15.1 Motion of an Object Attached to a Spring As a model for simple harmonic motion, consider a block of mass m attached to the end of a spring, with the block free to move on a horizont
Trang 153. The true weight of an object can be measured in a
vac-uum, where buoyant forces are absent An object of
vol-ume V is weighed in air on an equal-arm balance with the
use of counterweights of density r Representing the
den-sity of air as rairand the balance reading as F g, show that
the true weight F gis
54. Water is forced out of a fire extinguisher by air pressure
as shown in Figure P14.54 How much gauge air pressure
in the tank (above atmospheric) is required for the water
jet to have a speed of 30.0 m/s when the water level is
0.500 m below the nozzle?
F g F ¿ g a V F ¿ g
rgb r airg
57. As a 950-kg helicopter hovers, its horizontal rotor pushes a column of air downward at 40.0 m/s What can you say about the quantity of this air? Explain your answer You may model the air motion as ideal flow.
58. Evangelista Torricelli was the first person to realize that
we live at the bottom of an ocean of air He correctly mised that the pressure of our atmosphere is attributable
sur-to the weight of the air The density of air at 0°C at the Earth’s surface is 1.29 kg/m 3 The density decreases with increasing altitude (as the atmosphere thins) On the other hand, if we assume the density is constant at 1.29 kg/m 3up to some altitude h and is zero above that altitude, then h would represent the depth of the ocean
of air Use this model to determine the value of h that
gives a pressure of 1.00 atm at the surface of the Earth Would the peak of Mount Everest rise above the surface
60. In about 1657, Otto von Guericke, inventor of the air pump, evacuated a sphere made of two brass hemi- spheres Two teams of eight horses each could pull the hemispheres apart only on some trials and then “with greatest difficulty,” with the resulting sound likened to a
cannon firing (Fig P14.60) (a) Show that the force F
required to pull the thin-walled evacuated hemispheres
apart is pR2(P0 P), where R is the radius of the spheres and P is the pressure inside the hemispheres, which is much less than P0 (b) Determine the force for
55. A light spring of constant k 90.0 N/m is attached
verti-cally to a table (Fig P14.55a) A 2.00-g balloon is filled with
helium (density 0.180 kg/m 3 ) to a volume of 5.00 m 3
and is then connected to the spring, causing the spring to
stretch as shown in Figure P14.55b Determine the
exten-sion distance L when the balloon is in equilibrium.
L
Figure P14.55
56. We can’t call it Flubber Assume a certain liquid, with
density 1 230 kg/m 3 , exerts no friction force on spherical
objects A ball of mass 2.10 kg and radius 9.00 cm is
dropped from rest into a deep tank of this liquid from a
height of 3.30 m above the surface (a) Find the speed at
which the ball enters the liquid (b) What two forces are
exerted on the ball as it moves through the liquid?
(c) Explain why the ball moves down only a limited
dis-tance into the liquid and calculate this disdis-tance (d) With
what speed does the ball pop up out of the liquid?
(e) How does the time interval t down , during which the
ball moves from the surface down to its lowest point,
com-pare with the time interval t up for the return trip
between the same two points? (f) What If? Now modify
the model to suppose the liquid exerts a small friction
force on the ball, opposite in direction to its motion In
this case, how do the time intervals t down and tup
com-pare? Explain your answer with a conceptual argument
rather than a numerical calculation.
Figure P14.60 The colored engraving, dated 1672, illustrates Otto von Guericke’s demonstration of the force due to air pressure as it might have been performed before Emperor Ferdinand III.
Trang 261. A 1.00-kg beaker containing 2.00 kg of oil (density
916.0 kg/m 3 ) rests on a scale A 2.00-kg block of iron
sus-pended from a spring scale is completely submerged in
the oil as shown in Figure P14.61 Determine the
equilib-rium readings of both scales.
67. An incompressible, nonviscous fluid is initially at rest in the vertical portion of the pipe shown in Figure P14.67a,
where L 2.00 m When the valve is opened, the fluid flows into the horizontal section of the pipe What is the speed of the fluid when it is all in the horizontal section
as shown in Figure P14.67b? Assume the cross-sectional area of the entire pipe is constant.
