Tài liệu Physics exercises_solution: Chapter 15 pdf

d For the waves to exist, the water level cannot be level horizontal, and the boat would tend to move along a wave toward the lower level, alternately in the direction of and opposed to

15.1: a) The period is twice the time to go from one extreme to the other, and

sm1.2

or s,m20.1s)(5.0m)00.6

is half the total vertical distance, 0.310m c) The amplitude does not affect the wave speed; the new amplitude is 0.150m d) For the waves to exist, the water level cannot be level (horizontal), and the boat would tend to move along a wave toward the lower level, alternately in the direction of and opposed to the direction of the wave motion

15.2: fv

1.5 10 Hz

m001.0

sm

10

00

3

6 8 3

e) x

15.7: a) f  v (8.00m s) (0.320m)25.0Hz,

m

rad19.6m)320.0()2(2

s,104.00Hz)0.25

cosm)0700.0()

s0.1550m))

(0.320m)

(0.360Hz)(5

0.251(

15.9: a) sin( ) 2 cos( )

2

2

ωt kx Ak

x

y ωt

kx Ak x

t

y ωt

kx Aω t

2

t x y t

y ω

x

y ωt

kx Ak

t

y ωt

kx Aω

2

t x y t

y ω

15.10: a) The relevant expressions are

ω t

v t

in the figure labeled 1-7.) (i) v yωAcos(0)ωA, and the particle is moving upward (in

the positive y-direction) a yω2Asin(0)0, and the particle is instantaneously not accelerating (ii)v y ωAcos(π 4)ωA 2, and the particle is moving up

,2)

4

ω

a y    and the particle is speeding up

(iii) v y ωAcos(π 2)0, and the particle is instantaneously at rest

,)

2

ω

a y    and the particle is speeding up

(iv) v y ωAcos(3π 4)ωA 2, and the particle is moving down

,2)

43



a y and the particle is instantaneously not accelerating

(vi) v y ωAcos(5π 4)ωA 2 and the particle is moving down

2)

45

ω

a y    and the particle is speeding up (v and y a have the y

same sign) (vii) v y ωAcos(3π 2)0, and the particle is instantaneously at rest

A ω π

A

ω

a y  2 sin(3 2) 2 and the particle is speeding up

(viii) v y ωAcos(7π 4)ωA 2, and the particle is moving upward

2)

47

15.11: Reading from the graph, a) A4.0mm, b)T 0.040s c) A displacement of 0.090 m corresponds to a time interval of 0.025 s; that is, the part of the wave represented

by the point where the red curve crosses the origin corresponds to the point where the

blue curve crosses the t-axis (y0) at t0.025s, and in this time the wave has traveled 0.090 m, and so the wave speed is 3.6m s and the wavelength is

m14.0)s040.0)(

t x π

2

vt x

π A

πv t

15.13: a) t 0:

x(cm) 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00 y(cm) 0.000 0.212 0.300 0.212 0.000 0.212 0.300 0.212 0.000

15.14: Solving Eq (15.13) for the force F,

  ((40.0Hz (0.750m)) 43.2

m2.50

kg120

15.15: a) Neglecting the mass of the string, the tension in the string is the weight of

the pulley, and the speed of a transverse wave on the string is

s

m3.16)

mkg0550.0(

)sm80.9)(

kg50.1

b) v f (16.3m s) (120Hz)0.136m c) The speed is proportional to the square root of the tension, and hence to the square root of the suspended mass; the answers change by a factor of 2, to23.1m sand0.192m

kg50.7(

)m0.14)(

kg800.0()

m Mg

L v

L t

15.18: a) The tension at the bottom of the rope is due to the weight of the load, and

the speed is the same 88.5m s as found in Example 15.4 b) The tension at the middle

of the rope is (21.0 kg (9.80m s2)205.8 N(keeping an extra figure) and the speed of the rope is 90.7m s c) The tension at the top of the rope

is(22.0kg)(9.80m s2)215.6m s and the speed is 92.9m s (See Challenge Problem (15.80) for the effects of varying tension on the time it takes to send signals.)

15.19: a)v F μ  (5.00N) (0.0500kg m)10.0m s

b)v f (10.0m s) (40.0Hz)0.250m

c)y(x,t) Acos(kxωt)(Note:y(0.0) A,asspecified.)

srad0.802

m;

rad00

)cm00

1

A ω μF

P 

3 (25.0N (2 (120.0Hz))2(1.6 10 3m)2

m0.80

kg1000.32

(sin)

,(0,

max 2

2

ω μF t

x P

The graph is shifted by T 4 but is otherwise the same The instantaneous power

is still never negative and 21 max, thesameasat 0

m W 0 1 m W 11 0 1

15.23: a) 2

2 2

2 1

1r Ι r

Ι 

2 2

2 2

2 1

mkg10504

1

av (2.50 10 kg m)(28.3N (742rad s))

g

W

39.0)m

15.26: a) The wave form for the given times, respectively, is shown.

b)

15.27: a) The wave form for the given times, respectively, is shown.

b)

15.28:

or 0sin

is, that );

1.15(Eq

k x

ω t

15.33: a) The amplitude of the standing wave is Asw 0.85cm,the wavelength is twice the distance between adjacent antinodes, and so Eq (15.28) is

)

cm0.302sin(

))s0.0752

sin((

)cm85.0(),

15.34: y1y2  A[cos(kxωt)cos(kxωt)]

 A[coskxcosωtsinkxsinωtcoskxcosωtsinkxsinωt]

2Asinkxsinωt

15.35: The wave equation is a linear equation, as it is linear in the derivatives, and

differentiation is a linear operation Specifically,

.)

x

y x

y x

y y x

x

y x

t

y t

x

y x

1

2 2

t

y t



)kg1000.3(

)m400.0)(

N800(

10 24.5,so the24 harmonic may be heard, but not the 25th

15.37: a) In the fundamental mode,

s

m0.96)m60.1)(

Hz0.60(so

andm60

22

0.50(SW max

v y

f) f3 7.96Hz3f1,so f1 2.65Hz is the fundamental

srad1332

Hz;

2.21

f

cm.rad0906.02

andcm3.69Hz)(21.2s)cm1470

rad([133sin)cm]

rad([0.0906sin

cm)60.5()

,

15.40: (a) (4.44mm) 2.22mm.(b) 2 32.52radm 0.193m.

2

1 SW 2

k π A

A

( c)f  2ω 7542rad m 120Hz.(d) v ω k  32754.5radradmm 23.2m s.(e) If the wave traveling in

the x direction is written as y1(x,t) Acos(kxt),then the wave traveling in the

mrad5.32and(a),frommm22.2 where),cos(

),(is

mrad98.6 mm;

60.4so

mm30

rad)sin([742m]

radsin([6.98mm)

60.4(



),2

(

3



L so this is the 3rd harmonic

c) For this 3rd harmonic, f ω 2π118Hz

Hz39.33Hz)118(so

15.43: a) The product of the frequency and the string length is a constant for a given

string, equal to half of the wave speed, so to play a note with frequency

cm

45.0Hz)(587Hz)(440cm)0.60

15.44: a) (i)x 2 is a node, and there is no motion (ii) x 4 is an antinode,

4 2

2 max max

max  A πf  πfA a  πf v  π f A π 

multiplies the results of (ii), so v π fA a π2f2A

max max  2 , 2 2 b) The amplitude is

2(iii),(ii) ,0)

15.45: a) 2 3.00 m, (248(1..050mm)) 16.0Hz

2 1

b)3 1 31.00m, f2 3f1 48.0Hz c)4 1 40.75m, f3 4f164.0Hz

15.46: a) For the fundamental mode, the wavelength is twice the length of the string,

and v f2fL2(245Hz)(0.635m)311m s b) The frequency of the fundamental mode is proportional to the speed and hence to the square root of the tension;

Hz

2461.01

2 



ω v π

k

b)y(x,t) Acos(kxωt)(2.50mm)cos(3.49rad m)x(126rad s)t

c)At x0,y(0,t)Acosωt(2.50mm)cos(126rad s)t From this form it may be seen that at 0, 0, 0

t y t

15.48: a) From comparison with Eq.15.4,A0.75cm,0.4002cm 5.00cm,

s

m25.6and

s00800.0Hz,

μ F

μ

L t L

t L

1

μ F

μ F

μ F

15.50: The amplitude given is not needed, it just ensures that the wave disturbance is

small Both strings have the same tension F, and the same length L1.5m.The wave takes different times t1and t2 to travel along each string, so the design requirements is

dt

dy

πfA Aω

v y,max  2

FL f

L m

F v

b) To double v y,max increase Fby a factor of 4

15.52: The maximum vertical acceleration must be at least g Because

ω g A A

μ g

A  

15.53: a) See Exercise 15.10; 2 ,

a t

πv πf

also contains a factor of μ,and so the effective spring constant per unit length is

independent of μ

15.54: a), b)

c) The displacement is a maximum when the term in parentheses in the denominator

is zero; the denominator is the sum of two squares and is minimized when xvt,and the

maximum displacement is A At x4.50cm,the displacement is a maximum at

s

102.25)sm(20.0m)10



maximum when (xvt)2  A2, or t(x A) v1.7510 3sand2.7510 3s

d) Of the many ways to obtain the result, the method presented saves some algebra and minor calculus, relying on the chain rule for partial derivatives Specifically, let

,)

u du

dg t

f du

dg t

u du

dg x

f g(u t x

3 2 2

2 2 3 2

2 2

2 2

3

)(

)3(2,

)(

2

u A

u A A x

y u

A

u A x

)3(2,

)(

2

2 2 2

2 2 3 2 2

2 2

2 2

3

u A

u A A v t

y u

A

u A v t

15.55: a) and b) (1): The curve appears to be horizontal, and v = 0 As the wave y

moves, the point will begin to move downward, and a y  0 (2): As the wave moves in

the + x -direction (to the right in Fig (15.34)), the particle will move upward so v y> 0 The portion of the curve to the left of the point is steeper, so a y > 0 (3) The point is moving down, and will increase its speed as the wave moves; v y < 0, a y< 0 (4) The curve appears to be horizontal, and v y = 0 As the wave moves, the point will move away from the x -axis, and a y>0 (5) The point is moving downward, and will increase its speed as the wave moves; v y 0,a y 0.(6) The particle is moving upward, but the curve that represents the wave appears to have no curvature, so v y  anda y 0 c) The accelerations, which are related to the curvatures, will not change The transverse

velocities will all change sign

15.56: (a ) The wave travels a horizontal distance d in a time

0.600m40.0Hz 0.333s.

m 00

d t

(b) A point on the string will travel a vertical distance of 4A each cycle

Although the transverse velocity v y x,t is not constant, a distance of h8.00m

corresponds to a whole number of cycles,

,400))m1000.5(4((

)m00.8(

b)v y dy dt Aωsin(kxωt),v z dz dtAωcos(kxωt)

v v2yv z2  Aω, so the speed is constant



 is tangent to the circular path

c) a y dv y dtAω2cos(kxω t),a z dv z dtAω2sin(kxω t)

r a a

and180so



 ra   is opposite in direction to a r 

; is radially inward

y2 z2  A2, so the path is again circular, but the particle rotates in the opposite sense compared to part (a )

15.58: The speed of light is so large compared to the speed of sound that the travel time

of the light from the lightning or the radio signal may be neglected Them, the distance from the storm to the dorm is (344m s)(4.43s)1523.92m and the distance from the storm to the ballpark is (344m s)(3.00s)1032m The angle that the direction from the storm to the ballpark makes with the north direction is found from these distances using the law of cosines;

)m1120()m1032(2

)m1120()m1032()m1523.92(

so the storm can be considered to be due west of the park

15.59: a) As time goes on, someone moving with the wave would need to move in such

a way that the wave appears to have the same shape If this motion can be described by

with

c

vt

x  a constant (not the speed light),then y(x,t) f(c),

and the waveform is the same to such observer b) See Problem 15.54 The derivation is completed by taking the second partials,

,,

1

2

2 2

2 2

2 2 2

2

du

f d t

y du

f d v x

)

,

(

soy x t  f tx v is a solution to the wave equation with wave speed v c) This is

of the form y(x,t) f(u), with utx vand

2

2 ( ( ) ))

(u De C t B c x

and the result of part (b) may be used to determine the speed vC Bimmediately

0,

cos     d) To identify  uniquely, the quadrant in which  is must be known In physical terms, the signs of both the position and velocity, and the magnitude

of either, are necessary to determine  (within additive multiples of π)

