PSE9e ISM chapter15 final kho tài liệu bách khoa

792 15 Oscillatory Motion CHAPTER OUTLINE 15.1 Motion of an Object Attached to a Spring 15.2 Analysis Model: Particle in Simple Harmonic Motion 15.3 Energy of the Simple Harmonic Oscil

792

15

Oscillatory Motion CHAPTER OUTLINE

15.1 Motion of an Object Attached to a Spring

15.2 Analysis Model: Particle in Simple Harmonic Motion

15.3 Energy of the Simple Harmonic Oscillator

15.4 Comparing Simple Harmonic Motion with Uniform Circular Motion 15.5 The Pendulum

15.6 Damped Oscillations

15.7 Forced Oscillations

* An asterisk indicates a question or problem new to this edition

ANSWERS TO OBJECTIVE QUESTIONS

OQ15.1 Answer (d) The period of a simple pendulum is T = 2π  g, and its

frequency is f = 1 T = 1 2π( ) g  Thus, if the length is doubled so

′ = 2, the new frequency is

g

2 = 1

2

12π

OQ15.2 Answer (c) The equilibrium position is 15 cm below the starting

point The motion is symmetric about the equilibrium position, so the two turning points are 30 cm apart

OQ15.3 Answer (a) In this spring-mass system, the total energy equals the

elastic potential energy at the moment the mass is temporarily at rest

at x = A = 6 cm (i.e., at the extreme ends of the simple harmonic

motion) Thus, E = kA2 2 and we see that as long as the spring constant k and the amplitude A remain unchanged, the total energy

is unchanged

OQ15.4 Answer (c) The total energy of the object-spring system is

OQ15.5 Answer (d) When the object is at its maximum displacement, the

magnitude of the force exerted on it by the spring is F s = k xmax = 8.0 N m( ) (0.10 m)= 0.80 N This force will give the mass an acceleration of a = F s m= 0.80 N 0.40 kg = 2.0 m s2

OQ15.6 Answer (a) The car will continue to compress the spring until all of

the car’s original kinetic energy has been converted into elastic potential energy within the spring, i.e., until

OQ15.7 Answer (c) When an object undergoes simple harmonic motion, the

position as a function of time may be written as

x = Acosωt = Acos 2( πft) Comparing this to the given relation, we

see that the frequency of vibration is f = 3 Hz, and the period is

T = 1/f = 1/3

*OQ15.8 Answer (b) The frequency of vibration is

f = ω2π =

12π

k m

Thus, increasing the mass by a factor of 9 will decrease the

frequency to 1/3 of its original value

OQ15.9 Answer (a) Higher frequency When it supports your weight, the

center of the diving board flexes down less than the end does when it supports your weight—this is similar to a spring that stretches a smaller distance for the same force: its spring constant is greater because the displacement is smaller Therefore, the stiffness constant describing the center of the board is greater than the stiffness

constant describing the end And then

f = 12π

⎛

⎝⎜ ⎞⎠⎟ m k is greater for you bouncing on the center of the board

OQ15.10 (i) Answer (c) At 40 cm we have the midpoint between the turning

points, so it is the equilibrium position and the point of maximum speed, and therefore, maximum momentum

(ii) Answer (c) The position of maximum speed is also the position

of maximum kinetic energy

(iii) Answer (e) The total energy of the system is conserved, so it is the same at every position

OQ15.11 The ranking is (c) > (e) > (a) = (b) > (d) The amplitude does not affect

the period in simple harmonic motion; neither do constant forces that offset the equilibrium position Thus (a) and (b) have equal periods The period is proportional to the square root of mass divided by spring constant So (c), with larger mass, has a larger period than (a) And (d) with greater stiffness has smaller period In situation (e) the motion is not quite simple harmonic, but has slightly smaller angular frequency and so a slightly longer period

OQ15.12 (a) Yes In simple harmonic motion, one-half of the time, the

velocity is in the same direction as the displacement away from

OQ15.13 Answer (d) We assume that the coils of the spring do not hit one

another When the spring with two blocks is set into oscillation in space, the coil in the center of the spring does not move We can imagine clamping the center coil in place without affecting the motion We can effectively duplicate the motion of each individual block in space by hanging a single block on a half-spring here on Earth The half-spring with its center coil clamped—or its other half cut off—has twice the spring constant as the original uncut spring because an applied force of the same size would produce only one-half the extension distance Thus the oscillation frequency in space is