2 = intermediate; 3 = challenging; = SSM/SG; = ThomsonNOW; = symbolic reasoning; = qualitative reasoning
Figure P14.61 Problems 61 and 62.
62. A beaker of mass m b containing oil of mass m oand density
ro rests on a scale A block of iron of mass mFesuspended
from a spring scale is completely submerged in the oil as
shown in Figure P14.61 Determine the equilibrium
read-ings of both scales.
63. In 1983, the United States began coining the cent piece
out of copper-clad zinc rather than pure copper The
mass of the old copper penny is 3.083 g and that of the
new cent is 2.517 g Calculate the percent of zinc (by
vol-ume) in the new cent The density of copper is
8.960 g/cm 3 and that of zinc is 7.133 g/cm 3 The new and
old coins have the same volume.
64. Show that the variation of atmospheric pressure with
alti-tude is given by P P0e ay, where a r 0g/P0, P0is
atmo-spheric pressure at some reference level y 0, and r 0 is
the atmospheric density at this level Assume the decrease
in atmospheric pressure over an infinitesimal change in
altitude (so that the density is approximately uniform) is
given by dP rg dy and that the density of air is
propor-tional to the pressure.
65 Review problem. A uniform disk of mass 10.0 kg and
radius 0.250 m spins at 300 rev/min on a low-friction
axle It must be brought to a stop in 1.00 min by a brake
pad that makes contact with the disk at an average
dis-tance of 0.220 m from the axis The coefficient of friction
between the pad and the disk is 0.500 A piston in a
cylin-der of diameter 5.00 cm presses the brake pad against the
disk Find the pressure required for the brake fluid in the
cylinder.
66. A cube of ice whose edges measure 20.0 mm is floating in
a glass of ice-cold water, and one of the ice cube’s faces is
parallel to the water’s surface (a) How far below the
water surface is the bottom face of the ice cube? (b)
Ice-cold ethyl alcohol is gently poured onto the water surface
to form a layer 5.00 mm thick above the water The
alco-hol does not mix with the water When the ice cube again
attains hydrostatic equilibrium, what is the distance from
the top of the water to the bottom face of the block?
(c) Additional cold ethyl alcohol is poured onto the
water’s surface until the top surface of the alcohol
coin-cides with the top surface of the ice cube (in hydrostatic
equilibrium) How thick is the required layer of ethyl
alcohol?
Valve closed
Valve opened
(a)
L L
69. A U-tube open at both ends is partially filled with water (Fig P14.69a) Oil having a density 750 kg/m 3 is then
poured into the right arm and forms a column L 5.00 cm high (Fig P14.69b) (a) Determine the differ-
ence h in the heights of the two liquid surfaces (b) The
right arm is then shielded from any air motion while air is blown across the top of the left arm until the surfaces of the two liquids are at the same height (Fig P14.69c) Determine the speed of the air being blown across the left arm Take the density of air as 1.29 kg/m 3
Water
h L
Trang 3flowing without friction (a) Show that the time interval
required to empty the tank is
(b) Evaluate the time interval required to empty the tank
if it is a cube 0.500 m on each edge, taking A 2
and d 10.0 m.
¢t Ah
A¿ 22gd
wing Its area projected onto a horizontal surface is A.
When the boat is towed at sufficiently high speed, water
of density r moves in streamline flow so that its average
speed at the top of the hydrofoil is n times larger than its speed v b below the hydrofoil (a) Ignoring the buoyant force, show that the upward lift force exerted by the water
on the hydrofoil has a magnitude
(b) The boat has mass M Show that the liftoff speed is
(c) Assume an 800-kg boat is to lift off at 9.50 m/s
Evalu-ate the area A required for the hydrofoil if its design yields n 1.05.
Answers to Quick Quizzes
14.1 (a) Because the basketball player’s weight is distributed
over the larger surface area of the shoe, the pressure
(F/A) he applies is relatively small The woman’s lesser
weight is distributed over the very small cross-sectional
area of the spiked heel, so the pressure is high.
14.2 (a) Because both fluids have the same depth, the one
with the smaller density (alcohol) will exert the smaller
pressure.
14.3 (c) All barometers will have the same pressure at the
bottom of the column of fluid: atmospheric pressure.
Therefore, the barometer with the highest column will
be the one with the fluid of lowest density.
