-
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14-1
Conceptual Questions
14.1. The period of a block oscillating on a spring is 2 /T m k=
. We are told that 1 2 0 sT = . . (a) In this case the mass is
doubled: 2 12m m= .
22 2 1
1 1 11
2 / 2 22 /
m kT m mT m mm k
= = = =
So 2 12 2(2 0 s) 2 8 sT T= = . = . . (b) In this case the spring
constant is doubled: 2 12k k= .
22 1 1
1 2 11
2 / 122 / 2
m kT k kT k km k
= = = =
So 2 1/ 2 (2 0 s)/ 2 1 4 sT T= = . = . . (c) The formula for the
period does not contain the amplitude; that is, the period is
independent of the amplitude. Changing (in particular, doubling)
the amplitude does not affect the period, so the new period is
still 2.0 s. It is equally important to understand what doesnt
appear in a formula. It is quite startling, really, the first time
you realize it, that the amplitude doesnt affect the period. But
this is crucial to the idea of simple harmonic motion. Of course,
if the spring is stretched too far, out of its linear region, then
the amplitude would matter.
14.2. The period of a simple pendulum is 2 /T L g= . We are told
that 1 2 0 sT = . . (a) In this case the mass is doubled: 2 12m m=
. However, the mass does not appear in the formula for the period
of a pendulum; that is, the period does not depend on the mass.
Therefore the period is still 2.0 s. (b) In this case the length is
doubled: 2 12L L= .
22 2 1
1 1 11
2 / 2 22 /
L gT L LT L LL g
= = = =
So 2 12 2(2 0 s) 2 8 sT T= = . = . . (c) The formula for the
period of a simple small-angle pendulum does not contain the
amplitude; that is, the period is independent of the amplitude.
Changing (in particular, doubling) the amplitude, as long as it is
still small, does not affect the period, so the new period is still
2.0 s. It is equally important to understand what doesnt appear in
a formula. It is quite startling, really, the first time you
realize it, that the amplitude max( ) doesnt affect the period. But
this is crucial to the idea of simple harmonic motion. Of course,
if the pendulum is swung too far, out of its linear region, then
the amplitude would matter. The amplitude does appear in the
formula for a pendulum not restricted to small angles because the
small-angle approximation is not valid; but then the motion is not
simple harmonic motion.
OSCILLATIONS
14
-
14-2 Chapter 14
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14.3. (a) The amplitude is the maximum displacement obtained.
From the graph, 10 cmA = .
(b) From the graph, the period 2 0 sT = . . The angular
frequency is thus 2 3 1 rad/sT
= = . .
(c) The phase constant specifies at what point the cosine
function starts. From the graph, the starting point looks like
12
A. So at 0,t = 1cos2
A A= means 0 ( 60 )3= . Since the oscillator is moving to the
left at 0,t = it is in the
upper half of the circular-motion diagram, and we choose 0 3 = +
.
14.4. (a) A position-vs-time graph plots 0( ) cos( )x t A wt = +
. The graph of x(t) starts at 12 A and is increasing. So
at 0,t = 0 01 cos2 3
A A= = . We choose 0 3= since the particle is moving to the
right, indicating that it is
in the bottom half of the circular-motion diagram. (b) The phase
at each point can be determined in the same manner as for part (a).
For points 1 and 3, the amplitude is
again 12
A. At point 1, the particle is moving to the right so 1 3 = . At
point 3, the particle is moving left, so
3 .3 = + At point 2, the amplitude is A, so 2 2cos 1 0= = .
14.5. (a) A velocity-vs-time graph is a plot of 0( ) sin ( )v t
A t= + . At 0,t = the amplitude is max1 12 2 =v A
0sin , = A so 10 1 7sin or 2 6 6
= = .
Since ( 0 s)v t = is increasing, we need 0sin ( )t + increasing
at
0,t = so we choose 076
= .
(b) The phase is 0wt= + . At points 1 and 3, max1( ) ,2v t v= so
1 1 5sin or
2 6 6
= = . 2 At point 1, we need
1sin , increasing, so 1 56 = . At point 3, 3sin is decreasing,
so 3 6
= . At point 2, the amplitude is max ,v so
2sin 1, = thus 2 .2 =
14.6. Energy is transferred back and forth between all potential
energy at the extremes 212
kA and all kinetic
energy at the equilibrium point(s) 2max12
mv . The equation does not say that the particle ever has
amplitude A and speed maxv at the same time. The equation relates
expressions for the energy at two different times.
14.7. (a) The equilibrium length is 20 cm since that is where
the potential energy 21 ( ) 02 e
U k x x= = .
(b) The turning points occur where the potential energy is equal
to the total energy. These points are at 14 cmx = and 26 cm. (c)
The maximum kinetic energy of 7 J is obtained when the potential
energy is minimal, at the equilibrium point. (d) The total energy
will be around 14 J. If a horizontal line is drawn at that energy,
it intersects the potential energy curve at the turning points 12
cmx = and 28 cm.
-
Oscillations 14-3
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14.8. One expression for the total energy is 21 ,2
E kA= which occurs at the turning points when ( ) 0v t = . If
the total
energy is doubled, 2 21 12 ( ) ( 2 )
2 2E k A k A= =
Thus the new amplitude is 2 2(20 cm) 28 cmA A = = = .
14.9. One expression for the total energy is 2max1 ,2
E mv= which occurs at the equilibrium point when 0U = . If
the
total energy is doubled, 2 2
max max1 12 ( ) ( 2 )2 2
E m v m v= =
Thus the new maximum speed is max max2 2(20 cm/s) 28 cm/sv v = =
=
14.10. The time constant mb
= decreases as b increases, meaning energy is removed from the
oscillator more
quickly since /0( ) .tE t E e =
(a) The medium is more resistive. (b) The oscillations damp out
more quickly. (c) The time constant is decreased.
14.11. (a) T, the period, is the time for each cycle of the
motion, the time required for the motion to repeat itself. , the
damping time constant, is the time required for the amplitude of a
damped oscillator to decrease to 1 37%e
of its original value. (b) , the damping time constant, is the
time required for the amplitude of a damped oscillator to decrease
to
1 37%e of its original value, while 1/2,t the half-life, is the
time required for the amplitude of a damped oscillator to decrease
to 50% of its original value.
14.12. Natural frequency is the frequency that an oscillator
will oscillate at on its own. You may drive an oscillator at a
frequency other than its natural frequency. For example, if you are
pushing a child in a swing, you build up the amplitude of the
oscillation by driving the oscillator at its natural frequency. You
can achieve resonance by driving an oscillator at its natural
frequency.
Exercises and Problems
Section 14.1 Simple Harmonic Motion
14.1. Solve: The frequency generated by a guitar string is 440
Hz. The period is the inverse of the frequency, hence 31 1 2 27 10
s 2 27 ms
440 HzT
f
= = = . = .
14.2. Model: The air-track glider oscillating on a spring is in
simple harmonic motion. Solve: The glider completes 10 oscillations
in 33 s, and it oscillates between the 10 cm mark and the 60 cm
mark.
(a) 33 s 3 3 s/oscillation 3 3 s10 oscillations
T = = . = .
(b) 1 1 0 303 Hz 0 30 Hz3 3 s
fT
= = = . .
.
(c) 2 2 0 303 Hz 1 90 rad/sf= = ( . ) = .
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14-4 Chapter 14
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(d) The oscillation from one side to the other is equal to 60 cm
10 cm 50 cm 0 50 m = = . . Thus, the amplitude is 12 (0 50 m) 0 25
mA = . = . .
(e) The maximum speed is
max2 (1 90 rad/s)(0 25 m) 0 48 m/sv A AT
= = = . . = .
14.3. Model: The air-track glider attached to a spring is in
simple harmonic motion. Visualize: The position of the glider can
be represented as ( ) cosx t A t= . Solve: The glider is pulled to
the right and released from rest at 0 st = . It then oscillates
with a period 2 0 sT = . and a maximum speed max 40 cm/s 0 40 m/sv
= = . .
(a) maxmax2 2 0 40 m/s and rad/s 0 127 m 12 7 cm 13 cm
2 0 s rad/svv A A
T.
= = = = = = = . = . =.
(b) The gliders position at 0 25 st = . is 0 25 s (0 127 m)cos[(
rad/s)(0 25 s)] 0 090 m 9 0 cmx . = . . = . = .
Section 14.2 Simple Harmonic Motion and Circular Motion
14.4. Model: The oscillation is the result of simple harmonic
motion. Solve: (a) The amplitude 20 cmA = . (b) The period 4 0 s,T
= . thus
1 1 0 25 Hz4 0 s
fT
= = = .
