Download Part2 - Physics 3

Example 14-7
A block securely attached to a spring oscillates vertically with a frequency of 4.00 Hz and an
amplitude of 7.00 cm. A tiny bead is placed on top of the oscillating block just as it reaches
its lowest point. Assume that the bead's mass is so small that its effect on the motion of the
block is negligible. At what displacement from the equilibrium position does the bead lose
contact with the block?
PICTURE The forces on the bead are its weight mg downward and the upward normal force
exerted by the block. The magnitude of this normal force changes as the acceleration
changes. As the block moves upward from equilibrium, its acceleration and the acceleration of
the bead are downward and increasing in magnitude. When the acceleration reaches g downward, the normal force will be zero. If the block's downward acceleration becomes even
slightly larger, the bead will leave the block.
Y
SOLVE
1. Draw a sketch of the system (Figure 14-12). Include a y coordinate axis with its origin at
the equilibrium position and with down as the positive direction:
2. We are looking for the value of y when the
acceleration is g downward. Use Equation 14-7:
ay =
-w2y
g = -w2y
g
so
=
-(27Tf?y
y
= -
g
(27Tf?
2
9.81 m/s
= - [27T(4.00Hz)]2 = -0.0155 m =
I
-1.55 cm
I
CHECK The bead leaves the block when y is negative, which is when the bead is above the
equilibrium position because down was chosen as the positive y direction. This is as
expected.
A simple pendulum consists of a string of length L and a bob of mass m. When the
bob is released from an initial angle 1>0with the vertical, it swings back and forth
with some period T. The units of length, mass, and g, are m, kg, and m/s2, respectively. If we divide L by g, the meters cancel and we are left with seconds squared,
suggesting the form YfJi. If the formula for the period contains the mass, then the
unit kg must be canceled by some other quantity. But there is no combination of L
and g that can cancel mass units. So the period cannot depend on the mass of the
bob. Because the initial angle 1>0is dimensionless, we cannot tell whether or not it
is a factor in the period. We will see below that for small 1>0'the period is given by
T
=
271'YfJi.
We might expect the period of a
simple pendulum to depend on
the mass m of a pendulum bob,
the length L of the pendulum, the
acceleration due to gravity g, and
the initial angle 1>0'Find a simple
combination of some or all of
these quantities that gives the correct dimensions for the period.
A Foucault pendulum at the University of Louisville. In
1851,Leon Foucault suspended a 67-m-long pendulum
from the ceiling of the Pantheon in Paris. Because of the
rotation of Earth about its axis, the Pantheon rotates
about the pendulum. (If the Pantheon were at the North
Pole, it would rotate once every 24 hours.) The
observation of the building rotating about the plane of the
pendulum captured the imagination of the world.
(Courtesy of John Kielkopf/Universih) of Louisville,)
mg
The forces on the bob are its weight
and the string tension T (Figure 14-13).
_-\ an angle 4> with the vertical, the weight has components mg cos 4> along the
:::ring and mg sin 4> tangential to the circular arc in the direction of decreasing 4>.
:':sing tangential components, Newton's second law (LFt = mat) gives
-mg sin 4>
=
\
d2s
m- 2
dt
-,'here the arc length s is related to the angle 4> by s
. th sides of s = L4> gives
=
L
L4>. Repeatedly differentiating
:\Totethat the mass m does not appear in Equation 14-25-the motion of a pendulum
does not depend on its mass. For small 4>, sin 4> = 4>, and
FIG U R E 1 4 - 1 3
Equation 14-26 is of the same form as Equation 14-2 for an object on a spring. Thus,
the motion of a pendulum approximates simple harmonic motion for small angular
displacements.
Equation 14-26 can be written
where
T
= 27r = 21T
w
g
is the angular frequency-not
the angular speed-of the motion
of the pendulum.
w
w2 =-
L
[f
'Yg
where 4>0 is the maximum angular displacement.
According to Equation 14-28, the greater the length of a pendulum, the greater
the period, which is consistent with experimental observation. The period and
therefore the frequency are independent of the amplitude of oscillation (as long as
the amplitude is small). This statement is a general feature of simple harmonic
motion.
PRACTICE
PROBLEM
14-5
Find the period of a simple pendulum
of length 1.00 m undergoing
Forces on a pendulum
bob.
small oscillations.
The acceleration due to gravity can be measured using a simple pendulum undergoing small oscillations. We need only measure the length L and period T of the pendulum, and using Equation 14-28, solve for g. (To measure T, we usually measure the
time for n oscillations and then divide by n, which minimizes measurement error.)