14.4 (b) or (c) In all three cases, the weight of the treasure chest causes a downward force on the raft that makes the raft sink into the water In (b) and (c), however, the treasure chest also displaces water, which provides a buoyant force in the upward direction, reducing the effect of the chest’s weight.
14.5 (a) The high-speed air between the balloons results in low pressure in this region The higher pressure on the outer surfaces of the balloons pushes them toward each other.
Figure P14.71
71. The hull of an experimental boat is to be lifted above the
water by a hydrofoil mounted below its keel as shown in
Figure P14.71 The hydrofoil is shaped like an airplane
Trang 4We begin this new part of the text by
studying a special type of motion
called periodic motion, the repeating
motion of an object in which it
contin-ues to return to a given position after a
fixed time interval The repetitive
move-ments of such an object are called
oscil-lations We will focus our attention on a
special case of periodic motion called
simple harmonic motion All periodic motions can be modeled as combinations of simple monic motions.
har-Simple harmonic motion also forms the basis for our understanding of mechanical waves Sound waves, seismic waves, waves on stretched strings, and water waves are all produced by some source of oscillation As a sound wave travels through the air, elements of the air oscillate back and forth; as a water wave travels across a pond, elements of the water oscillate up and down and backward and forward The motion of the elements of the medium bears a strong resemblance to the periodic motion of an oscillating pendulum or an object attached to a spring.
To explain many other phenomena in nature, we must understand the concepts of oscillations and waves For instance, although skyscrapers and bridges appear to be rigid, they actually oscil- late, something the architects and engineers who design and build them must take into account.
To understand how radio and television work, we must understand the origin and nature of tromagnetic waves and how they propagate through space Finally, much of what scientists have learned about atomic structure has come from information carried by waves Therefore, we must first study oscillations and waves if we are to understand the concepts and theories of atomic physics.
elec-Oscillations and Mechanical
In Part 2 of the text, we will explore the principles related to oscillations and waves (Don Bonsey/Getty Images)
Trang 5Periodic motion is motion of an object that regularly returns to a given position
after a fixed time interval With a little thought, we can identify several types ofperiodic motion in everyday life Your car returns to the driveway each afternoon.You return to the dinner table each night to eat A bumped chandelier swingsback and forth, returning to the same position at a regular rate The Earth returns
to the same position in its orbit around the Sun each year, resulting in the tion among the four seasons
varia-In addition to these everyday examples, numerous other systems exhibit odic motion The molecules in a solid oscillate about their equilibrium positions;electromagnetic waves, such as light waves, radar, and radio waves, are character-ized by oscillating electric and magnetic field vectors; and in alternating-currentelectrical circuits, voltage, current, and electric charge vary periodically with time
peri-A special kind of periodic motion occurs in mechanical systems when the forceacting on an object is proportional to the position of the object relative to someequilibrium position If this force is always directed toward the equilibrium posi-
tion, the motion is called simple harmonic motion, which is the primary focus of this
chapter
To reduce swaying in tall buildings because of the wind, tuned dampers
are placed near the top of the building These mechanisms include an
object of large mass that oscillates under computer control at the same
frequency as the building, reducing the swaying The large sphere in the
photograph on the left is part of the tuned damper system of the building
in the photograph on the right, called Taipei 101, in Taiwan The building,
also called the Taipei Financial Center, was completed in 2004, at which
time it held the record as the world’s tallest building (left, Courtesy of
Motioneering, Inc.; right, © Simon Kwang/Reuters/CORBIS)
Oscillatory Motion
15
418
15.6 Damped Oscillations 15.7 Forced Oscillations
Trang 615.1 Motion of an Object Attached
to a Spring
As a model for simple harmonic motion, consider a block of mass m attached to
the end of a spring, with the block free to move on a horizontal, frictionless
sur-face (Active Fig 15.1) When the spring is neither stretched nor compressed, the
block is at rest at the position called the equilibrium position of the system, which
we identify as x 0 We know from experience that such a system oscillates back
and forth if disturbed from its equilibrium position
We can understand the oscillating motion of the block in Active Figure 15.1
qualitatively by first recalling that when the block is displaced to a position x, the
spring exerts on the block a force that is proportional to the position and given by
Hooke’s law(see Section 7.4):
(15.1)
We call F s a restoring force because it is always directed toward the equilibrium
position and therefore opposite the displacement of the block from equilibrium.
