Tài liệu Physics exercises_solution: Chapter 13 ppt

13.6: The period will be twice the time given as being between the times at which the glider is at the equilibrium position see Fig... a The mass would decrease by a factor of 1 33 127

13.2: a) Since the glider is released form rest, its initial displacement (0.120 m) is the

amplitude b) The glider will return to its original position after another 0.80 s, so the period is 1.60 s c) The frequency is the reciprocal of the period (Eq (13.2)),

0    and the angular frequency is



 T π

ω 2 5.53103rad s

13.4: (a) From the graph of its motion, the object completes one full cycle in 2.0 s; its

period is thus 2.0 s and its frequency 1period0.5s 1 (b) The displacement varies from 0.20m to0.20m, so the amplitude is 0.20 m (c) 2.0 s (see part a)

13.5: This displacement is 41 of a period

T 1 f 0.200s,sot0.0500s

13.6: The period will be twice the time given as being between the times at which the

glider is at the equilibrium position (see Fig (13.8));

m

N292.0kg)200.0(s)60.2(2

2kg)600.0(

2 2

13.9: From Eq (13.12) and Eq (13.10), 2 0.375s, 1 2.66Hz,

m N 140 kg 500

t x x ω β ωt A ω

a d dt x

x    is a solution to Eq (13.4) if

ω A a

ω2 m k.b) 2 a constant, so Eq (13.4) is not satisfied c) i(ωt β),

13.11: a) x(3.0mm)cos((2π)(440Hz)t) b)(3.010 3m)(2π)(440Hz)8.29m s,

),Hz))(440sin((2)sm1034.6()(c) .sm1029.2Hz)440()

v ω v

m)(0.383

,58.5rad02.1kgN/m/2.00300

m)(0.200

m/s)4.00(

.rad)715.0rad/s)((15.7cos)cm/s359(

rad)715.0rad/s)((15.7sin )scm9.22(

rad)715.0rad/s)((15.7coscm)46.1(

t v

t x

x x

13.15: The equation describing the motion is xA sin ωt;this is best found from either inspection or from Eq (13.14) (Eq (13.18) involves an infinite argument of the

arctangent) Even so, x is determined only up to the sign, but that does not affect the

result of this exercise The distance from the equilibrium position is

 

2  0.600m sin 4 5 0.353m

A

13.16: Empty chair: T 2π m k

N/m993s)

(1.30

kg)5.42(44

2

2 2

With person in chair:

kg120kg5.42kg162

kg1624

N/m)993(s)54.2(42

person

2

2 2

k T m

k m π T

13.17: T 2π m k,m0.400kg

s09.22

N/m60.3m

300.0

)m/s70.2kg)(

400.0(gives

:calculate to

m/s70

2

Use

2 2

π

T

x

ma k

ma

kx

k a

x x

cm/s60.31

π ω ωA

scm60.3scm60.3

1 



ω A

max ω A(4.71s ) (0.764cm)16.9cm s

a

13.19: a)x(t)(7.40cm)cos((4.16rad s)t2.42rad)

2 2

2 2

2 2 max

max

2 2 1 2

50.10.26(

.sm0.303

is

Speed

sm303.0sm)0125.0()0740.0(50.10.26

m0125.0givess00.1

at evaluated

)

(

e)

N92.1so

d)

sm308.0gives

m0740.0cm

2

b)

s1.51so

2s)rad(4.16

m

k

v

x t

t

x

kA F

kx

F

m k A v

kA kx

mv

A

T π m k k m

π

T

T π T T

t

13.20: See Exercise 13.15;

s

0.0871))

(20.36))(

1.5arccos(

ω

v A

x

Squaring and adding,

,2 2

2 0 2

12

12

12

12

0

2 0 2

2 0

2 0

ω

k kx

max max

1mv  kx in Eq (13.21) and solving for x gives x A2

Eliminating x in favor of v with the same relation gives 2 2 ωA2

v   b) This happens four times each cycle, corresponding the four possible combinations of + and –

in the results of part (a) The time between the occurrences is one-fourth of a period or

8 4

3 4

1 2

4

,

,c)

4

13.24: a) From Eq (13.23),

m/s

1.20m)040.0(kg0.500

mN450

m

k v

b) From Eq (13.22),

m/s

11.1m)015.0(m)040.0(kg0.500

m)N/m)(0.040450

d) From Eq (13.4), 13.5m/s2

kg) (0.500

m) 0.015 N/m)(

2 2

2 max ω A (2 f) A 2 (0.85Hz) (18.0 10 m) 5.13m/s v

m/s961.0

2 2 2

2

2 2

c) The fraction of one period is (1 2π arcsin ) (12.018.0), and so the time is

1kA  mv  kx We are given amplitude, frequency in Hz, and various values of x We could calculate velocity

from this information if we use the relationship k mω2 4π2f2and rewrite the

conservation equation as 2

2

1 4

2 1 2 2

1

2 2 2

x

A  π v f  Using energy principles is generally a good approach when we are dealing with velocities and positions as opposed to accelerations and time when using dynamics is often easier