In a physics lab on one-dimensional kinematics, Liz and Bob are tasked with measuring the
time it takes for a glider released from rest on an inclined 2.00-m-Iong air track to travel various distances. (An air track is a virtually frictionless track.) They tilt the track by putting a
2.0-cm-thick notebook under the legs at one end of the track. They release the glider from
the middle of the track and find the time for it to accelerate half the length of the track to be
4.8 s. They then release the glider from the high end of the track and find that the time it
takes for the glider to accelerate the entire length of the track is 4.8 s-the same time it took
to accelerate half the length of the track. Thinking that the times for the two distances cannot be equal, they repeat both measurements, only to obtain the same results. Confused, they
ask the instructor for an explanation. Can you think of a plausible explanation?
PICTURE If the track were perfectly straight, the acceleration would be the same everywhere along the track and the time for the glider to accelerate the entire length of the track,
starting from rest, would be greater than the time for it to accelerate only half the length of
the track. If the track sagged slightly, however, then the acceleration would be greatest at the
high end of the track where the slope is steepest. What would the assumption that the track
is sagging predict?
SOLVE
1. Suppose the track sags slightly, in such a way that the track
forms a circular arc whose center of curvature is directly above
the low end of the track:
If the track sags as supposed, then the glider would move like the
bob of a simple pendulum of length L = R, where R is the radius
of curvature of the track.
2. The period T of a pendulum is independent
small amplitudes:
The times measured by Liz and Bob would equal ~ the period T of
the pendulum, given by Equation 14-28. Because the period of a
pendulum is independent of amplitude (for small amplitudes), the
times measured by Liz and Bob would be expected to be equal.
of amplitude for
Is the amplitude of the pendulum sufficiently small when the glider is released
from the high end of the track? It is if the R is much greater than 2.00 m. Equation 14-28 tells
us that the length of the pendulum is given by L = gTZ/(47T2). Substituting 4 x (4.8 s) for T
gives R = L = 92 m, justifying the supposition that the amplitudes were small.
CHECK
Pendulum in an accelerated reference frame Figure 14-14a shows a simple
pendulum suspended from the ceiling of a boxcar that has acceleration o' relative
to the ground, to the right, and
is the acceleration of the bob relative to the
ground. Applying Newton's second law to the bob gives
a
a
2:.F = T
+
mg = ma
14-29
If the bob remains at rest relative to the boxcar, then
2:.F = T sin eo = mao
2:.Fy = T cos eo - mg
a = ao and
This clock keeps time by using a torsional
oscillator. (Courtesy of Bill MasterlAlibaba.
http://yuning.en.aliba ba.com.)
t
=
a
where eo is the equilibrium angle. Thus, eo is given by tan eo
•....•
\....,u\....>w"--''-''---'
(a)
=
ao/g. If the bob is
FIG U R E 1 4 - 1 4 (a) Simple
pendulum in apparent equilibrium
in an accelerating boxcar. Forces
are those as seen from a separate
stationary frame. (b) Forces on the
bob as seen in the accelerated
frame. Adding the pseudoforce
-mao is equivalent to replacing g
by of.
.=-cwingrelative to the boxcar, then a' = a
_ :..relative to the boxcar. Substituting for
"2.F
="
tracting
mao from both
T + mg
=
- aD' where a'
a in Equation
=
is the acceleration of the
14-29 gives
m(a' + ao)
sides of this equation and rearranging terms gives
T
+ mg'
a
=
ma'
g' and a by a'
ere g' = g - o' Thus, by replacing g by
in Equatio~ 14-29 we
solve for the motion of the bob relative to the boxcar. The vectors T and mg'
.=.-c shown in Figure 14-14b. If the string breaks so that T = 0, then our equation
~yes a' = g', which means that g' is the free-fall acceleration in the reference
.=-ameof the boxcar. If the bob is displaced slightly from equilibrium, it will oscil2.:e with a period T given by Equation 14-28 with g replaced by g'.
~
PRACTICE
PROBLEM
14-6
A.simple pendulum of length 1.00 m is in a boxcar that is accelerating
acceleration ao = 3.00 m/s2• Find g' and the period T.
horizontally
with
1.05
1.04
IIIo
_Jrge-AmpJitude oscillations When the amplitude of a pendulum's os:illation becomes large, its motion continues to be periodic, but it is no longer
:3. simple harmonic. For an angular amplitude cPo' the period can be shown to
given by
1.03
1.02
1.01
T
=
1 2-A.1
T [ 1 + -sin
o
22
2 '1'0
1 (3)2
1
+ -22
-4 sin4-A.