That is, when the block is displaced to the right of x 0 in Active Figure 15.1a,
the position is positive and the restoring force is directed to the left Figure 15.1b
shows the block at x 0, where the force on the block is zero When the block is
displaced to the left of x 0 as in Figure 15.1c, the position is negative and the
restoring force is directed to the right
Applying Newton’s second law to the motion of the block, with Equation 15.1
providing the net force in the x direction, we obtain
(15.2)
That is, the acceleration of the block is proportional to its position, and the
direc-tion of the acceleradirec-tion is opposite the direcdirec-tion of the displacement of the block
from equilibrium Systems that behave in this way are said to exhibit simple
har-monic motion.An object moves with simple harmonic motion whenever its
accel-eration is proportional to its position and is oppositely directed to the
displace-ment from equilibrium
If the block in Active Figure 15.1 is displaced to a position x A and released
from rest, its initial acceleration is kA/m When the block passes through the
equi-librium position x 0, its acceleration is zero At this instant, its speed is a
maxi-mum because the acceleration changes sign The block then continues to travel to
the left of equilibrium with a positive acceleration and finally reaches x –A, at
which time its acceleration is kA/m and its speed is again zero as discussed in
Sec-tions 7.4 and 7.9 The block completes a full cycle of its motion by returning to the
original position, again passing through x 0 with maximum speed Therefore,
sur-rium (x 0), the force exerted by the spring acts to the left.
(b) When the block is at its equilibrium position (x 0), the force exerted by the spring is zero (c) When the block
is displaced to the left of equilibrium (x 0), the force exerted by the spring acts to the right.
Sign in at www.thomsonedu.comand go to ThomsonNOW
to choose the spring constant and the initial position and velocity of the block and see the resulting simple harmonic motion
Hooke’s law
PITFALL PREVENTION 15.1
The Orientation of the Spring
Active Figure 15.1 shows a
horizon-tal spring, with an attached block
sliding on a frictionless surface Another possibility is a block hang-
ing from a vertical spring All the
results we discuss for the horizontal spring are the same for the vertical spring with one exception: when the block is placed on the vertical spring, its weight causes the spring
to extend If the resting position of
the block is defined as x 0, the results of this chapter also apply to this vertical system.
Trang 7the block oscillates between the turning points x A In the absence of friction,
this idealized motion will continue forever because the force exerted by the spring
is conservative Real systems are generally subject to friction, so they do not oscillateforever We shall explore the details of the situation with friction in Section 15.6
Quick Quiz 15.1 A block on the end of a spring is pulled to position x A and
released from rest In one full cycle of its motion, through what total distance does
it travel? (a) A/2 (b) A (c) 2A (d) 4A
The motion described in the preceding section occurs so often that we identify the
particle in simple harmonic motionmodel to represent such situations To develop
a mathematical representation for this model, first recognize that the block is aparticle under a net force as described in Equation 15.1 We will generally choose
x as the axis along which the oscillation occurs; hence, we will drop the subscript-x notation in this discussion Recall that, by definition, a dv/dt d2x/dt2, and so
we can express Equation 15.2 as
Let’s now find a mathematical solution to Equation 15.5, that is, a function x(t)
that satisfies this second-order differential equation and is a mathematical sentation of the position of the particle as a function of time We seek a functionwhose second derivative is the same as the original function with a negative signand multiplied by v2 The trigonometric functions sine and cosine exhibit thisbehavior, so we can build a solution around one or both of them The followingcosine function is a solution to the differential equation:
repre-(15.6)
where A, v, and f are constants To show explicitly that this solution satisfies
Equa-tion 15.5, notice that
(15.7) (15.8)
Comparing Equations 15.6 and 15.8, we see that d2x/dt2 v2x and Equation
15.5 is satisfied
The parameters A, v, and f are constants of the motion To give physical
signif-icance to these constants, it is convenient to form a graphical representation of
the motion by plotting x as a function of t as in Active Figure 15.2a First, A, called
the amplitude of the motion, is simply the maximum value of the position of the
particle in either the positive or negative x direction.The constant v is called the
The acceleration of a particle in
simple harmonic motion is not
constant Equation 15.3 shows that
its acceleration varies with position
x Therefore, we cannot apply the
kinematic equations of Chapter 2
in this situation.
