + (1)

Since the amplitude decreases to 1 after n periods, we have e

1

21nT n

πβ β

ω= = (2)

Substituting this relation into the equation connecting 1ω and 0ω (the frequency of undamped oscillations), 2 2

1 02ω ω β= − , we have

2

2 2 210 1 1 2 2

11

2 4n nω

ω ω ωπ π

= + = + (3)

Therefore,

1 2

12 2

0

11

4 nωω π

= + (4)

so that

12 2

2

11

8 nωω π

3-11. The total energy of a damped oscillator is

( ) ( ) ( )21 12 2

E t mx t kx t= +� 2 (1)

where

( ) ( )1costx t Ae tβ ω δ−= − (2)

( ) ( ) ( )1 1 1cos sintAe t tβx t β ω δ ω ω δ−= − − − −� (3)

Page 5: Homework 3 Solutions 1. - blogs.umass.edu · Homework 3 Solutions 84 1. CHAPTER 3 3-5. Diffe rentiating th e equation of motion for a simple har m onic oscill ator, xA sin Z 0 t (1)

5

NEWTONIAN MECHANICS—SINGLE PARTICLE 71

The equilibrium point (where 0ddU θ = ) that we wish to look at is clearly θ = 0. At that point,

we have ( )2 2 2d U d mg R bθ = − , which is stable for 2R b and unstable for R b . We can

use the results of Problem 2-46 to obtain stability for the case

2

2R b= , where we will find that the first non-trivial result is in fourth order and is negative. We therefore have an equilibrium at θ = 0 which is stable for 2R b and unstable for 2R b .

2-43. 3 2F kx kx α= − +

( )4

22

1 12 4

xU x F dx kx k

α= − = −∫

To sketch U(x), we note that for small x, U(x) behaves like the parabola 212

kx . For large x, the

behavior is determined by 4

2

14

xkα

U(x)

E0

E1

E2

E3 = 0

E4

x1 x2 x3

x4 x5 x

( )212

E mv U x= +

For E , the motion is unbounded; the particle may be anywhere. 0E=

For E (at the maxima in U(x)) the particle is at a point of unstable equilibrium. It may remain at rest where it is, but if perturbed slightly, it will move away from the equilibrium.

1E=

What is the value of ? We find the x values by setting 1E 0dUdx

= .

3 20 kx kx α= −

x = 0, ± α are the equilibrium points

( ) 2 21

1 1 12 4 4

U E k k k 2α α α± = = − = α

For E , the particle is either bounded and oscillates between 2E= 2x− and ; or the particle comes in from ± to ± and returns to ± .

2x

3x

4.

72 CHAPTER 2

For E , the particle is either at the stable equilibrium point x = 0, or beyond . 3 0= 4x x= ±

For E , the particle comes in from ± to 4 5x± and returns.

2-44.

m1

T

m1g

m2

T T

m2g

θ

From the figure, the forces acting on the masses give the equations of motion

11 1m m gx T= −�� (1)

22 2 2 cosm m g Tx θ= −�� (2)

where is related to by the relation 2x 1x

( )2

1 22 4

b xx

−d= − (3)

and ( )1cos 2d b xθ = − . At equilibrium, 1 2 0x x= =�� �� and T m1g= . This gives as the equilibrium values for the coordinates

110 2

1 2

4

4

m dx b

m m= −

− 2 (4)

220 2

1 24

m dx

m m=

− 2 (5)

We recognize that our expression is identical to Equation (2.105), and has the same requirement that

10x

2 1 2m m for the equilibrium to exist. When the system is in motion, the descriptive equations are obtained from the force laws:

( )2 1

112

( ) (4

m b xm gx

x−

2 )gx− =�� �� −

0

(6)

To examine stability, let us expand the coordinates about their equilibrium values and look at their behavior for small displacements. Let 1 1 1x x− and 2 2 2x x 0− . In the calculations, take terms in 1 and 2 , and their time derivatives, only up to first order. Equation (3) then becomes 2 1 2(m m 1)−� . When written in terms of these new coordinates, the equation of motion becomes

( )

( )

3 22 21 2

111 2 1 2

4

4

g m m

m m m m d

−= −

+�� (7)

Page 6: Homework 3 Solutions 1. - blogs.umass.edu · Homework 3 Solutions 84 1. CHAPTER 3 3-5. Diffe rentiating th e equation of motion for a simple har m onic oscill ator, xA sin Z 0 t (1)

6

NEWTONIAN MECHANICS—SINGLE PARTICLE 75

when 2=v c , we have 0.55 year10 3

ct = =

when v = 99% c, we have 99

6.67 years10 199

c= =t

2-51.

a) 2 02

0

( )mvdv dv b

m bv dt v tdt v m btv m

= − = − =+∫ ∫

Now let v(t) = v0/1000 , one finds 0

999138.7 hours

mt

v b= = .

t

v

b) 0

0

( ) lnt btv mm

x t vdtb m

+= =∫

We use the value of t found in question a) to find the corresponding distance

( ) ln(1000) 6.9 kmm

x tb

= =

2-52.

a) 2

02 2

4( ) 1

U xdU xF x

dx a a= − = − −

b)

x

U

When F = 0, there is equilibrium; further when U has a local minimum (i.e. 0dF dx ) it is stable, and when U has a local maximum (i.e. 0dF dx ) it is unstable.

5.

76 CHAPTER 2

So one can see that in this problem x = a and x = –a are unstable equilibrium positions, and x = 0 is a stable equilibrium position.

c) Around the origin, 0 02 2

4 4U x UkF kx

a mω− − = =

ma

d) To escape to infinity from x = 0, the particle needs to get at least to the peak of the potential,

2

0minmax 0 min

22

UmvU U v

m= = =

e) From energy conservation, we have

2 22 2

0 0min2 2

21

2 2U x Umv dx x

va dt m

+ = = = −mv

a

We note that, in the ideal case, because the initial velocity is the escape velocity found in d), ideally x is always smaller or equal to a, then from the above expression,

022

20 00 0

2 2

8exp 1

ln ( )2 8 81 exp 1

xU

a tmam dx ma a x

tU U a xx U

ta ma

−+

= = =−

− +∫t x

t

x

2-53.

F is a conservative force when there exists a non-singular potential function U(x) satisfying F(x) = –grad(U(x)). So if F is conservative, its components satisfy the following relations

yxFF

y x

∂∂=

∂ ∂

and so on.

a) In this case all relations above are satisfied, so F is indeed a conservative force.

2

1( , )2x

U bxayz bx c U ayzx cx f y z

x∂

= + + = − − − +∂

F = − (1)

where is a function of only y and z 1( , )f y z

2( , )yU

axz bz U ayzx byz f x zy

∂= + = − − +

∂F = − (2)