+ ... ]
2 '1'0
1
0
27TVLii
"'here To =
is the period for very small amplitudes. Figure 14-15 shows
-=- To as a function of amplitude cPo'
___
0.2
0.4
0.6
Amplitude I/Jo, rad
FIG U R E 1 4· 1 5 Note that the values on
the vertical axis range from 1 to 1.06. Over a
range of c/> from 0 to 0.8 rad (46°), the period
varies by about 5 percent.
A Pendulum Clock
A.simple pendulum clock is calibrated to keep accurate time at an amplitude of cPo = 10.0°. When
:he amplitude has decreased to the point where it is very small, does the clock gain or lose time?
.-\bout how much time will the clock gain or lose in one day if the amplitude remains very small.
PICTURE To calculate the period when the angular amplitude
correction term to Equation 14-30. That is, use
T
=
I[l +
o
is 10°, retain only the first
.!.A.]
.!.Sin2
22
2 '1"0
This equation provides sufficient accuracy because 10° is a fairly small amplitude.
amplitude of the pendulum slowly decreases due to the effects of air drag.
The
SOLVE
Cover the column to the right and try these on your own before looking at the answers.
Steps
1. Use Equation 14-30 to determine
Answers
if To is greater or less than I.
T decreases as cPo decreases, so
I
2. Use Equation 14-30 to find the percentage change [(T - To)/T] X 100%
for cP = 10°. Use only the first correction term.
0.190%
3. Find the number of minutes in a day.
There are 1440 minutes in a day.
4. Combine steps 2 and 3 to find the change in the number of
minutes in a day.
The gain is
12.73 min/d I
the clock gains time.
I
CHECK The first correction term in Equation 14-30 is bin2(1O.0o /2) = 1.90 x 10-3, so
T = 1.00190To and (T - To)/T = (1.00190To - To)/1.00190To = 0.00190. This value agrees
with our step-2 result.
TAKING IT FURTHER To avoid this gain, pendulum-clock
keep the amplitude fairly constant.
mechanisms
are designed
to
A system that undergoes rotational oscillations in a variation of simple-harmonic
motion is called a torsional oscillator. Figure 14-16 shows a torsional oscillator
consisting of a solid disk suspended from a steel wire. If the angular displacement
of the disk from the equilibrium position is c/> then the wire exerts a linear restoring torque T on the disk given by
where K is the torsional constant of the wire. Substituting -Kc/> for
tion T = Ia (Newton's second law for rotational motion) gives
-Kc/>
where the angular acceleration a
ranging gives
=
=
d2c/>/dt2.
T
FIG U R E 1 4 - 1 6 This torsional pendulum
consists of a solid disk suspended by a steel wire.
in the equa-
Ia
Substituting d2c/>/dt2 for a and rear-
which is identical to Equation 14-2, except with I in place of m, K in place of k,
and c/> in place of x. Thus, the solution to Equation 14-32 can be written by directly
substituting into Equation 14-4. Doing so gives
where w = v;ji is the angular frequency-and
motion. The period is therefore
T
=
27r
w
= 21T
not the angular speed-of
the
f!..
\j-;
A rigid object free to rotate about a horizontal axis that is not through its center of
mass will oscillate when displaced from equilibrium. Such a system is called a
physical pendulum. Consider a plane figure with a rotation axis a distance D
from the figure's center of mass and displaced from equilibrium by the angle c/>
(Figure 14-17). The torque about the axis has a magnitude MgD sin c/>. For sufficiently small values of C/>, we can simplify our expression for the torque using the
small-angle approximation (sin c/> = c/». Thus, for small angles the torque is a linear
restoring torque given by
Comparing this with T = -Kc/> (Equation 14-31), we can see that for small angular
displacements the physical pendulum is a torsional oscillator with a torsional
constant given by
All mechanical clocks keep time because the
period of the oscillating part of the
mechanism remains constant. The period of
any pendulum changes with changes in
amplitude. However, the driving mechanism
of a pendulum clock maintains the amplitude
at a constant value. (Richard Menga/
Fundamental Photographers.)
motion of the pendulum
is therefore
:::i7
-= -
=
w
is described by Equation 14-33 with
K =
MgD.
fi£
21T -MgD
:.arge amplitudes, the period is given by Equation 14-30, with To given by
. 14-36. For a simple pendulum of length L, the moment of inertia is
and D = L. Then, Equation 14-36 gives T = 27TVMUI(MgL) = 21TVLii,
.e as Equation 14-28.