Position versus time for an
object in simple harmonic
motion
PITFALL PREVENTION 15.3
Where’s the Triangle?
Equation 15.6 includes a
trigono-metric function, a mathematical
function that can be used whether it
refers to a triangle or not In this
case, the cosine function happens
to have the correct behavior for
representing the position of a
parti-cle in simple harmonic motion.
Trang 8angular frequency, and it has units1 of rad/s It is a measure of how rapidly the
oscillations are occurring; the more oscillations per unit time, the higher the value
of v From Equation 15.4, the angular frequency is
(15.9)
The constant angle f is called the phase constant (or initial phase angle) and,
along with the amplitude A, is determined uniquely by the position and velocity of
the particle at t 0 If the particle is at its maximum position x A at t 0, the
phase constant is f 0 and the graphical representation of the motion is as
shown in Active Figure 15.2b The quantity (vt + f) is called the phase of the
motion Notice that the function x(t) is periodic and its value is the same each
time vt increases by 2p radians.
Equations 15.1, 15.5, and 15.6 form the basis of the mathematical
representa-tion of the particle in simple harmonic morepresenta-tion model If you are analyzing a
situa-tion and find that the force on a particle is of the mathematical form of Equasitua-tion
15.1, you know the motion is that of a simple harmonic oscillator and the position
of the particle is described by Equation 15.6 If you analyze a system and find that
it is described by a differential equation of the form of Equation 15.5, the motion
is that of a simple harmonic oscillator If you analyze a situation and find that the
position of a particle is described by Equation 15.6, you know the particle
under-goes simple harmonic motion
Quick Quiz 15.2 Consider a graphical representation (Fig 15.3) of simple
har-monic motion as described mathematically in Equation 15.6 When the object is at
point on the graph, what can you say about its position and velocity? (a) The
position and velocity are both positive (b) The position and velocity are both
neg-ative (c) The position is positive, and its velocity is zero (d) The position is
nega-tive, and its velocity is zero (e) The position is posinega-tive, and its velocity is negative
(f) The position is negative, and its velocity is positive
Quick Quiz 15.3 Figure 15.4 shows two curves representing objects undergoing
simple harmonic motion The correct description of these two motions is that the
simple harmonic motion of object B is (a) of larger angular frequency and larger
amplitude than that of object A, (b) of larger angular frequency and smaller
ampli-tude than that of object A, (c) of smaller angular frequency and larger ampliampli-tude
than that of object A, or (d) of smaller angular frequency and smaller amplitude than
that of object A
Let us investigate further the mathematical description of simple harmonic
motion The period T of the motion is the time interval required for the particle
to go through one full cycle of its motion (Active Fig 15.2a) That is, the values of
x and v for the particle at time t equal the values of x and v at time t T Because
the phase increases by 2p radians in a time interval of T,
Simplifying this expression gives vT 2p, or
(15.10)
T 2pv
3v 1t T 2 f4 1vt f2 2p
vBm k
1 We have seen many examples in earlier chapters in which we evaluate a trigonometric function of an
angle The argument of a trigonometric function, such as sine or cosine, must be a pure number The
radian is a pure number because it is a ratio of lengths Angles in degrees are pure numbers because
the degree is an artificial “unit”; it is not related to measurements of lengths The argument of the
trigonometric function in Equation 15.6 must be a pure number Therefore, v must be expressed in
rad/s (and not, for example, in revolutions per second) if t is expressed in seconds Furthermore,
other types of functions such as logarithms and exponential functions require arguments that are pure
(a)
ACTIVE FIGURE 15.2
(a) An x–t graph for an object
under-going simple harmonic motion The
amplitude of the motion is A, the period (defined in Eq 15.10) is T (b) The x–t graph in the special case
in which x A at t 0 and hence
f 0.
Sign in at www.thomsonedu.comand
go to ThomsonNOW to adjust the graphical representation and see the resulting simple harmonic motion of the block in Active Figure 15.1.
t x
Figure 15.3 (Quick Quiz 15.2) An
x–t graph for an object undergoing
simple harmonic motion At a ular time, the object’s position is indi- cated by in the graph.
partic-t x
t x
Object A
Object B
Figure 15.4 (Quick Quiz 15.3) Two
x–t graphs for objects undergoing
simple harmonic motion The tudes and frequencies are different for the two objects.