.
(c) The position of an object undergoing simple harmonic motion
is 0( ) cos( )x t A t= + . At 00 s, 10 cmt x= = . Thus, 1 1
0 010 cm 110 cm (20 cm)cos cos cos rad 60 .20 cm 2 3
= = = = = Because the object is moving to the right at 0 s,t =
it is in the lower half of the circular motion diagram and thus
must have a phase constant between and 2 radians. Therefore, 0
rad 603 = = .
14.5. Model: The oscillation is the result of simple harmonic
motion. Solve: (a) The amplitude 10 cmA = . (b) The time to
complete one cycle is the period, hence 2 0 sT = . and
1 1 0 50 Hz2 0 s
fT
= = = .
.
(c) The position of an object undergoing simple harmonic motion
is 0( ) cos( )x t A t= + . 0At 0 s, 5 cm,t x= = thus 0
10 0
5 cm (10 cm)cos[ (0 s) ]
5 cm 1 1 2cos cos rad or 12010 cm 2 2 3
= +
= = = =
Since the oscillation is originally moving to the left, 0 120 =
+ .
14.6. Visualize: The phase constant 23 has a plus sign, which
implies that the object undergoing simple harmonic motion is in the
second quadrant of the circular motion diagram. That is, the object
is moving to the left. Solve: The position of the object is
20 0 3( ) cos( ) cos(2 ) (4 0 cm)cos[(4 rad/s) rad]x t A t A ft
t = + = + = . +
The amplitude is 4 cmA = and the period is 1/ 0 50 sT f= = . . A
phase constant 0 2 /3 rad 120 = = (second quadrant) means that x
starts at 12 A and is moving to the left (getting more
negative).
-
Oscillations 14-5
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Assess: We can see from the graph that the object starts out
moving to the left.
14.7. Visualize: A phase constant of 2
implies that the object that undergoes simple harmonic motion is
in the
lower half of the circular motion diagram. That is, the object
is moving to the right. Solve: The position of the object is given
by the equation
0 0( ) cos( ) cos(2 ) (8 0 cm)cos rad/s rad2 2x t A t A ft t = +
= + = .
The amplitude is 8 0 cmA = . and the period is 1/ 4 0 sT f= = .
. With 0 /2 rad, = x starts at 0 cm and is moving to the right
(getting more positive).
Assess: As we see from the graph, the object starts out moving
to the right.
14.8. Solve: The position of the object is given by the
equation
0 0( ) cos( ) cos(2 )x t A t A ft= + = + We can find the phase
constant 0 from the initial condition:
1 10 0 0 20 cm (4 0 cm)cos cos 0 cos (0) rad
= . = = = Since the object is moving to the right, the object is
in the lower half of the circular motion diagram. Hence,
10 2 rad = . The final result, with 4 0 Hz,f = . is
12( ) (4 0 cm)cos[(8 0 rad/s) rad]x t t = . .
-
14-6 Chapter 14
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material is protected under all copyright laws as they currently
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14.9. Solve: The position of the object is given by the
equation
0( ) cos( )x t A t= + The amplitude is 8 0 cmA = . . The angular
frequency 2 2 0 50 Hz rad/sf = = ( . ) = . Since at 0t = it has its
most negative position, it must be about to move to the right, so 0
= . Thus
( ) 8 0 cm cos[( rad/s) rad]x t t= ( . )
14.10. Model: The air-track glider is in simple harmonic motion.
Solve: (a) We can find the phase constant from the initial
conditions for position and velocity:
0 0 0 0cos sinxx A v A= = Dividing the second by the first, we
see that
0 00
0 0
sin tancos
xvx
= =
The glider starts to the left 0( 5 00 cm)x = . and is moving to
the right 0( 36 3 cm/s)xv = + . . With a period of 321 5 s s,.
=
the angular frequency is 432 / rad/sT= = . Thus
1 1 20 3 3
36 3 cm/stan rad (60 ) or rad ( 120 )(4 /3 rad/s)( 5 00 cm)
.= =
.
The tangent function repeats every 180, so there are always two
possible values when evaluating the arctan function. We can
distinguish between them because an object with a negative position
but moving to the right is in the third quadrant of the
corresponding circular motion. Thus 20 3 rad, = or 120 . (b) At
time t, the phase is 4 20 3 3( rad/s) radt t = + = . This gives 23
rad,= 0 rad, 23 rad, and 43 rad at, respectively, 0 s,t = 0.5 s,
1.0 s, and 1.5 s. This is one period of the motion.
Section 14.3 Energy in Simple Harmonic Motion
Section 14.4 The Dynamics of Simple Harmonic Motion
14.11. Model: The block attached to the spring is in simple
harmonic motion. Solve: The period of an object attached to a
spring is
02 2 0 smT Tk
= = = .
where m is the mass and k is the spring constant. (a) For mass 2
,m=
022 ( 2) 2 8 smT Tk
= = = .
(b) For mass 12 ,m
12
02 / 2 1 41 sm
T Tk
= = = .
(c) The period is independent of amplitude. Thus 0 2 0 sT T= = .
(d) For a spring constant 2 ,k=
02 / 2 1 41 s2mT Tk
= = = .
-
Oscillations 14-7
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14.12. Model: The air-track glider attached to a spring is in
simple harmonic motion. Solve: Experimentally, the period is (12 0
s)/(10 oscillations) 1 20 sT = . = . . Using the formula for the
period,
2 22 22 (0 200 kg) 5 48 N/m1 20 s
mT k mk T
= = = . = .
.
14.13. Model: The mass attached to the spring oscillates in
simple harmonic motion. Solve: (a) The period 1/ 1/2 0 Hz 0 50 sT
f= = . = . .
(b) The angular frequency 2 2 2 0 Hz 4 rad/sf= = ( . ) = . (c)
Using energy conservation
2 2 21 1 10 02 2 2 xkA kx mv= +
Using 0 5 0 cm,x = . 0 30 cm/sxv = and 2 2(0 200 kg)(4 rad/s) ,k
m= = . we get 5 54 cmA = . .
(d) To calculate the phase constant 0, 0 0
10
cos 5 0 cm5 0 cmcos 0 45 rad
5 54 cm
A x
= = .
. = = . .
(e) The maximum speed is max (4 rad/s)(5 54 cm) 70 cm/sv A= = .
= .
(f) The maximum acceleration is 2 2
max (4 rad/s)(70 cm/s) 8 8 m/sa A A= = ( ) = = . (g) The total
energy is 2 21 1max2 2 (0 200 kg)(0 70 m/s) 0 049 JE mv= = . . = .
.
(h) The position at 0 40 st = . is
0 4 s (5 54 cm)cos[(4 rad/s)(0 40 s) 0 45 rad] 3 8 cmx . = . . +
. = + .
14.14. Model: The oscillating mass is in simple harmonic motion.
Solve: (a) The amplitude 2 0 cmA = . . (b) The period is calculated
as follows:
2 210 rad/s 0 63 s10 rad/s
TT
= = = = . (c) The spring constant is calculated as follows:
2 2(0 050 kg)(10 rad/s) 5 0 N/mk k mm
= = = . = .
(d) The phase constant 10 4 rad= . (e) The initial conditions
are obtained from the equations
1 14 4( ) (2 0 cm)cos(10 ) and ( ) (20 0 cm/s)sin (10 )xx t t v
t t= . = .
At 0 s,t = these equations become 1 1
0 04 4(2 0 cm)cos( ) 1 41 cm and (20 cm/s)sin ( ) 14 1cm/sxx v=
. = . = = .
In other words, the mass is at 1 41 cm+ . and moving to the
right with a velocity of 14.1 cm/s. (f) The maximum speed is max (2
0 cm)(10 rad/s) 20 cm/sv A= = . = .
(g) The total energy 2 2 31 12 2 (5 0 N/m)(0 020 m) 1 00 10 JE
kA
= = . . = . .
(h) At 0 41 s,t = . the velocity is 1
0 4(20 cm/s)sin[(10 rad/s)(0 40 s) ] 1 46 cm/sxv = . = .