The period of a physical pendulum
depends on the distribution of the
mass, but not on the total mass M. The
moment of inertia I is proportional to
M, so the ratio 11M is independent
ofM.
-=.:
-;'m that the pace of a comfortable
-" pendulum.
=.aim correct?
walk can be predicted if we model each leg as a
Your teacher is skeptical about this claim and asks you to back it up. Is
E A simple model of each leg is that of a uniform
_
_
rod pivoted at one end. Each leg
ack and forth once every two steps, so the time required to walk 10 steps is 5T,
=- is the period of the "pendulum." How long will it take you to complete 10 steps at
Iy pace if your prediction is correct? Model your leg as a 0.90-m-long uniform rod
, about an axis through one end.
p
t
Ll2
=.
==aw and
label a uniform thin rod pivoted
t one end (Figure 14-18):
j
Ll2
_ :--e period
=-
of a physical pendulum
= 2r.VIIMgD (Equation 14-36):
is given by
!
T = 21T) M~D
: about the end is found in Table 9-1 and D is
- 1 the length of the rod:
-= e length
L = 0.90 m and the time for
:J steps is 5T:
FIG U R E 1 4 • 1 8 The distance between
the rotation axis and the center of mass is L/2.
5T = 5' 21T
L
- = 101T
~ 3g
2(0.90 m)
3(9.81 m/s2)
. [y hypothesis has merit. My hip joint is about 90 cm above
:he floor and it took me almost 6.7 s to complete 10 leisurely
steps. The upper half of my leg is more massive than the lower
nalf, so modeling my leg as a uniform rod is not completely
appropriate. In addition, what is a leisurely pace is subject to
:nterpretation.
:-::CK
Long-legged animals, like elephants and giraffes, seem to walk at a slow, lumber.: pace, and short-legged animals, like mice and sandpipers, walk at a fast pace. This
- usion is predicted by this model, because the period of a long pendulum is greater than
-=: of a short pendulum.
=
7.8 s
Example 14-11
i
A uniform rod of mass M and length L is free to swing about a horizontal axis perpendicular
to the rod and a distance x from the rod's center. Find the period of oscillation for small
angular displacements of the rod.
L/2
'x J~
'.
i 1\cm
!
PICTURE The period is given by Equation 14-36. The center of mass is at the center of the
rod, so the distance from the center of mass to the rotation axis is x (Figure 14-19). The moment of inertia of a uniform rod can be found from the parallel-axis theorem 1 = lern + MD2
(Equation 9-13), where lern can be found in Table 9-1.
L/2
SOLVE
1. The period is given by Equation 14-36:
2. D = x, and the moment of inertia is given by
the parallel-axis theorem. The moment of
inertia about a parallel axis through the center
of mass is found in Table 9-1:
I
t
FIG U R E 1 4 - 1 9 The distance between
the rotation axis and the center of mass is x.
T = 27r~ M~D
D=x
1 = lern
+
+
MD2 = f2ML2
Mx2
+ Mx2)
(f2MU
T = 21T~ M~D = 21T
Mgx
As x ~ 0, T ~ '" as expected. (If the rotation axis of the rod passes through its center of mass, we do not expect gravity to exert a restoring torque.) Also, if x = L/2 we get
T = 21TY2L/3g,
the same result as found in step 4 of Example 14-10. In addition, if x» L
the expression for the period approaches T = 21T%,
which is the expression for the period of a simple pendulum of length x (Equation 14-28).
CHECK
IT FURTHER The period T versus distance x from the center of mass for a rod of
length 1.00 m is shown in Figure 14-20.
TAKING
PRACTICE
period for x
14-7 Show that the step-3 expression
L/6 as for x = L/2.
PROBLEM
=
for the period gives the same
FIG U R E 1 4 - 2 0 Plot of the period versus
the distance from the pivot to the center of
mass. For x > 0.5 m the pivot point is beyond
the end of the rod.
Example 14-12
SOLVE
Cover the column to the right and try these on your own before looking at the answers.
Steps
Answers
1. The period, given by the Example 14-11 result, is
T = 21TVZfg, where Z = (f2U + x2)/x. Find the period both
as x approaches zero and as x approaches infinity.
T
=
21T
(f2 U
gx
where Z = (f2 U
As x~
2
+x) =
g
'V-g
21T
+ x2)/x
0, Z ~ "', and T ~"'.