15.61: a) μF F μ F F vF k ωand substituting this into Eq (15.33) gives the result

b) Quadrupling the tension for F to F4Fincreases the speed v F μ by a

factor of 2, so the new frequency ωand new wave number kare related to

k ω k

ω

k

ωand by (  )2( ) For the average power to be the same, we must have

4and

4so

L vt x L

vt x L h

vt x L L

vt x L h

L vt x t

x

y

)( for

0

)(0for )

(

0)(for

)(

)(for 0),(

L vt x L

h Fv L hv L h F

vt x L L

h Fv L hv L h F

L vt x F

t

t x y x

t x y F

t

x

P

)(for

0)0)(

0(

)(0 for)())(

(

0)(for

)())(

(

)(for 0)0)(

0()

,(),()

,

(

2 2

Thus the instantaneous power is zero except for L(xvt)L, where it has the constant value Fv(h L)2

15.63: a) 2 2

2 1

15.64: a) , d)

b)The power is a maximum where the displacement is zero, and the power is a minimum of zero when the magnitude of the displacement is a maximum c) The direction of the energy flow is always in the same direction d) In this case,

),sin(kx ωt

,(x t Fk A2 2 kx ωt

The power is now negative (energy flows in the  -direction), but the qualitative x

relations of part (b) are unchanged

μ

Τ YΑ F μ

F v μ

F

2 1 2 1

v v Τ μ) Α Y

v v

)((

2 2

2 1

2 2 2 1

15.66: (a) The string vibrates through 1 cycle in 2 450001 min,so

s109.6min106.1min

5000

42

f 1T 19.610 2s10.4Hz   L50.0cm0.50m (b) Second harmonic

at maximum speed, sinkxsinωt1

cm)Hz)(1.5(10.4

2cm)5.1(2cm)

sm2.5(

m)N)(0.50000

.1

2 2

=18.5 g

15.67: There is a node at the post and there must be a node at the clothespin There

could be additional nodes in between The distance between adjacent nodes is  2, so the distance between any two nodes isn( 2)for n1,2,3,

,),

2(cm

cm0.90(

f

15.68: (a) The displacement of the string at any point is y(x,t)(ASWsinkx)sint For the fundamental mode 2L, so at the midpoint of the string

,1)2)(

2

sin(

sinkx π  L  and

.sin

s,rad102.21s)m80.3()sm1040.8

),(

SW 2

2 2

2

ωt kx

A t t

t x y

),(

SW 2

2

ωt kx A

t

ω t

t x y

sw

2 2

2

ω kx Α

ω x

t x

y 



The displacement of the harmonic oscillator is periodic in both time and space.b) Yes, the travelling wave is also a solution of the wave equation

15.70: a) The wave moving to the left is inverted and reflected; the reflection means that

the wave moving to the left is the same function of x,and the inversion means that the function is f ( x ).More rigorously, the wave moving to the left in Fig (15.17) is

obtained from the wave moving to the right by a rotation of 180, so both the coordinates

)

and

(f x have their signs changed b) The wave that is the sum is f(x) f(x) (an inherently odd function), and for any f,f(0)  f(0)0 c) The wave is reflected but not inverted (see the discussion in part (a) above), so the wave moving to the left in Fig (15.18) is  f ( x )

d)

dx

x d x d

x df dx

x df dx

x df dx

x df x f x f dx

d dx

)(

)()()()())()(

df dx

df







At x0, the terms are the same and the derivatives is zero (See Exercise 20-2 for a

situation where the derivative of f is not finite, so the string is not always horizontal at the

cm)(0.400(ii)

node),(a0)(cm)sin

sin(

cm)(0.400

(iii)

s

1008

1 f    3 c) In each case, the maximum velocity is the amplitude

multiplied by ω 2 πf and the maximum acceleration is the amplitude multiplied by

.sm106.43,sm4.27(iii)

;sm109.10 s,m6.03(ii) 0;

0,(i)

15.73: The plank is oscillating in its fundamental mode, so2L10.0m,with a frequency of 2.00Hz.a)v  f20.0m s b) The plank would be its first overtone, with twice the frequency, or 4jumps s.

15.74: (a) The breaking stress is 7.0 108 N m2,

2 8





π r

The mass and density are fixed, πr2L,

M

ρ so the minimum radius gives the maximum

) m kg 7800 ( m) 10 (6.4

kg 10 0 4

3 2

4 3

15.75: a) The fundamental has nodes only at the ends, x0andxL b) For the second harmonic, the wavelength is the length of the string, and the nodes are at

.and2