12π

⎛

⎝⎜ ⎞⎠⎟⎛⎝⎜2k m⎞⎠⎟

1 2

= 2 f The absence of a force required to support the

vibrating system in orbital free fall has no effect on the frequency of its vibration

OQ15.14 Answer (d) is the only false statement At the equilibrium position,

x = 0, the elastic potential energy of the system

OQ15.15 (i) Answer (e) We have

OQ15.16 (i) Answer (b) The upward acceleration has the same effect as an

increased gravitational acceleration

(ii) Answer (a) The downward acceleration has the same effect as a decreased gravitational acceleration

(iii) Answer (c) The absence of acceleration means that the effective gravitational field is the same as that for a stationary elevator

OQ15.17 (i) Answer (c) At 120 cm we have the midpoint between the

turning points, so it is the equilibrium position and the point of maximum speed

(ii) Answer (a) In simple harmonic motion the acceleration is maximum when the displacement from equilibrium is maximum

(iii) Answer (a), by the same logic as in part (ii)

ANSWERS TO CONCEPTUAL QUESTIONS

CQ15.1 An imperceptibly slight breeze blowing over the edge of a leaf can

produce fluttering in the same way that a breeze can cause a flag to flap As a leaf twists in the wind, the fibers in its stem provide a restoring torque If the frequency of the breeze matches the natural frequency of vibration of one particular leaf as a torsional pendulum, that leaf can be driven into a large-amplitude resonance vibration

Note that it is not the size of the driving force that sets the leaf into resonance, but the frequency of the driving force If the frequency

changes, another leaf will be set into resonant oscillation

CQ15.2 (a) No Since the acceleration is not constant in simple harmonic

motion, none of the equations in Table 2.2 are valid

(b) Equation Information given by equation

x t( )= Acos(ωt + φ) position as a function of time

v t( )= −ωAsin ωt + φ( ) velocity as a function of time

v x( )= ±ω A( 2− x2)1 2

velocity as a function of position

a t( )= −ω2Acos(ωt+φ) acceleration as a function of time

a t( )= −ω2x t( ) acceleration as a function of

(c) The angular frequency ω appears in every equation

CQ15.3 (a) The general equation of position is x t( )= Acos(ωt + φ) If

x = −Acos ωt( ), then φ = π, or equally well, φ = −π

(b) At t = 0, the particle is at its turning point on the negative side of equilibrium, at x = –A

CQ15.4 We assume the diameter of the bob is not very small compared to the

length of the cord supporting it As the water leaks out, the center of mass of the bob moves down, increasing the effective length of the pendulum and slightly lowering its frequency As the last drops of water dribble out, the center of mass of the bob moves back up to the center of the sphere, and the pendulum frequency quickly increases

to its original value

CQ15.5 (a) No force is exerted on the particle The particle moves with

constant velocity

(b) The particle feels a constant force toward the left It moves with constant acceleration toward the left If its initial push is toward the right, it will slow down, turn around, and speed up in motion toward the left If its initial push is toward the left, it will just speed up

(c) A constant force toward the right acts on the particle to produce constant acceleration toward the right

(d) The particle moves in simple harmonic motion about the lowest point of the potential energy curve

CQ15.6 Most everyday vibrations are damped, they eventually die down as

their energy is transferred to their surroundings However, as you will learn later, atoms in the molecules have vibration modes that do not damp out

CQ15.7 The mechanical energy of a damped oscillator changes back and

forth between kinetic and potential while it gradually and permanently decreases and transforms to internal energy

CQ15.8 Yes An oscillator with damping can vibrate at resonance with

amplitude that remains constant in time Without damping, the amplitude would increase without limit at resonance

CQ15.9 No If the resistive force is large compared to the restoring force of

the spring (in particular, if b2 > 4mk ), the system will be overdamped

and will not oscillate

CQ15.10 The period of a pendulum depends on the acceleration of gravity:

T= 2π L

g If the acceleration of gravity is different at the top of the mountain, the period is different and the pendulum does not keep perfect time Two things can effect the acceleration of gravity, the top

of the mountain is farther from the center of the Earth, and the nearby large mass of the mountain under the pendulum