As x ~ "', Z ~ "', and T ~ "'.
-
~ e period
~?proaches
?E'riod is a
set it equal
goes to infinity as x approaches zero and as x
infinity. Somewhere in the range 0 < x < 00 the
minimum. To find the minimum, evaluate dT / dx,
to zero, and solve for x.
dT
dx
dT dZ
dZ dx
=
~Z-1/2dZ
vg
dx
Z > 0 throughout the range 0
dT
dZ
dx = 0 => dx = O.
dZ
-
dx
= - ECK
=
We expect an answer between 0 and O.5L. The step-2 result of x
=tation.
=
=
0 => x
=
•
L
t.::.
v12
=
<x <
00,
so
O.289L
O.289L meets that
0 itself, a spring or a pendulum eventually stops oscillating because the me-:tical energy is dissipated by frictional forces. Such motion is said to be damped.
_ ..:e damping is large enough, as, for example, a pendulum submerged in
--,' ses, the oscillator fails to complete even one cycle of oscillation. Instead, it just
-es toward the equilibrium position with a speed that approaches zero as the
approaches the equilibrium position. This type of motion is referred to as
damped. If the damping is small enough that the system oscillates with an am- --.: de that decreases slowly with time-like a child on a playground swing when
:G....~ntstops providing a push each cycle-the motion is said to be underdamped.
":on with the minimum damping for nonoscillatory motion is said to be
. ally damped. (With any less damping, the motion would be underdamped.)
~_
erdamped motion The damping force exerted on an oscillator such as the
,; shown in Figure 14-21a can be represented by the empirical expression
Fd
=
-bv
b is a constant. Such a system is said to be linearly damped. The discussion
_ is for linearly damped motion. Because the damping force is opposite to
c direction of motion, it does negative work and causes the mechanical energy of
Om?
FIG U R E 1 4 - 2 1 (a) A damped oscillator suspended in a viscous liquid. The motion of
the cylinder is damped by drag forces. (b) Damped oscillation curve.
the system to decrease. This energy is proportional to the square of the amplitude
(Equation 14-17), and the square of the amplitude decreases exponentially with increasing time. That is,
where A is the amplitude, Ao is the amplitude at t = 0, and T is the decay time or
time constant. The time constant is the time for the energy to change by a factor
of e-1.
The motion of a damped system can be obtained from Newton's second law. For
an object of mass m on a spring that has a force constant k, the net force is
-kx - b(dx/dt). Setting the net force equal to the mass times the acceleration
d2x/dt2, we obtain
dx
-kx - bdt
d2x
m- 2
dt
=
d2x
m-2
dt
dx
dt
+ b- + kx = 0
The exact solution of this equation can be found using standard methods for solving
differential equations. The solution for the underdamped case is
where Ao is the initial amplitude. The frequency w' is related to the natural frequency
W (the frequency with no damping) by
o
W'
= w
o
)1 _ (_b
2mw o
)2
For a mass on a spring Wo = ~.
For weak damping, b/(2mwo) « 1 and w' is
nearly equal to WOo The dashed curves in Figure 14-21b correspond to x = A and
x = - A, where A is given by
By squaring both sides of this equation
Equation 14-37, we have
and comparing
the results
with
m
b
T =-
If the damping constant b is gradually increased, the angular frequency w' decreases
until it becomes zero at the critical value
When b is greater than or equal to be' the system does not oscillate. If b > be' the
system is overdamped. The smaller b is, the more rapidly the object returns to equi:ibrium. If b = be' the system is said to be critically damped and the object returns
:0 equilibrium
(without oscillation) very rapidly. Figure 14-22 shows plots of the
. placement versus time for a critically damped and an overdamped oscillator.
We often use critical damping when we want a system to avoid oscillations and yet
return to equilibrium quickly.
Critically damped
Overdamped
',I,
,,
' ... ...-
--
FIG U R E 1 4 - 2 2 Plots of displacement
versus time for a critically damped and an
overdamped oscillator, each released from rest.
__
Sprung Mass of a Passenger Car
The sprung mass of an automobile is the mass that is supported by the springs. (It does not
include the mass of the wheels, axles, brakes, and so on.) A passenger car has a sprung mass
of 1100 kg and an unsprung mass of 250 kg. If the four shock absorbers are removed, the car
bounces up and down on its springs with a frequency of 1.0 Hz. What is the damping constant provided by the four shocks if the car, with shocks, is to return to equilibrium as
quickly as possible without passing it after hitting a speed bump?
PICTURE Because the car returns to equilibrium as quickly as possible without passing it,
we know the car is a critically damped oscillator. Use be = 2mwo (Equation 14-43) to solve for
the damping constant for critical damping.
SOLVE
1. The damping constant for critical damping is related to the
natural frequency by be = 2mwo (Equation 14-43):
2. With the tires in contact with the pavement,
the sprung mass enters the picture:
3. The natural frequency
only the inertia of
o is given in the problem statement:
o = 1.0 Hz
W
W
b = be = 2(1100 kg)/(1.0 Hz) = 12.2
X
103 kg/s
I
CHECK The damping force is given by F = -bv, so bv has 51 units of newtons. Our step-4
value for b has units of kg/s, so bv has units of (kg/s)(m/s) = kg' m/s2, which are the 51 units
for mass times acceleration. Thus, kg/s are appropriate units for b.
TAKING IT FURTHER The optimal shock absorber for any vehicle is a shock absorber that
has a damping constant such that the oscillations are critically damped. Thus, the optimal
choice for the critical damping constant be is determined by the sprung mass of the vehicle
and the force constant k of the suspension springs.)
Because the energy of an oscillator is proportional to the square of its amplitude,
the energy of an underdamped
oscillator (averaged over a cycle) also decreases
exponentially with time:
!0'0
V
1-. -;:.
61-
W"£\¥
(
LA?
where Eo = ~mw2A6 and T = m/b.
(7- K
0
J
A damped oscillator is often Jdes~rifed by its Q factor (for quality factor),
Q ~
W,T
~
LJ" ( ~
l
vv
~
W0
/-
DEFtTION-Q
v~
14-45
FACTOR
Weights are placed in automobile wheels
when the wheels are "balanced." The purpose
of balancing the wheels is to prevent
vibrations that will drive oscillations of the
wheel assembly. (David Wrobel/ Visuals
Unlimited.)
The Q factor is dimensionless. (Because wa has dimensions
is without dimension.) We can relate Q to the fractional
Differentiating Equation 14-44 gives
dE
dt =
-(1/r)Eae-t/
T
= -(1/r)E
or
dE
E=
of reciprocal time, war
energy loss per cycle.
dt
--;;
If the damping is weak so that the energy loss per cycle is a small fraction of the
energy E, we can replace dE by IlE and dt by the period T. Then IIlEI!E in one cycle
(one period) is given by
C~EI)cYcle
T
27T
27T
r
war
Q
so
27T
Q
=
(1IlEI/E)cycle
IIlEI « 1
E
Example 14-14
When middle C on a piano (frequency 262 Hz) is struck, it loses half its energy after 4.00 s.
(a) What is the decay time T? (b) What is the Q factor for this piano wire? (e) What is the fractional energy loss per cycle?
PICTURE
(a) We use E = Eoe-tlT and set E equal to !Eo' (b) The Q value can then be found
from the decay time and the frequency.
SOLVE
(a) 1. Set the energy at time t
energy:
=
4.00 s equal to half the original
so
E = Eoe-tlT
!=
!Eo = Eoe-(4.00
s/T)
e-(4.00s/T)
In.!- = _ 4.00 s
2
so
Q
T
4.00 s
r;:;;:;-:-,
T = 1n2 = 5.771 = ~
=
WOT
=
27TfT
= 27T(262 Hz)(5.771 s) = 9.500
(e) The fractional energy loss in a cycle is given by Equation 14-46
and the frequency f = liT:
(I~EI)
E
T
cycle
=
WOT
6.614
CHECK Q can also be calculated from Q = 27T/(~EIE)cYcle
= 27T/(6.61 X 10-4) = 9.50 X 103.
Note that the fractional energy loss after 4.00 s is not just the number of cycles (4.00 X 262)
times the fractional energy loss per cycle, because the energy decreases exponentially, not
linearly.
IT FURTHER Figure 14-23 shows the relative amplitude AI Ao versus time and the
relative energy EIEo versus time for the oscillation of a piano string after middle C is struck.
After 4.00 s, the amplitude has decreased to about 0.7 times its initial value, and the energy,
which is proportional to the amplitude squared, drops to about half its initial value.
TAKING
FIG U R E 1 4 - 2 3 Plots of AI Ao and EIEo
for a struck piano string.
X
fT
10-4
X
103 = 19.50
(262 Hz)(5.771 s)
=
16.61
X
10-41
x
1031
:--Jotethat the value of Q in Example 14-4 is relatively large. You can estimate T
.md Q for various oscillating systems. Tap a crystal wine glass and see how long it
~gs. The longer it rings, the greater the value of T and Q and the lower the
amping. Glass beakers from the laboratory may also have a high Q. Try tapping
a plastic cup. How does the damping compare to that of the glass beaker?
In terms of Q, the exact frequency of an underdamped oscillator is
w' = w
o\j
(_b )2
/1 -
2mw
)1
= w
_
0
o
1
4Q2
14-48
Because b is quite small (and Q is quite large) for a weakly damped oscillator
(Example 14-14), we see that Wi is nearly equal to WoWe can understand much of the behavior of a weakly damped oscillator by
considering its energy. The power dissipated by the damping force equals the
instantaneous rate of change of the total mechanical energy
= -dE = F~ . v- = -bv·- v- = -bv
P
dt
2
d
For a weakly damped oscillator with linear damping, the total mechanical energy
decreases slowly with time. The average kinetic energy per cycle equals half the
total energy
2)
('!mv
2
If we substitute (v2)av
=
=.! E
(v2)
or
2
av
~
=~
m
Elm for v2 in Equation 14-49, we have
dE
-
dt
= -bv2 = -b(v2)
av
b
m
= --E
Rearranging Equation 14-50 gives
dE
-
E
=
b
--dt
m
which upon integration gives
E
14-5
= E e-(bll1l)t
o
= Eoe-II'
DRIVEN OSCillATIONS AND RESONANCE
To keep a damped system going indefinitely, mechanical energy must be put into
the system. When this is done, the oscillator is said to be driven or forced. When
Mom (or Dad) kept your swing going by pushing on it once each cycle, she was
driving an oscillator. Likewise, when you keep a swing going by "pumping," you
are driving an oscillator. If the driving mechanism
puts energy into the system at a greater rate than it
is dissipated, the system's mechanical energy increases with time, and the amplitude increases. If the
driving mechanism puts energy in at the same rate it
~
is being dissipated, the amplitude remains constant
over time. The motion of the oscillator is then said to
be steady-state motion.
Figure 14-24 shows a system consisting of an object on a spring that is being driven by moving the
point of support up and down with simple harmonic
motion of frequency w. At first the motion is complicated, but eventually steady-state
motion is
reached in which the system oscillates with the same
By pumping the swing, the young woman is
transferring her internal energy into the
mechanical energy of the oscillator.
(Eye Wire! Gettlj,)
FIG U R E 1 4 - 2 4 An object on a vertical
spring can be driven by moving the support
up and down.
frequency as that of the driver and with a constant amplitude and, therefore, at
constant energy. In the steady state, the energy put into the system per cycle by the
driving force equals the energy dissipated per cycle due to the damping.
The amplitude, and therefore the energy, of a system in the steady state depends
not only on the amplitude of the driving force, but also on its frequency. The
natural frequency of an oscillator, wo' is its frequency when no driving or damping
forces are present. (In the case of a spring, for example, Wo = Yk/m.) If the driving
frequency is sufficiently close to the natural frequency of the system, the system
will oscillate with a relatively large amplitude. For example, if the support in
Figure 14-24 oscillates at a frequency close to the natural frequency of the massspring system, the mass will oscillate with a much greater amplitude than it would
if the support oscillates at significantly higher or lower frequencies. This phenomenon is called resonance. When the driving frequency equals the natural frequency
of the oscillator, the energy per cycle transferred to the oscillator is maximum. The
natural frequency of the system is thus called the resonance
frequency.
(Mathematically, the angular frequency w is more convenient to use than the
frequency f(f = W/27T). Because wand
f are proportional, most statements
concerning angular frequency also hold for frequency. In verbal descriptions,
we usually omit the word angular when the omission will not cause confusion.)
Figure 14-25 shows plots of the average power delivered to an oscillator as a
function of the driving frequency for two different values of damping. These
curves are called resonance curves. When the damping is weak (large Q), the
wid th of the peak of the resonance curve is correspondingly narrow, and we speak
of the resonance as being sharp. For strong damping, the resonance curve is broad.
The width of each resonance curve b..w, indicated in the figure, is the width at half
the maximum height. For weak damping, the ratio of the width of the resonance
to the resonant frequency can be shown to equal the reciprocal of the Q factor
(see Problem 106):
Thus, the Q factor is a direct measure of the sharpness of resonance.
You can do a simple experiment to demonstrate resonance. Hold a meterstick at
one end between two fingers so that it acts like a pendulum. (If a meterstick is not
available, use whatever is convenient. A golf club works fine.) Release the stick
from some initial angular displacement and observe the natural frequency of its
motion. Then, move your hand back and forth horizontally, driving it at its natural
frequency. Even if the amplitude of the motion of your hand is small, the stick will
oscillate with a substantial amplitude. Now move your hand back and forth at a
frequency two or three times the natural frequency and note the decrease in amplitude of the oscillating stick.
There are many familiar examples of resonance. When you sit on a swing, you
learn intuitively to pump with the same frequency as the natural frequency of the
swing. Many machines vibrate because they have rotating parts that are not in perfect balance. (Observe a washing machine in the spin cycle, for example.) If such a
machine is attached to a structure that can vibrate, the structure becomes a driven
oscillatory system that is set in motion by the machine. Engineers pay great attention to balancing the rotary parts of such machines, damping their vibrations, and
isolating them from building supports.
A crystal goblet with weak damping can be broken by an intense sound wave
at a frequency equal to or very nearly equal to the natural frequency of vibration
of the goblet. The breaking of the goblet is often done in physics demonstrations
using an audio oscillator, a loudspeaker and an amplifier.
Weak damping,
high Q
Heavy damping,
lowQ
FIG U R E 1 4 • 2 5 Resonance for an
oscillator. The width I1w of the resonance peak
for a high-Q oscillator (the orange curve) is
small compared to the natural frequency of wOo
The resonance peak of the low-Q oscillator
(the blue curve) with the same natural
frequency has a width that is considerably
larger that that for the high-Q oscillator.
Extended objects have more than one
resonance frequency. When plucked, a guitar
string transmits its energy to the body of the
guitar. The body's oscillations, coupled to
those of the air mass it encloses, produce the
resonance patterns shown. (Royal Swedish
Academy of Music.)
" -e can treat a driven oscillator mathematically
by assuming that, in addition to the
_ - oring force and a damping force, the oscillator is subject to an external driving
:-orcethat varies harmonically with time:
ere Fa and ware the amplitude and angular frequency of the driving force. This
::-equency is generally not related to the natural angular frequency of the system wa'
_ ewton's second law applied to an object that has a mass m attached to a spring
=-ut has a force constant k and subject to a damping force -bvx and an external
:-'rce fa cos wt gives
-,- ere we have used ax
:earranging gives
=
d2x/dt2•
Substituting
mW6 for k (Equation 14-8) and
We now discuss the general solution of Equation 14-53 qualitatively. It consists of two parts, the transient solution and the steady-state solution. The
transient part of the solution is identical to that for a damped oscillator given in
Equation 14-39. The constants in this part of the solution depend on the initial
conditions. Over time, this part of the solution becomes negligible because of the
exponential decrease of the amplitude. We are then left with the steady-state
solution, which can be written as
e-1fve'7l\'l.
\-1'\ r\-~
where the angular frequency
amplitude A is given by
tan 0
=
w
is the same as that of the driving force. The
bw
(2
m Wo - w 2)
Comparing Equations 14-52 and 14-54, we can see that the displacement and the
driving force oscillate with the same frequency, but they differ in phase by o. When
the driving frequency w approaches zero, 0 approaches zero, as can be seen from
Equation 14-56. At resonance, w equals Wo and 0 equals 90°, and when w is much
greater than Ww 0 approaches 180°. At the beginning of this chapter, the
displacement of a particle undergoing simple harmonic motion is written
x = A cos(wt + 0) (Equation 14-4). This equation is identical to Equation 14-54
except for the sign preceding the phase constant o. The phase of a driven oscillator always lags behind the phase of the driving force. The negative sign in
Equation 14-54 ensures that 0 is always positive (rather than always negative).
In your simple experiment to drive a meterstick by moving your hand back and
forth (see discussion immediately following Equation 14-51), you should note that
at resonance the oscillation of your hand is neither in phase nor 180° out of phase
with the oscillation of the stick. If you move your hand back and forth at a
frequency several times the natural frequency of the pendulum, the stick's steadystate motion will be almost 180° out of phase with your hand.
The velocity of the object in the steady state is obtained by differentiating x with
respect to t:
v
v
x
=
dx
= x
dt
=
-wA sin(wt
-wAsin(wt
-
7:.)
2
=
- 0)
+wA coswt
Thus, at resonance, the object is always moving in the direction of the driving
force, as would be expected for maximum power input. The velocity amplitude wA
is maximum at w = Wo-
At resonance, the object is always
moving in the direction of the
driving force, as would be expected for
maximum power input.