Trang 9ampli-The inverse of the period is called the frequency f of the motion Whereas the
period is the time interval per oscillation, the frequency represents the number of
oscillations the particle undergoes per unit time interval:
f As we might expect, the frequency is larger for a stiffer spring (larger value of
k) and decreases with increasing mass of the particle.
We can obtain the velocity and acceleration2 of a particle undergoing simpleharmonic motion from Equations 15.7 and 15.8:
(15.15) (15.16)
From Equation 15.15 we see that, because the sine and cosine functions late between 1, the extreme values of the velocity v are vA Likewise, Equation 15.16 shows that the extreme values of the acceleration a are v2A Therefore, the maximum values of the magnitudes of the velocity and acceleration are
oscil-(15.17) (15.18)
Figure 15.5a plots position versus time for an arbitrary value of the phase stant The associated velocity–time and acceleration–time curves are illustrated inFigures 15.5b and 15.5c They show that the phase of the velocity differs from the
con-phase of the position by p/2 rad, or 90° That is, when x is a maximum or a mum, the velocity is zero Likewise, when x is zero, the speed is a maximum Fur-
mini-thermore, notice that the phase of the acceleration differs from the phase of the
position by p radians, or 180° For example, when x is a maximum, a has a
maxi-mum magnitude in the opposite direction
Quick Quiz 15.4 An object of mass m is hung from a spring and set into tion The period of the oscillation is measured and recorded as T The object of
Two Kinds of Frequency
We identify two kinds of frequency
for a simple harmonic oscillator:
f, called simply the frequency, is
measured in hertz, and v, the
angular frequency, is measured in
radians per second Be sure you
are clear about which frequency is
being discussed or requested in a
given problem Equations 15.11
and 15.12 show the relationship
between the two frequencies.
2 Because the motion of a simple harmonic oscillator takes place in one dimension, we denote velocity
as v and acceleration as a, with the direction indicated by a positive or negative sign as in Chapter 2.
velocity and acceleration in
simple harmonic motion
Trang 10mass m is removed and replaced with an object of mass 2m When this object is
set into oscillation, what is the period of the motion? (a) 2T (b) (c) T
(d) (e) T/2
Equation 15.6 describes simple harmonic motion of a particle in general Let’s
now see how to evaluate the constants of the motion The angular frequency v is
evaluated using Equation 15.9 The constants A and f are evaluated from the
ini-tial conditions, that is, the state of the oscillator at t 0
Suppose the particle is set into motion by pulling it from equilibrium by a
dis-tance A and releasing it from rest at t 0 as in Active Figure 15.6 We must then
require our solutions for x(t) and v(t) (Eqs 15.6 and 15.15) to obey the initial
conditions that x(0) A and v(0) 0:
These conditions are met if f 0, giving x A cos vt as our solution To check
this solution, notice that it satisfies the condition that x(0) A because cos 0 1.
The position, velocity, and acceleration versus time are plotted in Figure 15.7a
for this special case The acceleration reaches extreme values of v2A when the
position has extreme values of A Furthermore, the velocity has extreme values
of vA, which both occur at x 0 Hence, the quantitative solution agrees with
our qualitative description of this system
Let’s consider another possibility Suppose the system is oscillating and we
define t 0 as the instant the particle passes through the unstretched position of
the spring while moving to the right (Active Fig 15.8) In this case, our solutions
for x(t) and v(t) must obey the initial conditions that x(0) 0 and v(0) v i:
The first of these conditions tells us that f p/2 With these choices for f, the
second condition tells us that A v i/v Because the initial velocity is positive and
the amplitude must be positive, we must have f p/2 Hence, the solution is
The graphs of position, velocity, and acceleration versus time for this choice of t
0 are shown in Figure 15.7b Notice that these curves are the same as those in
Fig-ure 15.7a, but shifted to the right by one fourth of a cycle This shift is described
mathematically by the phase constant f p/2, which is one fourth of a full
Sign in at www.thomsonedu.comand
go to ThomsonNOW to compare the oscillations of two blocks starting from different initial positions and see that the frequency is independent
Figure 15.7 (a) Position, velocity, and acceleration versus time for a block undergoing simple
har-monic motion under the initial conditions that at t 0, x(0) A and v(0) 0 (b) Position, velocity,
and acceleration versus time for a block undergoing simple harmonic motion under the initial
condi-tions that at t 0, x(0) 0 and v(0) v.
The block–spring system is
undergo-ing oscillation, and t 0 is defined
at an instant when the block passes through the equilibrium position
x 0 and is moving to the right with
speed v i.
Sign in at www.thomsonedu.comand
go to ThomsonNOW to compare the oscillations of two blocks with differ-
ent velocities at t 0 and see that the frequency is independent of the amplitude.
Trang 11Use Equation 15.9 to find the angular frequency of
the block-spring system:
hori-(A)Find the period of its motion
SOLUTION
Conceptualize Study Active Figure 15.6 and imagine the block moving back and forth in simple harmonic motiononce it is released Set up an experimental model in the vertical direction by hanging a heavy object such as a staplerfrom a strong rubber band
Categorize The block is modeled as a particle in simple harmonic motion We find values from equations oped in this section for the particle in simple harmonic motion model, so we categorize this example as a substitu-tion problem
(C)What is the maximum acceleration of the block?
SOLUTION
Use Equation 15.17 to find vmax: vmax vA 15.00 rad>s2 15.00 102 m2 0.250 m>s
(D) Express the position, velocity, and acceleration as functions of time
SOLUTION
Use Equation 15.18 to find amax: amax v2A 15.00 rad>s2215.00 102 m2 1.25 m>s2
Find the phase constant from the initial condition that
x A at t 0: x 102 A cos f A S f 0
Use Equation 15.6 to write an expression for x(t): x A cos 1vt f2 10.050 0 m2 cos 5.00t Use Equation 15.15 to write an expression for v(t): v vA sin 1vt f2 10.250 m>s2 sin 5.00t
Use Equation 15.16 to write an expression for a(t): a v2A cos 1vt f2 11.25 m>s22cos 5.00t
What If? What if the block were released from the same initial position, x i 5.00 cm, but with an initial velocity of
v i 0.100 m/s? Which parts of the solution change and what are the new answers for those that do change?
Answers Part (A) does not change because the period is independent of how the oscillator is set into motion Parts(B), (C), and (D) will change
Write position and velocity expressions for the initial
conditions:
(1)(2) v 102 vA sin f v
x 102 A cos f x i
Trang 12As we saw in Chapters 7 and 8, many problems are easier to solve using an energy approach rather than one based
on variables of motion This particular What If? is easier to solve from an energy approach Therefore, we shall
inves-tigate the energy of the simple harmonic oscillator in the next section
Divide Equation (2) by Equation (1) to find the phase
SOLUTION
Conceptualize Think about your experiences with automobiles When you sit in a car, it moves downward a smalldistance because your weight is compressing the springs further If you push down on the front bumper and release
it, the front of the car oscillates a few times
Categorize We imagine the car as being supported by a single spring and model the car as a particle in simple monic motion
har-Analyze First, let’s determine the effective spring constant of the four springs combined For a given extension x of
the springs, the combined force on the car is the sum of the forces from the individual springs
Watch Out for Potholes!
Find an expression for the total force on the car: Ftotal a 1kx2 1a k2x
In this expression, x has been factored from the sum because it is the same for all four springs The effective spring
constant for the combined springs is the sum of the individual spring constants
Evaluate the effective spring constant: keff a k 4 20 000 N>m 80 000 N>m
Use Equation 15.14 to find the frequency of vibration: f 1
Trang 1315.3 Energy of the Simple
Harmonic Oscillator
Let us examine the mechanical energy of the block-spring system illustrated inActive Figure 15.1 Because the surface is frictionless, the system is isolated and weexpect the total mechanical energy of the system to be constant We assume amassless spring, so the kinetic energy of the system corresponds only to that of theblock We can use Equation 15.15 to express the kinetic energy of the block as
(15.19)
The elastic potential energy stored in the spring for any elongation x is given by
(see Eq 7.22) Using Equation 15.6 gives
(15.20)
We see that K and U are always positive quantities or zero Because v2 k/m, we
can express the total mechanical energy of the simple harmonic oscillator as
From the identity sin2u cos2u 1, we see that the quantity in square brackets isunity Therefore, this equation reduces to
(15.21)
That is, the total mechanical energy of a simple harmonic oscillator is a constant
of the motion and is proportional to the square of the amplitude. The totalmechanical energy is equal to the maximum potential energy stored in the spring
when x A because v 0 at these points and there is no kinetic energy At the equilibrium position, where U 0 because x 0, the total energy, all in the form
of kinetic energy, is again Plots of the kinetic and potential energies versus time appear in Active Figure15.9a, where we have taken f 0 At all times, the sum of the kinetic and poten-tial energies is a constant equal to , the total energy of the system
The variations of K and U with the position x of the block are plotted in Active
Figure 15.9b Energy is continuously being transformed between potential energystored in the spring and kinetic energy of the block
Active Figure 15.10 illustrates the position, velocity, acceleration, kinetic energy,and potential energy of the block-spring system for one full period of the motion.Most of the ideas discussed so far are incorporated in this important figure Study
What If? Suppose the car stops on the side of the road and the two people exit the car One of them pushes ward on the car and releases it so that it oscillates vertically Is the frequency of the oscillation the same as the value
down-we just calculated?
Answer The suspension system of the car is the same, but the mass that is oscillating is smaller: it no longerincludes the mass of the two people Therefore, the frequency should be higher Let’s calculate the new frequencytaking the mass to be 1 300 kg:
As predicted, the new frequency is a bit higher
simple harmonic oscillator
Total energy of a simple
harmonic oscillator
Kinetic energy of a simple
harmonic oscillator
Trang 14Finally, we can obtain the velocity of the block at an arbitrary position by
expressing the total energy of the system at some arbitrary position x as
(15.22)
When you check Equation 15.22 to see whether it agrees with known cases, you
find that it verifies that the speed is a maximum at x 0 and is zero at the turning
points x A.
You may wonder why we are spending so much time studying simple harmonic
oscillators We do so because they are good models of a wide variety of physical
phenomena For example, recall the Lennard–Jones potential discussed in
Exam-ple 7.9 This complicated function describes the forces holding atoms together
Figure 15.11a (page 428) shows that for small displacements from the equilibrium
position, the potential energy curve for this function approximates a parabola,
which represents the potential energy function for a simple harmonic oscillator
Therefore, we can model the complex atomic binding forces as being due to tiny
springs as depicted in Figure 15.11b
sim-tor In either plot, notice that K U constant.
Sign in at www.thomsonedu.comand go to ThomsonNOW to compare the physical oscillation of a block with energy graphs
in this figure as well as with energy bar graphs.
w
1 0
Total energy
Potential energy
Several instants in the simple harmonic motion for a block–spring system Energy bar graphs show the distribution of the energy of the system at
each instant The parameters in the table at the right refer to the block–spring system, assuming that at t 0, x A; hence, x A cos vt.
Sign in at www.thomsonedu.comand go to ThomsonNOW to set the initial position of the block and see the block–spring system and the gous energy bar graphs.
Trang 15The ideas presented in this chapter apply not only to block-spring systems andatoms, but also to a wide range of situations that include bungee jumping, tuning
in a television station, and viewing the light emitted by a laser You will see moreexamples of simple harmonic oscillators as you work through this book
r U
Figure 15.11 (a) If the atoms in a molecule do not move too far from their equilibrium positions, a graph of potential energy versus separation distance between atoms is similar to the graph of potential energy versus position for a simple harmonic oscillator (dashed blue curve) (b) The forces between atoms in a solid can be modeled by imagining springs between neighboring atoms.
Conceptualize The system oscillates in exactly the same way as the block in Active Figure 15.10
Categorize The cart is modeled as a particle in simple harmonic motion
Oscillations on a Horizontal Surface
Analyze Use Equation 15.21 to find the energy of the
When the cart is at x 0, the energy of the oscillator is
entirely kinetic, so set E 12mv2 :
The positive and negative signs indicate that the cart could be moving to either the right or the left at this instant
(C)Compute the kinetic and potential energies of the system when the position is 2.00 cm