CQ15.11 Neither are examples of simple harmonic motion, although they are

both periodic motion In neither case is the acceleration proportional

to the displacement from an equilibrium position Neither motion is

so smooth as SHM The ball’s acceleration is very large when it is in contact with the floor, and the student’s when the dismissal bell rings

CQ15.12 The motion will be periodic—that is, it will repeat, though it is not

harmonic at large angles The period is nearly constant as the angular amplitude increases through small values; then the period becomes noticeably larger as θ increases farther

CQ15.13 The angle of the crank pin is θ =ωt. Its x coordinate is x = A cos θ =

A cos ωt, where A is the distance from the center of the wheel to the crank pin This is of the form x = A cos θt + φ( ), so the yoke and piston move with simple harmonic motion

ANS FIG CQ15.13

SOLUTIONS TO END-OF-CHAPTER PROBLEMS

Section 15.1 Motion of an Object Attached to a Spring

P15.1 (a) Taking to the right as positive, the spring force acting on the block

at the instant of release is

or

a= 28 m s2 to the left P15.2 When the object comes to equilibrium (at distance y0 below the

unstretched position of the end of the spring), F∑ y = −k −y( )0 − mg = 0

and the force constant is

y0 = (4.25 kg) (9.80 m s2)

2.62× 10−2 m = 1.59 × 103 N= 1.59 kN/m

Section 12.2 Analysis Model: Particle in

Simple Harmonic Motion P15.3 The spring constant is found from

A= 4.00 m (d) φ = π rad

(e) x t = 0.250 s( )= 4.00 m( )cos 1.75π( )= 2.83 m

P15.6 From the information given, we write the equation for position as

x = Acosωt, with the amplitude given as A = 0.050 0 m Differentiating

gives us the piston’s velocity,

v = −Aωsinωt

and differentiating again gives its acceleration

a = −Aω2cosωt Then, if f = 3600 rev/min = 60 Hz, then ω = 2πf = 120π s−1

(a)

vmax=ω A = 120π( ) (0.050 0) m s= 18.8 m s (b)

f = 1

2.40 = 0.417 Hz(c)

ω = 2π f = 2π 0.417( )= 2.62 rad s

P15.9 An object hanging from a vertical spring moves with simple harmonic

motion just like an object moving without friction attached to a horizontal spring We are given the period, which is related to the

frequency of motion by T = 1/f Then, since

*P15.10 For a simple harmonic oscillator, the maximum speed occurs at the

equilibrium position and is given by Equation 15.17:

vmax= A k

m

F g = mg = 4.00 kg( ) (9.80 m/s2)= 39.2 N

*P15.11 The mass of the cube is

m =ρV = 2.70 × 10( 3 kg/m3)(0.015 m)3 = 9.11 × 10−3 kgThe spring constant of the strip of steel is

x = 1.43 N0.027 5 m= 52.0 N/m

P15.12 (a) The spring constant of this spring is

Distance moved = 70.569 0.72 m( )= 50.8 m

v x = v xi + a x t= 0.140 m s − 0.320 m s( 2) (4.50 s)= −1.30 m s (c) For simple harmonic motion we have

instead

x = A cos(ωt+φ) and v = −Aω sin(ωt+φ)

ANS FIG P15.13(c, d)

P15.14 (a) Since the collision is perfectly elastic, the ball will rebound to the

height of 4.00 m and then repeat the motion over and over again Thus, the motion is periodic

(b) To determine the period, we use

T= 2 0.904 s( )= 1.81 s (c)

The motion is not simple harmonic The net force action on the

ball is a constant given by F = −mg (except when it is in contact

with the ground), which is not in the form of Hooke’s law

P15.15 The period of the oscillation is

amax = Aω2 = 2.00 3.00( π)2 = 18.0π2 cm/s2 = 178 cm/s2

(e) This positive value of maximum acceleration first occurs when

the particle is reversing its direction on the negative x axis, quarters of a period after t = 0: at

t= 3

4T= 0.500 s

(a) The acceleration being a negative constant times position means

we do have SHM, and its angular frequency is ω At t = 0 the equations reduce to x = x i and v = v i, so they satisfy all the requirements