13.26: In the example, A2 A1 M Mm and now we want So21 ,

1 2

1

1 4

1 2 1 2

1

2 A,E E ,or E

heat

13.27: a) 2 0.0284 J

2 1 2 2

kg)(0.150N/m)

300(

m/s)300.0(m)

012.0(

2 2

2

2 0 2

ω

v x

m600.0)(

ω a

ωt ωA v

ωt A

x

Using the displacement and acceleration equations:

2 2

2 sin (  )(3.742s )(2.20m s)8.232m s

Next square both this new equation and the acceleration equation and add them:

m840.0

m7054.0)s742.3(

sm3.138s

m3.138

sm3.138s

m56.70sm77.67

)(

cos)(

sin

)sm40.8()sm232.8()(

cos)

(

sin

2 4

1

4 2 4

4 2 2

4 2 4

2 4

2 2

4

2 2

2 4

2 2 2

2 2

2 4 2

A ω

ωt ωt

A ω

ωt A

ω ωt

The object will therefore travel0.840m0.600m0.240mto the right before stopping

at its maximum amplitude

13.29: vmax A k m

sm509.0Then

m

0405.0)(so

:find

to

Use

s158)

2(so

2

:find

max

2 2

v

m k a A m

kA

a

A a

T π m k k

m

π

T

m k T

13.30: Using

0

0

L F

k  from the calibration data,

kg

00.6Hz))

60.2((2

m)10(1.25N)200()2(

)(

2

1 2

0

π πf

L F m

m)120.0(

)sm(9.80kg)650(Δ

sm9.80

m120.022

g

l π k

m π

13.32: a) At the top of the motion, the spring is unstretched and so has no potential

energy, the cat is not moving and so has no kinetic energy, and the gravitational potentialenergy relative to the bottom is 2mgA2(4.00kg)(9.80m/s2)(0.050m)3.92 J.This is the total energy, and is the same total for each part

b) Ugrav 0,K 0,soUspring 3.92 J

c) At equilibrium the spring is stretched half as much as it was for part (a), and so

J 98.0soandJ,1.96J)92.3(J,

0.98J)92

3

grav 4

mg k

w l

13.34: See Exercise 9.40 a) The mass would decrease by a factor of (1 3)3 127and so the moment of inertia would decrease by a factor of (1 27)(1 3)2 (1 243), and for the same spring constant, the frequency and angular frequency would increase by a factor of

6

15

243 b) The torsion constant would need to be decreased by a factor of 243, or

changed by a factor of 0.00412 (approximately)

13.35: a) With the approximations given, I mR2 2.7210 8kgm2,

)m)10kg)(2.2010

00.2)(

21((

s00.122

5

2 2 3

2 2

13.37:

2 (125) (265s) 0.0152kg m .

m/radN

450.0)2(

2 2





π πf

I

13.38: The equation θcos(ωtφ)describes angular SHM In this problem, φ0

2 2

t ω ω

t ω ω

dt θ d dt

b) When the angular displacement is ,cos(ω t), and this occurs at t0,so

1

cos(0)since,and

0,sin(0) since

since,2and

,2

3)sin(

since2

ω

ω dt

dθ

This corresponds to a displacement of 60

13.39: Using the same procedure used to obtain Eq (13.29), the potential may be

expressed as

]

)1

(2)1

0

12 0 0

2 0 0

0

2

766

1

2

2

131212

1

R x R

x

R x R

x U

36

R U

122

13.41: T 2 L g, so for a different acceleration due to gravity ,g

13.42: a) To the given precision, the small-angle approximation is valid The highest

speed is at the bottom of the arc, which occurs after a quarter period, T4 2 g L 0.25s b) The same as calculated in (a), 0.25 s The period is independent of amplitude

13.43: Besides approximating the pendulum motion as SHM, assume that the angle

is sufficiently small that the length of the spring does not change while swinging in thearc Denote the angular frequency of the vertical motion as 0  m k  kg and L g

 which is solved for L 4 w k But L is the length of the stretched

spring; the unstretched length is L0 Lw k3w k 31.00N 1.50N/m2.00m

2

2 2

13.46: From the parallel axis theorem, the moment of inertia of the hoop about the nail is

13.39.Eq

in with

,22so,

8 2

gT π

R

13.47: For the situation described, I mL2 andd L in Eq (13.39); canceling the factor

of m and one factor of L in the square root gives Eq (13.34).

13.48: a) Solving Eq (13.39) for I,

1.80kg 9.80m s  0.250m 0.0987kg m 2

s940.02

2 2

2 2

max

I

Expressing maxin terms of the period of small-angle oscillations, this becomes

s940.0

22cos

1

22

2 2

13.49: Using the given expression for I in Eq (13.39), with d=R (and of course m=M),

s

58.03

2

2 2

T mgd

mN50.2

b) b2 km2 2.50N m0.300kg1.73kg s

13.52: From Eq (13.42)A2 A1exp2b m t.Solvingfor b,

s

kg0220.0m100.0

m300.0ln)s00.5(

)kg050.0(2ln

m b

As a check, note that the oscillation frequency is the same as the undamped frequency to

valid

is(13.42)Eq

13.53: a) With 0,x(0) A.

b) cos sin ,

2

) 2 (

2 2

2 ) 2 (

m

b Ae

2 2

b A ω

m

b A

a x

(Note that this is (bv0 kx0) m.) This will be negative if

.2if

positiveand

2if

zero

,

be curved down, not curved, or curved up, respectively

13.54: At resonance, Eq (13.46) reduces to AFmax bd a) A31 b)2A1 Note that

the resonance frequency is independent of the value of b (see Fig (13.27)).

13.55: a) The damping constant has the same units as force divided by speed, or

kgm s2   m s  kg s. b)The units of km are the same as [[kg s2][kg]] 2 kg s,the same as those for c) 2 (i) d 0.2 , so max 0.2  5 max

ω

,5

.2)4.0(so

,0.4

ii)

( bd  k AFmax k  Fmax k as shown in Fig.(13.27)

13.56: The resonant frequency is

Hz,22.2srad139)kg108)mN102.1



m k

and this package does not meet the criterion

13.57: a)

.sm1072.6min

rev

srad30)minrev3500(2

m100

srad30)m05(

)minrev3500( c)N

1002

kg45)(

2 1 2

2

 mv

and so 2, thus 2 (3500rev min)( ) 3503 rad s,

min rev s rad

ω ω π

t π

W

101.76s)(J6.75

or ,Then

2(350)3)

13.58: Denote the mass of the passengers by m and the (unknown) mass of the car by M

The spring cosntant is then k mg l The period of oscillation of the empty car is

2 2

L E

2 2

E L

g

l π T k

m M π T

13.59: a) For SHM, the period, frequency and angular frequency are independent of

amplitude, and are not changed b) From Eq (13.31), the energy is decreased by a factor

of 14 c) From Eq (13.23), the maximum speed is decreased by a factor of 21 d) Initially, the speed at A1 4 was 415ωA1; after the amplitude is reduced, the speed is

unchanged From the result of part (d), the kinetic energy is decreased by a factor of 51

13.60: This distance LisLmg k; the period of the oscillatory motion is

,22

g

L k

m π

which is the period of oscillation of a simple pendulum of lentgh L.

13.61: a) Rewriting Eq (13.22) in terms of the period and solving,

s

68.1

b) Using the result of part (a),

m

0904.02

x

c) If the block is just on the verge of slipping, the friction force is its maximum,

.s

sn μ mg

μ

f   Setting this equal to mamA2π T2givesμ s  A2π T2 g0.143

13.62: a) The normal force on the cowboy must always be upward if he is not holding on

He leaves the saddle when the normal force goes to zero (that is, when he is no longer in contact with the saddle, and the contact force vanishes) At this point the cowboy is in free fall, and so his acceleration is g; this must have been the acceleration just before

he left contact with the saddle, and so this is also the saddle’s acceleration

b) xa (2π f)2 (9.80m s2) 2π(1.50Hz))2 0.110m c) The cowboy’s speed will

be the saddle’s speed, v(2πf) A2x2 2.11m s d) Taking t 0 at the time when the cowboy leaves, the position of the saddle as a function of time is given by Eq

(13.13), withcos 2 ;

A ω

x    Finding the time at which the cowboy and the saddle are again in contact involves a transcendental equation which must be solved numerically; specifically,

rad),11.1s)rad42.9((

cosm)25.0()sm90.4(s)m11.2(m)

13.63: The maximum acceleration of both blocks, assuming that the top block does not

slip, is amax kA (mM), and so the maximum force on the top block is

 m mM kA μsmg,andso themaximumamplitudeisAmax μs(mM)g k

13.64: (a) Momentum conservation during the collision: mv0 (2m)V

sm00.1s)m00.2(2

12

2

12

1

kx

)(amplitudem

500.0m

N0.80

s)m00.1kg)(

0.20

πf

Hz318.0kg0.20

mN0.802

12

k

π

f

s14.3Hz318

c) In example 13.5, the mass increased This means that T increases rather than decreases When the mass is added at x0,the energy and amplitude change When the mass is added at xA,the energy and amplitude remain the same This is the same as

in this problem

13.66: a)

For space considerations, this figure is not precisely to the scale suggested in the problem The following answers are found algebraically, to be used as a check on the graphical method

b) 0.200m

N/m)(10.0

J)200.0(2



k

E A

K

x

429.00

0 2

and  arctan  0.4290.580rad

13.67: a) The quantity l is the amount that the origin of coordinates has been moved from the unstretched length of the spring, so the spring is stretched a distance lx (see Fig (13.16 ( c ))) and the elastic potential energy is 2

1

0 2

2 0

1

0 2

sm80.92

12

12

g π m

k π f

13.69: a)2T πA0.150m s b) a2π T2x0.112m s2 The time to go from equilibrium to half the amplitude is sin ωt 12,or ωtπ 6rad, or one-twelfth of a period The needed time is twice this, or one-sixth of a period, 0.70 s

g ω

g

k

mg

l

13.70: Expressing Eq (13.13) in terms of the frequency, and with 0,and taking

2cossm2110.4s

50.1

2cosm240.0s50.12

s50.1

2sinsm00530.1s

50.1

2sins

50.1

m240.02

s50.1

2cosm240.0

2 2

π a

πt πt

π v

πt x

the steak hits, the pan is Mg k above the new equilibrium position The ratio

kg)m)(2.4N

400(

kg)m)(2.2)(0.40

sm80.9(2m

N400

m/s80.9kg2.2

2

2 2

2 2

2 2

ghM k

Mg A

(This avoids the intermediate calculation of the speed.) c) Using the total mass,

s

487.0)(

T

13.72: f 0.600Hz,m400kg; f  21 m k givesk 5685N/m.

This is the effective force constant of the two springs

a) After the gravel sack falls off, the remaining mass attached to the springs is 225

kg The force constant of the springs is unaffected, so f 0.800Hz

To find the new amplitude use energy considerations to find the distance

downward that the beam travels after the gravel falls off

Before the sack falls off, the amount x that the spring is stretched at equilibrium is 0

b) The remaining mass and the spring constant is the same in part (a), so the new frequency is again 0.800Hz

The sack falls off when the spring is stretched 0.6895 m And the speed of the beam

at this point is v A k m 0.400m 5685N/m 400kg1.508m/s Take y0

at this point The total energy of the beam at this point, just after the sack falls off, is

225kg 1.508m/s2 215695N/m0.6895m2 0 1608J

2

1 g

equilibrium position the spring is stretched 0.3879 m, so the new amplitude is 0.3789 m +

0.0380 m = 0.426 m Energy calculations show that v is also zero when the beam is 0.426

m below the equilibrium position

13.73: The pendulum swings through 21 cycle in 1.42 s, so T 2.84s.L1.85m

Use T to find g:

m/s055.92

m1018.8 som,1014.5

2

/

24

2 p p

6 p

7 p

2 p p

R GM

g

13.74: a) Solving Eq (13.12) for m, and using k  F l

kg

05.4m0.250

N0.402

12

2 2

b) t(0.35)T,andsoxAsin2π(0.35)0.0405m.Sincet T4,the mass has already passed the lowest point of its motion, and is on the way up

c) Taking upward forces to be positive, Fspring mg kx, wherexis the

displacement from equilibrium , so

N

5.44)m/skg)(9.80

(4.05m)

0.030m)(

N160

,)

(

E

3 E

E 2

3 E

E

R

g R

GM ω

x R

m GM dx

dU x

The period is then

s,5070m/s

9.80

m1038.622

π T

or 84.5 min

13.77: Take only the positive root (to get the least time), so that

,2

)1arcsin(

)(

or,

1 1

2 2

2 2

1

t m

k π

t m k

t m

k dt m

k x

A dx

dt m

k x

A dx

x A m

k dt dx

t A

13.78: a)

4

4 3



 x F dx c x x dx c x U

a) From conservation of energy, ( 4 4),

4

2 2

1mv  c A x and using the technique of Problem 13.77, the separated equation is

.24

m

c x

A

dx



Integrating from 0 to A with respect to x and from 0 to T 4 with respect to t,



A

T m

c x

A

dx

.42

To use the hint, let u A x, so that dxa du and the upper limit of the u integral is 

1



u Factoring A out of the square root,2

,32

31.11

u

du A

which may be expressed as 7 41

c

m A

T  c) The period does depend on amplitude, and the motion is not simple harmonic

13.79: As shown in Fig.13.5 b ,vvtansinθ With vtan Aωandθωt, this is

Eq.13.15

13.80: a) Taking positive displacements and forces to be upwad,

 2 ,,a πf 2x

g