-
14-8 Chapter 14
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14.15. Model: The block attached to the spring is in simple
harmonic motion. Visualize:
Solve: (a) The conservation of mechanical energy equation f sf i
siK U K U+ = + is 2 2 2 2 21 1 1 1 11 0 02 2 2 2 2
0
( ) 0 J 0 J 0 J
1 0 kg (0 40 m/s) 0 10 m 10 cm16 N/m
mv k x mv kA mv
mA vk
+ = + + = +. = = . = . =
(b) We have to find the velocity at a point where /2x A= . The
conservation of mechanical energy equation 2 s2 i siK U K U+ = +
is
22 2 2 2 2 2 2 22 0 2 0 0 0 0
2 0
1 1 1 1 1 1 1 1 1 1 3 10 J2 2 2 2 2 2 4 2 2 4 2 4 2
3 3 (0 40 m/s) 0 346 m/s4 4
Amv k mv mv mv kA mv mv mv
v v
+ = + = = =
= = . = .
The velocity is 35 cm/s.
Section 14.5 Vertical Oscillations
14.16. Model: The vertical oscillations constitute simple
harmonic motion. Visualize:
-
Oscillations 14-9
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Solve: (a) At equilibrium, Newtons first law applied to the
physics book is sp
2
( ) 0 N 0 N
/ (0 500 kg)(9 8 m/s )/( 0 20 m) 24 5 N/m 25 N/m
yF mg k y mg
k mg y
= = = = . . . = .
(b) To calculate the period: 24 5 N/m 2 2 rad7 0 rad/s and 0 90
s0 500 kg 7 0 rad/s
k Tm
.
= = = . = = = .
. .
(c) The maximum speed is max 0 10 m 7 0 rad/s 0 70 m/sv A= = ( .
)( . ) = .
Maximum speed occurs as the book passes through the equilibrium
position.
14.17. Model: The vertical oscillations constitute simple
harmonic motion. Visualize:
Solve: The period and angular frequency are 20 s 2 20 6667 s and
9 425 rad/s
30 oscillations 0 6667 sT
T= = . = = = .
.
(a) The mass can be found as follows:
2 215 N/m 0 169 kg 0 17 kg
(9 425 rad/s)k kmm
= = = = . ..
(b) The maximum speed max (9 425 rad/s)(0 060 m) 0 57 m/sv A= =
. . = . .
14.18. Model: The vertical oscillations constitute simple
harmonic motion. Solve: To find the oscillation frequency using 2 /
,f k m = = we first need to find the spring constant k. In
equilibrium, the weight mg of the block and the spring force k L
are equal and opposite. That is, /mg k L k mg L= = . The frequency
of oscillation f is thus given as
21 1 / 1 1 9 8 m/s 3 5 Hz2 2 2 2 0 020 m
k mg L gfm m L
.= = = = = .
.
Section 14.6 The Pendulum
14.19. Model: Assume a small angle of oscillation so there is
simple harmonic motion. Solve: The period of the pendulum is
00 2 4 0 s
LTg
= = .
-
14-10 Chapter 14
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material is protected under all copyright laws as they currently
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or by any means, without permission in writing from the
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(a) The period is independent of the mass and depends only on
the length. Thus 0 4 0 sT T= = . . (b) For a new length 02 ,L
L=
00
22 2 5 7 sLT Tg
= = = .
(c) For a new length 0/2,L L=
00
/2 12 2 8 s2
LT Tg
= = = .
(d) The period is independent of the amplitude as long as there
is simple harmonic motion. Thus 4 0 sT = . .
14.20. Model: Assume the small-angle approximation so there is
simple harmonic motion. Solve: The period is 12 s/10 oscillations 1
20 sT = = . and is given by the formula
2 221 20 s2 (9 8 m/s ) 36 cm
2 2L TT L gg
. = = = . =
14.21. Model: Assume a small angle of oscillation so there is
simple harmonic motion. Solve: (a) On the earth the period is
earth 21 0 m2 2 2 0 s
9 80 m/sLTg
.
= = = .
.
(b) On Venus the acceleration due to gravity is 11 2 2 24
2VenusVenus 2 6 2
Venus
Venus 2Venus
(6 67 10 N m /kg )(4 88 10 kg) 8 86 m/s(6 06 10 m)
1 0 m2 2 2 1 s8 86 m/s
GMgR
LTg
. . = = = .
.
. = = = ..
14.22. Model: Assume the pendulum to have small-angle
oscillations. In this case, the pendulum undergoes simple harmonic
motion. Solve: Using the formula 2/ ,g GM R= the periods of the
pendulums on the moon and on the earth are
2 2earth earth moon moon
earth moonearth moon
2 2 and 2L L R L RT Tg GM GM
= = =
Because earth moon ,T T= 22 2
earth earth moon moon moon earthmoon earth
earth moon earth moon222 6
24 6
2 2
7 36 10 kg 6 37 10 m (2 0 m) 33 cm5 98 10 kg 1 74 10 m
L R L R M RL LGM GM M R
= =
. .
= . = . .
14.23. Model: Assume a small angle of oscillation so that the
pendulum has simple harmonic motion. Solve: The time periods of the
pendulums on the earth and on Mars are
earth Marsearth Mars
2 and 2L LT Tg g
= =
Dividing these two equations, 2 2
2 2earth Mars earthMars earth
Mars earth Mars
1 50 s(9 8 m/s ) 3 67 m/s2 45 s
T g Tg gT g T
. = = = . = .
.
-
Oscillations 14-11
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14.24. Model: This is not a simple pendulum, but is a physical
pendulum. Model the bar as thin enough to use 21
3I ML= where L is the full length of the rod.
Visualize: For a small-angle physical pendulum 2 IMglT = where l
is the distance from the pivot to the center of
mass. Note that = /2.l L We seek L. Solve: Substitute in for I
and then solve for L.
2 21 23 3 22 2 2 2
/2 3ML LI LT
Mgl MgL gL g= = = =
2 22 2
2 23 3 3 (9.8 m/s ) (1.2 s) 0.54 m 54 cm
2 2 8 8T g gL T
= = = = =
Assess: It seems reasonable that a uniform bar 54 cm long would
have a period of 1.2 s.
Section 14.7 Damped Oscillations
Section 14.8 Driven Oscillations and Resonance
14.25. Model: The spider is in simple harmonic motion. Solve:
Your tapping is a driving frequency. Largest amplitude at ext 1 0
Hzf = . means that this is the resonance frequency, so 0 ext 1 0
Hzf f= = . . That is, the spiders natural frequency of oscillation
0f is 1.0 Hz and 0 02 f= =
2 rad/s. We have
2 20 0 (0 0020 kg)(2 rad/s) 0 079 N/m
k k mm
= = = . = .
14.26. Model: The motion is a damped oscillation. Solve: The
amplitude of the oscillation at time t is given by Equation 14.58:
/20( ) ,
tA t A e= where /m b= is the time constant. Using 0 368 Ax = .
and 10 0 s,t = . we get
10.0 s/2 10 s 10.0 s0.368 ln(0.368) 5.00 s2 2ln(0.368)
A Ae
= = = =
14.27. Model: The motion is a damped oscillation. Solve: The
position of a damped oscillator is ( /2 ) 0( ) cos( )
tx t Ae t= + . The frequency is 1.0 Hz and the damping time
constant is 4.0 s. Let us assume 0 0 = rad and 1A = with arbitrary
units. Thus,
t/(8 0 s) 0 125 t( ) cos[2 1 0 Hz ] ( ) cos(2 )x t e t x t e t .
.= ( . ) = where t is in s. Values of x(t) at selected values of t
are displayed in the following table:
t(s) x(t) t(s) x(t) t(s) x(t) 0 1 2.00 0.779 6.00 0.472 0.25 0
2.50 0.732 6.50 0.444 0.50 0.939 3.00 0.687 7.00 0.417 0.75 0 3.50
0.646 7.50 0.392 1.00 0.882 4.00 0.607 8.00 0.368 1.25 0 4.50 0.570
8.50 0.346 1.50 0.829 5.00 0.535 9.00 0.325 1.75 0 5.50 0.503 9.50
0.305
10.00 0.286
-
14-12 Chapter 14
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14.28. Model: The pendulum is a damped oscillator. Solve: The
period of the pendulum and the number of oscillations in 4 hours
are calculated as follows:
osc215 0 m 4(3600 s)2 2 7 773 s 1850
7 773 s9 8 m/sLT Ng
.
= = = . = =..
The amplitude of the pendulum as a function of time is /2( ) bt
mA t Ae= . The exponent of this expression can be calculated to
be
(0.010 kg/s)(4 3600 s) 0.65452 2(110 kg)
= =
btm
We have 0 6545( ) 1 50 m 0 78 mA t e .= ( . ) = . . 14.29.
Model: Assume the eye is a simple driven oscillator. Visualize:
Given the mass and the resonant frequency, we can determine the
effective spring constant using the relationship 2 /f k m= = .
Solve: Solving the above expression for the spring constant,
obtain
2 2 3(2 ) [2 29 Hz ] (7 5 10 kg) 249 N/m 250 N/mk f m = = ( ) .
= Assess: As spring constants go, this is a fairly large value,
however the musculature holding the eyeball in the socket is strong
and hence will have a large effective spring constant.
14.30. Solve: The position and the velocity of a particle in
simple harmonic motion are 0 0 max 0( ) cos( ) and ( ) sin ( ) sin
( )xx t A t v t A t v t = + = + = +
From the graph, 12 sT = and the angular frequency is 2 2
rad/s
12 s 6T= = =
(a) Because max 60 cm/s,v A= = we have 60 cm/s 60 cm/s 115
cm
/6 rad/sA = = =
(b) At 0 s,t =
0 0 0
1 510 6 6
sin 30 cm/s 60 cm/s sin 30 cm/s
sin (0 5 rad) rad (30 ) or rad (150 )
xv A
= = ( ) = = . =
Because the velocity at 0 st = is negative and the particle is
slowing down, the particle is in the second quadrant of the
circular motion diagram. Thus 50 6 rad = . (c) At 0 s,t = 50 6(115
cm)cos( rad) 100 cmx = = .
-
Oscillations 14-13
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14.31. Solve: The position and the velocity of a particle in
simple harmonic motion are 0 0 max 0( ) cos( ) and ( ) sin( ) sin (
)xx t A t v t A t v t= + = + = +
(a) At 0 s,t = the equation for x yields 1 2
0 0 3( 5 0 cm) (10 0 cm)cos( ) cos ( 0 5) rad
. = . = . = Because the particle is moving to the left at 0 s,t
= it is in the upper half of the circular motion diagram, and the
phase constant is between 0 and radians. Thus, 20 3 rad= . (b) The
period is 4.0 s. At 0 s,t =
0 02 2sin (10 0 cm) sin 13 6 cm/s
3xv A
T
= = . = . (c) The maximum speed is
max2 (10 0 cm) 15 7 cm/s
4 0 sv A = = . = .
.
Assess: The negative velocity at 0 st = is consistent with the
position-vs-time graph and the positive sign of the phase
constant.
14.32. Model: The vertical mass/spring systems are in simple
harmonic motion. Solve: Spring/mass A undergoes three oscillations
in 12 s, giving it a period A 4 0 sT = . . Spring/mass B undergoes
2 oscillations in 12 s, giving it a period B 6 0 sT = . . We
have
A B A A BA B
A B B B A
4 0 s 22 and 26 0 s 3
m m T m kT Tk k T m k
.= = = = =
.
If A B,m m= then
B A
A B
4 9 2 259 4
k kk k
= = = .
14.33. Solve: The objects position as a function of time is 0( )
cos( )x t A t= + . Letting 0 mx = at 0 s,t = gives 1
0 0 20 cosA= = Since the object is traveling to the right, it is
in the lower half of the circular motion diagram, giving a phase
constant between and 0 radians. Thus, 10 2= and
1 12 2( ) cos( ) ( ) sin (0 10 m)sin ( )x t A t x t A t t= = =
.
where we have used 0 10 mA = . and 2 2 rad rad/s
4 0 s 2T= = =
.
Let us now find t where 0 060 m:= .x 12 0 060 m0 060 m (0 10
m)sin sin 0 41 s
2 0 10 mt t . . = . = = .
.
Assess: The answer is reasonable because it is approximately 18
of the period.
14.34. Model: The block attached to the spring is in simple
harmonic motion. Visualize: The position and the velocity of the
block are given by the equations
0 0( ) cos( ) and ( ) sin ( )xx t A t v t A t = + = + Solve: (a)
To graph x(t) we need to determine , 0, and A. These quantities
will be found by using the initial ( 0 s)t = conditions on x(t) and
( )xv t . The period is
1 0 kg 2 2 rad2 2 1 405 s 1 4 s 4 472 rad/s20 N/m 1 405 s
mTk T
.
= = = . . = = = . .
-
14-14 Chapter 14
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At 0 s,t = 0 0 0 0cos and sinxx A v A= = . Dividing these
equations, 0
0 00
( 1 0 m s)tan 1 1181 0 841 rad(4 472 rad/s)(0 20 m)
xv /x
.
= = = . = .. .
From the initial conditions, 2 2
2 200
1 0 m/s(0 20 m) 0 30 m4 472 rad/s
xvA x . = + = . + = . .
14.35. Model: The astronaut attached to the spring is in simple
harmonic motion. Solve: (a) From the graph, 3 0 s,T = . so we
have
2 23 0 s2 (240 N/m) 55 kg2 2
m TT m kk
. = = = =
(b) Oscillations occur about an equilibrium position of 1.0 m.
From the graph, 1 02 (0 80 m) 0 40 m, 0 rad, andA = . = . = 2 2 2
1rad/s
3 0 sT= = = .
.
The equation for the position of the astronaut is ( ) cos 1 0 m
(0 4 m) cos[(2 1 rad/s) ] 1 0 m
1 2 m (0 4 m)cos[(2 1 rad/s) ] 1 0 m cos[(2 1 rad/s) ] 0 5 0 50
sx t A t t
t t t= + . = . . + .
. = . . + . . = . = .
The equation for the velocity of the astronaut is
0 5 s
( ) sin ( )(0 4 m)(2 1 rad/s)sin[(2 1 rad/s)(0 50 s)] 0 73
m/s
xv t A tv
.
=
= . . . . = .
Thus her speed is 0.73 m/s.
14.36. Model: The particle is in simple harmonic motion. Solve:
The equation for the velocity of the particle is
( ) (25 cm)(10 rad/s)sin (10 )= xv t t Substituting into 2K U=
gives
2 2 2 2
2 22 2
2 2
2 2 2 1
1 1 1( ) 2 ( ) [ (250 cm/s)sin (10 )] [(25 cm)cos(10 )]2 2 2
sin (10 ) (25 cm) 12 2 s100cos (10 ) (250 cm/s)
1 1tan (10 ) 2(10 rad/s) s 2.0 tan 2.0 0.09100 10
= =
= = = = = =
xmv t kx t m t k t
t kmt
t t 6 s
14.37. Model: The spring undergoes simple harmonic motion.
Solve: (a) Total energy is 212E kA= . When the displacement is
12 ,x A= the potential energy is
( ) ( )22 2 31 1 1 1 1 12 2 2 4 2 4 4U kx k A kA E K E U E= = =
= = = One quarter of the energy is potential and three-quarters is
kinetic. (b) To have 12U E= requires
( )2 21 1 1 12 2 2 2 2AU kx E kA x= = = = 14.38. Solve: Average
speed is avg /v x t= . During half a period 12( ),t T = the
particle moves from x A= to
( 2 )x A x A= + = . Thus
avg max max avg2 4 4 2 2( )/2 2 / 2
x A A Av A v v vt T T
= = = = = = =
-
Oscillations 14-15
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14.39. Model: The ball attached to a spring is in simple
harmonic motion. Solve: (a) Let 0 st = be the instant when 0 5 0
cmx = . and 0 20 cm/sv = . The oscillation frequency is
2 5 N/m 5 0 rad/s0 100 kg
km
.
= = = .
.
Solving Equation 14.26 for the amplitude gives 22
2 200
20 cm/s( 5 0 cm) 6 4 cm5 0 rad/s
vA x
= + = . + = . .
(b) The maximum acceleration is 2 2max 160 cm/sa A= = . (c) For
an oscillator, the acceleration is most positive max( )a a= when
the displacement is most negative
max( )x x A= = . So the acceleration is maximum when 6 4 cmx = .
. (d) We can use the conservation of energy between 0 5 0 cmx = .
and 1 3 0 cm:x = .
2 2 2 2 2 2 21 1 1 10 0 1 1 1 0 0 12 2 2 2 ( ) 0 283 m/s
kmv kx mv kx v v x xm
+ = + = + = . The speed is 28 cm/s. Because k is known in SI
units of N/m, the energy calculation must be done using SI units of
m, m/s, and kg.
14.40. Model: The block on a spring is in simple harmonic
motion. Solve: (a) The position of the block is given by 0( ) cos(
)x t A t= + . Because ( )x t A= at 0 s,t = we have
0 0 rad,= and the position equation becomes ( ) cosx t A t= . At
0 685 s,t = . 3 00 cm cos(0 685 )A. = . and at 0 886 s,t = . 3 00
cm cos(0 886 )A . = . . These two equations give
cos(0 685 ) cos(0 886 ) cos( 0 886 )0 685 0 886 2 00 rad/s
. = . = .
. = . = .
(b) Substituting into the position equation, 3 00 cm3 00 cm
cos((2 00 rad/s)(0 685 s)) cos(1 370) 0 20 15 0 cm
0 20A A A A .. = . . = . = . = = .
.
14.41. Model: The oscillator is in simple harmonic motion.
Energy is conserved. Solve: The energy conservation equation 1 2E
E= is
2 2 2 21 1 1 11 1 2 22 2 2 2
2 2 2 21 1 1 1(0 30 kg)(0 954 m/s) (0 030 m) (0 30 kg)(0 714
m/s) (0 060 m)2 2 2 2
44 48 N/m
mv kx mv kx
k k
k
+ = +
. . + . = . . + .
= .
The total energy of the oscillator is 2 2 2 2
total 1 11 1 1 1(0 30 kg)(0 954 m/s) (44 48 N/m)(0 030 m) 0 1565
J2 2 2 2
E mv kx= + = . . + . . = .
Because 21total max2 ,E mv=
2max max
10 1565 J (0 300 kg) 1 02 m/s2
v v. = . = . Assess: A maximum speed of 1.02 m/s is
reasonable.
14.42. Model: The transducer undergoes simple harmonic motion.
Solve: Newtons second law for the transducer is
3 8 2restoring max max max40,000 N (0 10 10 kg) 4 0 10 m/sF ma a
a
= = . = .
-
14-16 Chapter 14
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Because 2max ,a A= 8 2
5max2 6 2
4 0 10 m/s 1 01 10 m 10 1 m[2 1 0 10 Hz ]
aA . = = = . = .( . )
(b) The maximum velocity is 6 5
max 2 1 0 10 Hz 1 01 10 m 64 m/sv A
= = ( . )( . ) =
14.43. Model: The block attached to the spring is in simple
harmonic motion. Solve: (a) The frequency is
1 1 2000 N/m 3 183 Hz2 2 5 0 kg
kfm
= = = .
.
The frequency is 3.2 Hz. (b) From energy conservation,
2 22 200
1 0 m/s(0 050 m) 0 0707 m2 3 183 Hz
vA x . = + = . + = . .
The amplitude is 7.1 cm. (c) The total mechanical energy is
2 21 12 2 (2000 N/m)(0 0707 m) 5 0 JE kA= = . = .
14.44. Model: Model this situation as a simple harmonic
oscillator without damping. Visualize: The period is independent of
the amplitude. If the oscillator is undamped the amplitude will be
constant; the amplitude is just the arbitrary distance from the
equilibrium from which the mass was released. The column of
amplitude data is not needed.
The period is related to the spring constant however: 2 .mkT
=
Solve: The equation 2 24 mkT = leads us to believe that a graph
of 2T vs. m would give a straight line graph whose
slope is 24 /k . Note that the period is 1/10 of the times given
in the table.
From the spreadsheet we see that the linear fit is very good and
that the slope is 26.0997s /kg. 2
22
4slope 4 / 6.5 N/m6.0997s /kg
k k= = =
Assess: Check the units carefully in the last equation. 6.5 N/m
is a reasonable spring constant.
-
Oscillations 14-17
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14.45. Model: The block undergoes simple harmonic motion.
Visualize:
Solve: (a) The frequency of oscillation is 1 1 10 N/m 1 125
Hz
2 2 0 20 kgkfm
= = = .
.
The frequency is 1.1 Hz. (b) Using conservation of energy, 2 2 2
21 1 1 11 1 0 02 2 2 2 ,mv kx mv kx+ = + we find
2 2 2 2 2 21 0 0 1
0 20 kg( ) ( 0 20 m) ((1 00 m/s) 0 50 m/s )10 N/m
0 2345 m or 23 cm
mx x v vk
.
= + = . + . ( . )= .
(c) At time t, the displacement is 0cos( )x A t= + . The angular
frequency is 2 7 071 rad/sf= = . . The amplitude is 2 2
2 200
1 00 m/s( 0 20 m) 0 245 m7 071 rad/s
vA x . = + = . + = . .
The phase constant is 1 10
00 200 mcos cos 2 526 rad or 145
0 245 mxA
. = = = .
. A negative displacement (below the equilibrium point) and
positive velocity (upward motion) indicate that the corresponding
circular motion is in the third quadrant, so 0 2 526 rad= . . Thus
at 1 0 s,t = .
(0 245 m)cos((7 071 rad/s)(1 0 s) 2 526 rad) 0 0409 m 4 09 cmx =
. . . . = . = . The block is 4.1 cm below the equilibrium
point.
14.46. Model: The mass is in simple harmonic motion.
Visualize:
The high point of the oscillation is at the point of release.
This conclusion is based on energy conservation. Gravitational
potential energy is converted to the springs elastic potential
energy as the mass falls and stretches the spring, then the elastic
potential energy is converted 100% back into gravitational
potential energy as the mass rises, bringing the mass back to
exactly its starting height. The total displacement of the
oscillationhigh point to low pointis 20 cm. Because the
-
14-18 Chapter 14
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oscillations are symmetrical about the equilibrium point, we can
deduce that the equilibrium point of the spring is 10 cm below the
point where the mass is released. The mass oscillates about this
equilibrium point with an amplitude of 10 cm, that is, the mass
oscillates between 10 cm above and 10 cm below the equilibrium
point. Solve: The equilibrium point is the point where the mass
would hang at rest, with sp GF F mg= = . At the equilibrium
point, the spring is stretched by 10 cm 0 10 my = = . . Hookes
law is sp ,F k y= so the equilibrium condition is 2
2sp G
9 8 m/s[ ] [ ] 98 s0 10 m
k gF k y F mgm y
.
= = = = = = .
2
The ratio /k m is all we need to find the oscillation
frequency:
21 1 98 s 1 58 Hz 1 6 Hz2 2
kfm
= = = . .
14.47. Model: The spring is ideal, so the apples undergo simple
harmonic motion. Solve: The spring constant of the scale can be
found by considering how far the pan goes down when the apples are
added.
20 N 222 N/m0 090 m
mg mgL kk L
= = = = .
The frequency of oscillation is
21 1 222 N/m 1 66 Hz 1 7 Hz
2 2 (20 N/9 8 m/s )kfm
= = = . .
.
Assess: An oscillation of fewer than twice per second is
reasonable.
14.48. Model: The compact car is in simple harmonic motion.
Solve: (a) The mass on each spring is (1200 kg)/4 300 kg= . The
spring constant can be calculated as follows:
2 2 2 2 4(2 ) (300 kg)[2 (2 0 Hz)] 4 74 10 N/mk k m m fm
= = = = . = . The spring constant is 44 7 10 N/m. . (b) The car
carrying four persons means that each spring has, on the average,
an additional mass of 70 kg. That is,
300 kg 70 kg 370 kgm = + = . Thus, 41 1 4 74 10 N/m 1 8 Hz
2 2 2 370 kgkfm
. = = = = .
Assess: A small frequency change from the additional mass is
reasonable because frequency is inversely proportional to the
square root of the mass.
14.49. Model: Assume simple harmonic motion for the two-block
system without the upper block slipping. We will also use the model
of static friction between the two blocks. Visualize:
Solve: The net force on the upper block 1m is the force of
static friction due to the lower block 2m . The two blocks ride
together as long as the static friction doesnt exceed its maximum
possible value. The model of static friction gives the maximum
force of static friction as
s max s s 1 1 max max s2 22
max max maxs 2
( ) ( )
2 2 0 40 m 0 721 5 s 9 8 m/s
f n m g m a a g
a A Ag g T g
= = = = . = = = = = .
. .
Assess: Because the period is given, we did not need to use the
block masses or the spring constant in our calculation.
-
Oscillations 14-19
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14.50. Model: The DNA and cantilever undergo simple harmonic
motion. Solve: The cantilever has the same spring constant with and
without the DNA molecule. The frequency of oscillation without the
DNA is
1 13
kM
=
With the DNA, the frequency of oscillation is
2 13
kM m
=
+
where m is the mass of the DNA. Divide the two equations, and
express 2 1 ,= where 2 2 (50 Hz)f = = .
( )( )
1 131 1 1 3
12 2 1 3
13
kM M mf fkf f f M
M m
+= = = =
+
Thus
( ) ( )2 21 11 13 32 2 2
1 1 11 13 2 3 2
1 1
( )
( ( ) ) 1( ) ( )
f M f f M m
f f f fm M Mf f f f
= +
= =
Since 1,f f (50 Hz 12 MHz),2
2 2 21 1 1
1 1( ) 1 1 2f ff f f f
f f
= + .
Thus
1 23 3
1 11 2 1f fm M M
f f
= + =
The mass of the cantilever 3 9 9 16(2300 kg/m )(4000 10 m)(100
10 m) 3 68 10 kgM = = .
Thus the mass of the DNA molecule is
16 216
2 50 Hz(3 68 10 kg) 1 02 10 kg3 12 10 Hz
m = . = .
Assess: The mass of the DNA molecule is about 56 2 10. atomic
mass units, which is reasonable for such a large molecule.
14.51. Model: Assume that the swinging lamp makes a small angle
with the vertical so that there is simple harmonic motion.
Visualize:
-
14-20 Chapter 14
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Solve: (a) Using the formula for the period of a pendulum, 2
2
2 5 5 s2 (9 8 m/s ) 7 5 m2 2
L TT L gg
. = = = . = .
(b) The conservation of mechanical energy equation 0 g0 1 g1K U
K U+ = + for the swinging lamp is
2 2 21 1 10 0 1 1 max2 2 2
max
2
0 J 0 J
2 2 ( cos3 )
2(9 8 m/s )(7 5 m)(1 cos3 ) 0 45 m/s
mv mgy mv mgy mgh mv
v gh g L L
+ = + + = + = =
= . . = .
14.52. Model: Assume that the angle with the vertical that the
pendulum makes is small enough so that there is simple harmonic
motion. Solve: The angle made by the string with the vertical as a
function of time is
max 0( ) cos( )t t= + The pendulum starts from maximum
displacement, thus 0 0= . Thus, max( ) cost t= . To find the time t
when the pendulum reaches 4.0 on the opposite side:
1( 4 0 ) 8 0 cos cos ( 0 5) 2 094 radt t . = ( . ) = . = . Using
the formula for the angular frequency,
29 8 m/s 2 0944 rad 2 094 rad3 130 rad/s 0 669 s1 0 m 3 130
rad/s
g tL
. . .
= = = . = = = .. .
The time 0 67 st = . . Assess: Because 2 / 2 0 s,T = = . a value
of 0.67 s for the pendulum to cover a little less than half the
oscillation is reasonable.
14.53. Model: Assume the orangutan is a simple small-angle
pendulum. Visualize: One swing of the arm would be half a period of
the oscillatory motion. The horizontal distance traveled in that
time would be 2(0 90m)sin (20 ) 0 616m. = . from analysis of a
right triangle. Solve:
22 2 0 90 m 0 952s
2 2 2 9 8 m/sT L
g.
= = = .
.
dist 0 616 mspeed 0 65 m/stime 0 952 s
.
= = = .
.
Assess: This isnt very fast, but isnt out of the reasonable
range.
14.54. Model: The mass is a particle and the string is massless.
Solve: Equation 14.51 is
MglI
=
The moment of inertia of the mass on a string is 2,I Ml= where l
is the length of the string. Thus
2Mgl g
lMl= =
This is Equation 14.48 with L l= . Assess: Equation 14.48 is
really a specific case of the more general physical pendulum
described by Equation 14.51.
14.55. Model: The rod is thin and uniform with moment of inertia
described in Table 12.2. The clay ball is a particle located at the
end of the rod. The ball and rod together form a physical pendulum.
The oscillations are small.
-
Oscillations 14-21
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Visualize:
Solve: The moment of inertia of the composite pendulum formed by
the rod and clay ball is 2 2
rod ball rod ball rod ball
2 2 3 2
13
1 (0 200 kg)(0 15 m) 0 020 kg 0 15 m 1 95 10 kg m3
I I I m L m L+
= + = +
= . . + ( . )( . ) = .
The center of mass of the rod and ball is located at a distance
from the pivot point of
2cm
0 15(0 200 kg) m 0 020 kg 0 15 m2 8 18 10 m
(0 200 kg 0 020 kg)y
. . + ( . )( . )
= = . . + .
The frequency of oscillation of a physical pendulum is 2 2
3 21 1 (0 220 kg)(9 8 m/s )(8 18 10 m) 1 51 Hz
2 2 1 95 10 kg mMglf
I
. . . = = = .
.
The period of oscillation 1Tf
= = 0.66 s.
14.56. Model: Model the rod as a small-angle physical pendulum
without damping. Visualize: Figure 14.22 in the chapter shows that
l is the distance from the pivot to the center of mass; in this
case
/4.l L= Solve: We first need to compute the moment of inertia of
a thin rod about an axis 1/4 of the way from one end. Use the
parallel-axis theorem to do so:
22 2 2 2
c.m.1 1 1 7
12 4 12 16 48LI I Md ML M ML ML = + = + = + =
For the physical pendulum
( )427
48
1 1 1 12 1 3= =2 2 2 7 7
LMgMgl g gfI L LML
= =
Assess: The period 1/T f= is a little shorter than swinging the
rod from the end, which is what we expect.
14.57. Model: Model the sphere as a small-angle physical
pendulum without damping. Visualize: Figure 14.22 in the chapter
shows that l is the distance from the pivot to the center of mass;
in this case l = R. Solve: We first need to compute the moment of
inertia of a sphere about an axis tangent to the edge. Use the
parallel-axis theorem to do so:
2 2 2 2c.m.
2 75 5
I I Md MR MR MR= + = + =
For the physical pendulum
275
1 1 1 52 2 2 7
Mgl MgR gfI RMR
= = =
Assess: The frequency is less than for a simple pendulum of
length R as we expect.
-
14-22 Chapter 14
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material is protected under all copyright laws as they currently
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14.58. Model: Model this situation as a small-angle physical
pendulum without damping. The moment of inertia for a uniform meter
stick pivoted on one end is 213 .ML
Visualize: The frequency and length of a physical pendulum are
related by 2 ,Mgl
If 1
= where /2.l L=
Solve: Square both sides of the equation above and substitute in
for I.
2 22 2 2 2 21
3
1 1 3 3 14 4 8 8
LMgMgl g gfI L LML
= = = =
This leads us to believe that a graph of 2f vs. 1/L would
produce a straight line whose slope is 23
8.g
From the spreadsheet we see that the linear fit is excellent and
that the slope is 20.3707m/s .
2 22
23 (8 )(0.3707 m/s )slope 9.76 m/s
38g g= = =
Assess: If the uncertainties in our measurements were small
enough then our answer would show that g varies slightly from place
to place on the earth.
14.59. Model: Treat the lower leg as a physical pendulum.
Visualize: We can determine the moment of inertia by combining 2 /T
I mgL= and 1/T f= .
Solve: Combining the above expressions and solving for the
moment of inertia we obtain 2 2 2 2 2/(2 ) (5 0 kg)(9 80 m/s )(0 18
m)/[2 (1 6 Hz)] 8 7 10 kg mI mgL f = = . . . . = .
Assess: NASA determines the moment of inertia of the shuttle in
a similar manner. It is suspended from a heavy cable, allowed to
oscillate about its vertical axis of symmetry with a very small
amplitude, and from the period of oscillation one may determine the
moment of inertia. This arrangement is called a torsion
pendulum.
14.60. Model: A completely inelastic collision between the two
gliders resulting in simple harmonic motion.
-
Oscillations 14-23
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Visualize:
Let us denote the 250 g and 500 g masses as 1m and 2,m which
have initial velocities i1v and i2v . After 1m collides with and
sticks to 2,m the two masses move together with velocity fv .
Solve: The momentum conservation equation f ip p= for the
completely inelastic collision is 1 2 f 1 i1 2 i2( ) .m m v m v m
v+ = + Substituting the given values,
f f(0 750 kg) (0 250 kg)(1 20 m/s) (0 500 kg)(0 m/s) 0 400 m/sv
v. = . . + . = . We now use the conservation of mechanical energy
equation:
2 21 1s compressed s equilibrium 1 2 f2 2
1 2f
( ) 0 J ( ) 0 J
0 750 kg (0 400 m/s) 0 11 m10 N/m
K U K U kA m m v
m mA vk
+ = ( + ) + = + ++ . = = . = .
The period is
1 2 0 750 kg2 2 1 7 s10 N/m
m mTk+ .
= = = .
14.61. Model: The block attached to the spring is oscillating in
simple harmonic motion. Solve: (a) Because the frequency of an
object in simple harmonic motion is independent of the amplitude
and/or the maximum velocity, the new frequency is equal to the old
frequency of 2.0 Hz. (b) The speed 0v of the block just before it
is given a blow can be obtained by using the conservation of
mechanical energy equation as follows:
2 2 21 1 1max 02 2 2
0 (2 ) (2 )(2 0 Hz)(0 02 m) 0 25 m/s
kA mv mv
kv A A f Am
= =
= = = = . . = .
The blow to the block provides an impulse that changes the
velocity of the block: f 0
3f f( 20 N)(1 0 10 s) (0 200 kg) (0 200 kg)(0 25 m/s) 0 150
m/s
x xJ F t p mv mv
v v= = =
. = . . . = . Since fv is the new maximum velocity of the block
at the equilibrium position, it is equal to A . Thus,
0 150 m/s 0 150 m/s 0 012 m 1 19 cm 1 2 cm 2 (2 0 Hz)
A . . = = = . = . ..
Assess: Because fv is positive, the block continues to move to
the right even after the blow.
-
14-24 Chapter 14
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14.62. Model: Assume the small-angle approximation.
Visualize:
Solve: The tension in the two strings pulls downward at angle .
Thus Newtons second law is 2 siny yF T ma = =
From the geometry of the figure we can see that
2 2sin y
L y=
+
If the oscillation is small, then y L and we can approximate sin
/y L . Since /y L is tan , this approximation is equivalent to the
small-angle approximation sin tan if 1 rad. With this
approximation, Newtons second law becomes
2 2
2 22 22 sin yT d y d y TT y ma m yL mLdt dt
= = = This is the equation of motion for simple harmonic motion
(see Equations 14.32 and 14.46). The constants 2T/mL are equivalent
to k/m in the spring equation or g/L in the pendulum equation. Thus
the oscillation frequency is
1 22
TfmL
=
14.63. Solve: The potential energy curve of a simple harmonic
oscillator is described by 212 ( ) ,U k x= where
0x x x = is the displacement from equilibrium. From the graph,
we see that the equilibrium bond length is 0 0.13 nm.x = We can
find the bonds spring constant by reading the value of the
potential energy U at a displacement x and using the potential
energy formula to calculate k.
x (nm) x (nm) U (J) k (N/m)
0.11 0.02 190 8 10 J. 400 0.10 0.03 191 9 10 J. 422 0.09 0.04
193 4 10 J. 425
The three values of k are all very similar, as they should be,
with an average value of 416 N/m. Knowing the spring constant, we
can now calculate the oscillation frequency of a hydrogen atom on
this spring to be
1327
1 1 416 N/m 7 9 10 Hz2 2 1 67 10 kg
kfm
= = = . .
-
Oscillations 14-25
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14.64. Model: Assume that the size of the ice cube is much less
than R and that is a small angle. Visualize:
Solve: The ice cube is like an object on an inclined plane. The
net force on the ice cube in the tangential direction is 2 2
G 2 2sin sind dF ma mR mR mg mRdt dt
( ) = = = = where is the angular acceleration. With the
small-angle approximation sin , this becomes
22
2d g
Rdt= =
This is the equation of motion of an object in simple harmonic
motion with a period of 2 2 RT
g= =
14.65. Visualize:
Solve: (a) Newtons second law applied to the penny along the
y-axis is
net yF n mg ma= =
netFG
is upward at the bottom of the cycle (positive ),ya so n mg>
. The speed is maximum when passing through
equilibrium, but 0ya = so n mg= . The critical point is the
highest point. netFG
points down and ya is negative. If
ya becomes sufficiently negative, n drops to zero and the penny
is no longer in contact with the surface.
(b) When the penny loses contact ( 0),n = the equation for
Newtons law becomes maxa g= . For simple harmonic motion,
22
max9 8 m/s 15 65 rad0 040 m
15 65 rad/s 2 5 Hz2 2
ga AA
f
.
= = = = ..
. = = = .
-
14-26 Chapter 14
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or by any means, without permission in writing from the
publisher.
14.66. Model: The vertical oscillations constitute simple
harmonic motion. Visualize:
Solve: At the equilibrium position, the net force on mass m on
Planet X is: X
net X 0 Nk gF k L mgm L
= = =
For simple harmonic motion 2/ ,k m = thus 2 2
2 2X XX
2 2 2 (0 312 m) 5 86 m/s14 5 s/10
g g g LL L T T
= = = = = . = . .
Assess: Because ordinary tasks seemed easier than on earth we
expected an answer less than 29.8 m/s .
14.67. Model: The dolls head is in simple harmonic motion and is
damped. Solve: (a) The oscillation frequency is
2 2 21 (2 ) (0 015 kg)(2 ) (4 0 Hz) 9 475 N/m2
kf k m fm
= = = . . = .
The spring constant is 9.5 N/m. (b) Using /20( ) ,
bt mA t A e= we get (4 0 s)/(2 0 015 kg) (133 3 s/kg)(0 5 cm) (2
0 cm) 0 25
133 33 s/kg ln 0 25 0 0104 kg/s 0 010 kg/s
b be eb b
. . .. = . . = ( . ) = . = . .
14.68. Model: The oscillator is in simple harmonic motion.
Solve: The maximum displacement at time t of a damped oscillator
is
/2 maxmax
( )( ) ln2
t t x tx t AeA
= =
Using max 0 98x A= . at 0 50 s,t = . we can find the time
constant to be 0 50 s 12 375 s
2ln(0 98).
= = .
.
25 oscillations will be completed at 25 12 5 st T= = . . At that
time, the amplitude will be 12 5 s/(2)(12 375 s)
max, 12 5 s (10 cm) 6 0 cmx e . .
.
= = .
14.69. Model: The vertical oscillations are damped and follow
simple harmonic motion. Solve: The position of the ball is given by
( /2 ) 0( ) cos( )
tx t Ae t= + . The amplitude ( /2 )( ) tA t Ae= is a function of
time. The angular frequency is
(15 0 N/m) 25 477 rad/s 1 147 s0 500 kg
k Tm
.
= = = . = = ..
-
Oscillations 14-27
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Because the balls amplitude decreases to 3.0 cm from 6.0 cm
after 30 oscillations, that is, after 30 1 147 s 34.41 s, . = we
have
(34 414 s/2 ) (34 41 s/2 ) 34 41 s3 0 cm (6 0 cm) 0 50 ln(0 50)
25 s 2
e e . . .. = . . = . = =
14.70. Model: The motion is a damped oscillation. Solve: The
position of the air-track glider is ( /2 ) 0( ) cos( ),
tx t Ae t= + where /m b= and 2
24k bm m
=
Using 0 20 m,A = . 0 0 rad,= and 0 015 kg/s,b = . 2
42
4 0 N/m (0 015 kg/s) 16 9 10 rad/s 4 0 rad/s0 250 kg 4(0 250
kg)
. .
= = = . . .
Thus the period is 2 2 rad 1 57 s
4 0 rad/sT = = = .
.
The amplitude at 0 st = is 0x A= and the amplitude will be equal
to 1e A at a time given by
( /2 )1 2 2 33 3 st mA Ae te b
= = = = . The number of oscillations in a time of 33.3 s is (33
3 s)/(1 57 s) 21. . = .
14.71. Model: The oscillator is in simple harmonic motion.
Solve: The maximum displacement, or amplitude, of a damped
oscillator decreases as /2max ( ) ,
tx t Ae= where is the time constant. We know max/ 0 60x A = . at
50 s,t = so we can find as follows:
max ( ) 50 sln 48 9 s2 2ln(0 60)t x t
A
= = = . .
Now we can find the time 30t at which max/ 0 30:x A = .
max30
( )2 ln 2(48 9 s)ln(0 30) 118 sx ttA
= = . . =
The undamped oscillator has a frequency 2 Hz 2f = = oscillations
per second. Damping changes the oscillation frequency slightly, but
the text notes that the change is negligible for light damping.
Damping by air, which allows the oscillations to continue for well
over 100 s, is certainly light damping, so we will use 2 0 Hzf = .
. Then the number of oscillations before the spring decays to 30%
of its initial amplitude is
30 (2 oscillations/s) (118 s) 236 oscillationsN f t= = =
14.72. Solve: The solution of the equation 2
2 0d x b dx k x
m dt mdt+ + =
is /2 0( ) cos( ).bt mx t Ae t = + The first and second
derivatives of x(t) are
/2 /20 0
2 22 /2
0 02 2
cos( ) sin ( )2
cos( ) sin ( )4
bt m bt m
bt m
dx Ab e t A e tdt m
d x Ab AbA t t emdt m
= + +
= + + +
-
14-28 Chapter 14
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publisher.
Substituting these expressions into the differential equation,
the terms involving 0sin ( )t + cancel and we obtain the simplified
result
22
02 cos( ) 04b k t
mm + + =
Because 0cos( )t + is not equal to zero in general, 2 2
22 204 4
b k k bm mm m
+ = =
14.73. Model: The two springs obey Hookes law. Visualize:
Solve: There are two restoring forces on the block. If the
blocks displacement x is positive, both restoring forcesone
pushing, the other pullingare directed to the left and have
negative values:
net sp 1 sp 2 1 2 1 2 eff( ) ( ) ( ) ( )x x xF F F k x k x k k x
k x= + = = + =
where eff 1 2k k k= + is the effective spring constant. This
means the oscillatory motion of the block under the influence of
the two springs will be the same as if the block were attached to a
single spring with spring constant effk . The frequency of the
blocks, therefore, is
2 2eff 1 2 1 21 22 2
1 12 2 4 4
k k k k kf f fm m m m
+= = = + = +
14.74. Model: The two springs obey Hookes law. Assume massless
springs. Visualize: Each spring is shown separately. Note that 1 2x
x x = + .
Solve: Only spring 2 touches the mass, so the net force on the
mass is 2 on m mF F= . Newtons third law tells us that
2 on on 2m mF F= and that 2 on 1 1 on 2F F= . From net ,F ma=
the net force on a massless spring is zero. Thus w on 1F
2 on 1 1 1= = F k x and on 2 1 on 2 2 2mF F k x= = . Combining
these pieces of information,
1 1 2 2mF k x k x= =
The net displacement of the mass is 1 2,x x x = + so
1 21 2
1 2 1 2 1 2
1 1m mm m
F F k kx x x F Fk k k k k k
+ = + = + = + =
-
Oscillations 14-29
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Turning this around, the net force on the mass is
1 2 1 2eff eff
1 2 1 2 where m
k k k kF x k x kk k k k
= = =+ +
eff ,k the proportionality constant between the force on the
mass and the masss displacement, is the effective spring constant.
Thus the masss angular frequency of oscillation is
eff 1 2
1 2
1k k km m k k
= =
+
Using 21 1/k m= and 22 2/k m= for the angular frequencies of
either spring acting alone on m, we have
2 21 2 1 2
2 21 2 1 2
( / )( / )( / ) ( / )
k m k mk m k m
= =
+ +
Since the actual frequency f is simply a multiple of , this same
relationship holds for :f
2 21 2
2 21 2
f fff f
=
+
14.75. Model: The blocks undergo simple harmonic motion.
Visualize:
The length of the stretched spring due to a block of mass m is
1L . In the case of the two-block system, the spring is further
stretched by an amount 2L . Solve: The equilibrium equations from
Newtons second law for the single-block and double-block systems
are
1 1 2( ) and ( ) (2 )L k mg L L k m g = + =
Using 2 5 0 cm,L = . and subtracting these two equations, gives
us
1 2 1( ) (2 ) (0 05 m)L L k L k m g mg k mg + = . = 2
29 8 m/s 196 s0 05 m
km
. = =.
With both blocks attached, giving total mass 2m, the angular
frequency of oscillation is
21 1196 s 9 90 rad/s2 2 2k km m
= = = = .
Thus the oscillation frequency is /2 1 6 Hzf = = . .
-
14-30 Chapter 14
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material is protected under all copyright laws as they currently
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or by any means, without permission in writing from the
publisher.
14.76. Model: A completely inelastic collision between the
bullet and the block resulting in simple harmonic motion.
Visualize:
Solve: (a) The equation for conservation of energy after the
collision is
2 2b B f f
b B
1 1 2500 N/m( ) (0 10 m) 5 0 m/s2 2 1 010 kg
kkA m m v v Am m
= + = = . = . + .
The momentum conservation equation for the perfectly inelastic
collision after beforep p= is
b B f b b B B2
b b
( )
(1 010 kg)(5 0 m/s) (0 010 kg) 1 00 kg 0 m/s 5 0 10 m/s
m m v m v m v
v v
+ = +
. . = . + ( . )( ) = .
(b) No. The oscillation frequency b B/( )k m m+ depends on the
masses but not on the speeds.
14.77. Model: The block undergoes simple harmonic motion after
sticking to the spring. Energy is conserved throughout the motion.
Visualize:
Its essential to carefully visualize the motion. At the highest
point of the oscillation the spring is stretched upward. Solve:
Weve placed the origin of the coordinate system at the equilibrium
position, where the block would sit on the spring at rest. The
spring is compressed by L at this point. Balancing the forces
requires k L mg = . The
-
Oscillations 14-31
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angular frequency is 2 / / ,w k m g L= = so we can find the
oscillation frequency by finding L . The block hits the spring (1)
with kinetic energy. At the lowest point (3), kinetic energy and
gravitational potential energy have been transformed into the
springs elastic energy. Equate the energies at these points:
2 21 11 1g 3s 3g 12 2 ( ) ( )K U U U mv mg L k L A mg A+ = + + =
+ +
Weve used 1y L= as the block hits and 3y A= at the bottom. The
spring has been compressed by y L A = + .
Speed 1v is the speed after falling distance h, which from
free-fall kinematics is 21 2v gh= . Substitute this expression
for 21v and /mg L for k, giving
2( ) ( )2( )
mgmgh mg L L A mg AL
+ = + +
The mg term cancels, and the equation can be rearranged into the
quadratic equation 2 2( ) 2 ( ) 0L h L A + =
The positive solution is 2 2 2 2(0 030 m) 0 100 m 0 030 m 0 0744
mL h A h = + = . + ( . ) . = .
Now that L is known, we can find 29 80 m/s 11 48 rad/s 1 8
Hz
0 0744 m 2g fL
.
= = = . = = . .
14.78. Model: Assume that the pendulum is restricted to small
angles so sin . Visualize: Because the string is massless the
center of mass is at the center of the sphere, so l L= in the
formula for
a physical pendulum: 2 .IMglT =
Solve: (a) First we use the parallel axis theorem to find I for
the sphere around the pivot point at the top (L away):
2 2 2cm .I I Md MR ML25= + = +
2 2 2 22 25 52 2 2
MR ML R LITMgl MgL gL
+ += = =
This reduces to the traditional simple pendulum formula 2 LgT =
in the limit L R . (b)
2 225 2 2 2 22 2
5 5real
simple
2 (0.010 m) (1.0 m)1.00002
1.0 m2 Lg
R LR LgLT
T L
+
+ += = = =
Assess: Making the simple pendulum approximation introduces an
error of only 0.002% with these numbers.
-
14-32 Chapter 14
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14.79. Model: Assume a simple system with a massless spring: 2
.mkT = Visualize: The simplest way to find T is to consider T as a
function of m and use calculus to take the differential
dT and consider that a good stand-in for .T Also note that 222 4
.m mk TT k = =
Solve:
(a) Re-write 2 mkT = as a function of m.
1/22( )T m mk
=
Now take the differential of T with respect to m using the
exponent rule and chain rule.
1/22 12
dT m dmk
=
Now cancel the 2s and plug in 224 .m
Tk =
22
1 124 m
T
TdT dm dm dmmk m m
= = =
Or, in the notation given in the problem:
2T mT
m =
(b) The new period, ,T is given by
1 11 (2.000 s) 1 (0.001) 2.001s2 2 2T m mT T T T T
m m
= + = + = + = + =
Assess: We did not expect a 0.1% change in mass to change the
period by much.
14.80. Model: The rod is thin and uniform. Visualize:
Solve: We must derive our own equation for this combination of a
pendulum and spring. For small oscillations, sFG
remains horizontal. The net torque around the pivot point is
Gcos sin2net sLI F L F = =
With 2
2 ,ddt
=
G ,F mg= s sin ,F k x kL= = and 2
1 ,3
I mL=
2
23 3sin cos sin
2d k g
m Ldt=
-
Oscillations 14-33
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We can use 1sin cos sin 22
= . For small angles, sin and sin 2 2 . So 2
23 3
2d k g
m Ldt
= +
This is the same as Equations 14.32 and 14.46 with 3 3
2k g
m L= +
The frequency of oscillation is thus 21 3(3 0 N/m) 3(9 8 m/s ) 1
73 Hz
2 (0 200 kg) 2(0 20 m)f . .= + = .
. .
The period 1Tf
= = 0.58 s.
Assess: Fewer than two oscillations per second is reasonable.
The rods angle from the vertical must be small enough that sin 2 2
. This is more restrictive than other examples, which only require
that sin .