= x i2ω2 sin2 ωt − 2x i v iω sinωtcosωt + v i2 cos2 ωt

+ x i2ω2 cos2ωt + 2x i v iω cosωtsinωt

+ v i2 sin2ωt

= x i2ω2+ v i2

So the expression v2− ax is constant in time because all the

parameters in the final equivalent expression x i2ω2+ v i2 are

constant Because v2− ax must have the same value at all times, it must be equal to the value at t = 0, so v2− ax = v i2− a i x i If we

evaluate v2− ax at a turning point where v = 0 and x = A, it is

v2− ax = v2+ω2x2 = 02+ω2( )A2 =ω2A2 Thus it is proved

*P15.17 (a) The distance traveled in one cycle is four times the amplitude of

(b) vmax = Aω = 3.00 × 10( −2 m) (5.00 rad/s)= 0.150 m/s

0.500 kg = 4.00 s−1 Assuming the position of the object is

at the origin at t = 0, position is given by x = 10.0 sin 4.00t( ), where x is

in cm From this, we find that v = 40.0 cos 4.00t( ), where v is in cm/s, and a = −160 sin 4.00t( ), where a is in cm/s2

From our assumed expression for x, we solve for the time t:

t= 14.00 Hz

⎛

⎝⎜ ⎞⎠⎟ sin−1⎛⎝⎜10.0 cmx ⎞⎠⎟

When x = 6.00 cm,

t= 14.00 Hz

⎛

⎝⎜ ⎞⎠⎟ sin−1⎛⎝⎜6.00 cm10.0 cm⎞⎠⎟ = 0.161 s

We find then that at that time:

(c) v = 40.0 cm/s( )cos 4.00 Hz[ ( ) (0.161 s) ]= 32.0 cm/s and (d)

a= − 160 cm/s( 2)sin 4.00 Hz[ ( ) (0.161 s) ]= −96.0 cm/s2

(e) Using

t= 14.00 Hz

⎛

⎝⎜ ⎞⎠⎟ sin−1⎛⎝⎜10.0 cmx ⎞⎠⎟ we find that when x = 0,

t = 0, and when x = 8.00 cm, t = 0.232 s Therefore,

Δt = 0.232 s

P15.20 (a) Yes

(b) We assume that the mass of the spring is negligible and that we

are on Earth Let m represent the mass of the object Its hanging at

Section 15.3 Energy of the Simple Harmonic Oscillator

P15.21 Choose the car with its shock-absorbing bumper as the system; by

*P15.23 (a) Energy is conserved for the block-spring system between the

maximum-displacement and the half-maximum points:

*P15.24 (a) The mechanical energy of the system is equal to the potential

energy stored in the spring at maximum amplitude:

− 1.00 × 10( −2 m)2

= 1.02 m/s(c)

vmax =ωA = 50.0 0.200( )= 1.41 m s (d) Maximum speed occurs when the object passes through its equilibrium position, at x = 0

P15.29 (a) Energy is conserved by an isolated simple harmonic oscillator:

P15.30 (a) Particle under constant acceleration

is isolated (d) The system is isolated, so energy is conserved within the system Take the initial point where she steps off the bridge and the final point at the bottom of her motion

(g) Set x = 0 at the equilibrium position of the bungee jumper on the

spring Relative to the equilibrium position, the lowest part of the

drop corresponds to x = +16.3 m—we have taken down as

positive—and the point in the drop where the spring begins

to stretch is at x = –8.68 m Take the phase as zero at maximum downward extension (x = +16.3 m) We find that the phase, ωt,

was 25 m higher where x = –8.68 (above the equilibrium point):

x = A cos ωt: at time t = 0, x = 16.3 m( )cos 0= 16.3 m , and when

x = −8.68 = 16.3 cos ωt( ), ωt= ±122° = ±2.13 rad Which sign do

we pick for ωt? From

v= dx

dt = −ωAsinωt, at x = –8.68 m, v is downward, which means by our choice of positive direction, v is positive Pick ωt = −2.13 rad:

v = −ω A sin −2.13 rad( )= +ωA 0.848( ), which is positive

Therefore,

ωt = 1.06t = −2.13 rad → t = −2.13 rad

1.06 = −2.01 s,

meaning t = –2.01 s when the spring begins to stretch and t = 0

when the jumper reaches the bottom of the jump: then +2.01 s

is the time over which the spring stretches

(h) total time = 1.50 s + 2.01 s = 3.50 s

(c) While the block was held stationary at x = 5.46 cm,

F∑ x = −F s + F = 0, or the spring force was equal in magnitude and

oppositely directed to the applied force When the applied force is

suddenly removed, there is a net force F s = 4.58 N directed toward the equilibrium position acting on the block This gives

the block an acceleration having magnitude

a = F s

m = 4.58 N0.250 kg = 18.3 m s2

(d) At the equilibrium position, PE s = 0, so the block has kinetic

energy K = E = 0.125 J and speed

(f) The coefficient of kinetic friction between the block and surface

(g) The block will come to a stop after sliding through distance

P15.32 (a) At the equilibrium position, the total energy of the system is in

the form of kinetic energy and mvmax2 2= E, so the maximum speed is

(b) The period of an object-spring system is T = 2π m k, so the force

constant of the spring is

with Uniform Circular Motion P15.33 (a) The motion is simple harmonic because the tire is rotating with

constant angular velocity and you see the projection of the motion

of the bump in a plane perpendicular to the tire

(b) Since the car is moving with speed v = 3.00 m/s, and its radius

is 0.300 m, we have

ω = 3.00 m s

0.300 m = 10.0 rad sTherefore, the period of the motion is

( )= 0.628 s

P15.34 The period in Tokyo is

P15.35 The period of a pendulum is the time for one complete oscillation and

is given by T = 2π  g, where  is the length of the pendulum

P15.36 Referring to ANS FIG P15.36, we have

F = −mg sinθ and tanθ = x

R

For small displacements,

tanθ ≈ sinθ and F= −mg

R x = −kx

Since the restoring force is proportional to the displacement from equilibrium, the motion is simple harmonic motion

P15.38 Please see ANS FIG P15.37 For a physical pendulum,

We then solve ω = g

L for L= g

ω2 = 9.80 m/s24.43 rad/s

−ICM− md2 + 2md2 = 0

or

ICM = md2

vmax = Aω = 0.262 m( ) (3.13 s−1)

= 0.820 m/s(b) For simple harmonic motion, the maximum acceleration

(c) The maximum restoring force causes the maximum acceleration:

F = mamax = 0.25 kg 2.57 m/s( 2)= 0.641 N(d) (a) Applying energy conservation to the isolated pendulum-

vmax = 2gh = 2gL 1 − cosθ( )

= 2(9.80 m/s2)(1.00 m)(1− cos15.0°)

= 0.817 m s

ANS FIG P15.42 P15.42 (a) The parallel-axis theorem gives:

= 2.09 s(b) For the simple pendulum,

T = 2π 1.00 m

9.80 m s2 = 2.01 s difference= 2.09 s− 2.01 s

2.01 s = 4.08%

P15.43 (a) The string tension must support the weight of the bob, accelerate

it upward, and also provide the restoring force, just as if the elevator were at rest in a gravity field of (9.80 + 5.00) m/s2 Thus the period is

P15.45 The period of oscillation for the watch balance wheel is T = 0.250 s

Modeling the 20.0-g mass as a particle, we find the moment of inertia

from I = mr2 (a)

P15.46 We are given θ i = 15.0°, and θ t = 1 000 s( )= 5.50° We then use

Equation 15.32 for damped oscillations:

P15.47 If the oscillation was undamped, its frequency would be

Then, substituting from Equation 15.31,

and b2 < 4mk so that ω is real, [2]

we take x = Ae −bt 2m cos(ωt+φ) and compute [3]

The cosine term is

−k + b22m= −b

Section 15.7 Forced Oscillations

P15.50 (a) For resonance, her frequency must match:

f0 =ω0

2π =

12π

k

2π

7.00× 102 N m12.5 kg = 1.19 Hz

(b) From x = A cos ωt, v= dx dt = −Aω sinωt, and

a= dv

dt = −Aω2 cosωt, the maximum acceleration is Aω2 When

this becomes equal to the acceleration due to gravity, the normal force exerted on her by the mattress will drop to zero at one point

P15.51 The pendulum is resonating with the beeper The beeper must vibrate

at the frequency of a simple pendulum of frequency 1.50 Hz: