Chapter 13
Vibrations and Waves
Quick Quizzes
1.
(d). To complete a full cycle of oscillation, the object must travel distance 2A to position
x = − A and then travel an additional distance 2A returning to the original position at
x = +A .
2.
(c). The force producing harmonic oscillation is always directed toward the equilibrium
position, and hence, directed opposite to the displacement from equilibrium. The
acceleration is in the direction of the force. Thus, it is also always directed opposite to the
displacement from equilibrium.
3.
(b). In simple harmonic motion, the force (and hence, the acceleration) is directly
proportional to the displacement from equilibrium. Therefore, force and acceleration are
both at a maximum when the displacement is a maximum.
4.
(a). The period of an object-spring system is T = 2π m k . Thus, increasing the mass by a
factor of 4 will double the period of oscillation.
5.
(c). The total energy of the oscillating system is equal to 12 kA2 , where A is the amplitude of
oscillation. Since the object starts from rest at displacement A in both cases, it has the same
amplitude of oscillation in both cases.
6.
(d). The expressions for the total energy, maximum speed, and maximum acceleration are
E = 12 kA2 , vmax = A k m , and amax = A ( k m) where A is the amplitude. Thus, all are
changed by a change in amplitude. The period of oscillation is T = 2π m k and is
unchanged by altering the amplitude.
7.
(c), (b). An accelerating elevator is equivalent to a gravitational field. Thus, if the elevator
is accelerating upward, this is equivalent to an increased effective gravitational field
magnitude g, and the period will decrease. Similarly, if the elevator is accelerating
downward, the effective value of g is reduced and the period increases. If the elevator
moves with constant velocity, the period of the pendulum is the same as that in the
stationary elevator.
465
466
CHAPTER 13
8.
(a). The clock will run slow. With a longer length, the period of the pendulum will
increase. Thus, it will take longer to execute each swing, so that each second according to
the clock will take longer than an actual second.
9.
(b). Greater. The value of g on the Moon is about one-sixth the value of g on Earth, so the
period of the pendulum on the moon will be greater than the period on Earth.
Vibrations and Waves
467
Answers to Even Numbered Conceptual Questions
2.
Each half-spring will have twice the spring constant of the full spring, as shown by the
following argument. The force exerted by a spring is proportional to the separation of the
coils as the spring is extended. Imagine that we extend a spring by a given distance and
measure the distance between coils. We then cut the spring in half. If one of the halfsprings is now extended by the same distance, the coils will be twice as far apart as they
were for the complete spring. Thus, it takes twice as much force to stretch the half-spring,
from which we conclude that the half-spring has a spring constant which is twice that of
the complete spring.
4.
To understand how we might have anticipated this similarity in speeds, consider sound as
a motion of air molecules in a certain direction superimposed on the random, high speed,
thermal molecular motions predicted by kinetic theory. Individual molecules experience
billions of collisions per second with their neighbors, and as a result, do not travel very far
in any appreciable time interval. With this interpretation, the energy of a sound wave is
carried as kinetic energy of a molecule and transferred to neighboring molecules by
collision. Thus, the energy transmitted by a sound wave in, say, a compression, travels
from molecule to molecule at about the rms speed, or actually somewhat less, as observed,
since multiple collisions slow the process a bit.
6.
Friction. This includes both air-resistance and damping within the spring.
8.
No. The period of vibration is T = 2π L g and g is smaller at high altitude. Therefore, the
period is longer on the mountain top and the clock will run slower.
10.
Shorten the pendulum to decrease the period between ticks.
12.
The speed of the pulse is v = F µ , so increasing the tension F in the hose increases the
speed of the pulse. Filling the hose with water increases the mass per unit length µ , and
will decrease the speed of the pulse.
14.
Assume that the building has height h and that you wish the jumper to start stretching the
bungee cord when he reaches a height of h 2 above the ground. The unstretched length of
the bungee cord must then be A = h 2 .
Furthermore, assume that you wish the bungee cord to bring the jumper to rest just as he
reaches ground level (that is, when the cord is stretched a distance of ∆A = h 2 ). For this to
occur, the elastic potential energy in the cord at this point must equal the total
gravitational potential energy the jumper has lost, leaving him with zero kinetic energy
when he reaches ground level. This means that
2
1
1 h
2
k ( ∆A )max = k   = mgh
2
2 2
where mg is the weight of the jumper. The required spring constant of the elastic would
then be k = 8mg h . You must be careful to check that the cord can withstand a maximum
 8 mg  h 
tension of Fmax = k ( ∆A )max = 
   = 4 mg without breaking. Also realize that the
 h  2 
468
CHAPTER 13
jumper will experience a net upward force of 3mg when the cord has this maximum
tension. Thus, the jumper must withstand a 3g upward acceleration just as he is brought
to rest at ground level.
16.
If the tension remains the same, the speed of a wave on the string does not change. This
means, from v = λ f , that if the frequency is doubled, the wavelength must decrease by a
factor of two.
18.
The speed of a wave on a string is given by v = F µ . This says the speed is independent
of the frequency of the wave. Thus, doubling the frequency leaves the speed unaffected.
20.
(a) The wall exerts a force to the left, and the cart exerts an equal magnitude force to the
right. The total force acting on the spring is zero. (b) The experimenter exerts a force to the
right, and the spring exerts an equal magnitude force to the left. The total force acting on
the cart is zero. (c) After the cart is released, the only force acting on the cart is the spring
force. The cart then undergoes simple harmonic motion along the surface, with amplitude
equal to the original displacement from the equilibrium position.
Vibrations and Waves
Answers to Even Numbered Problems
2.
(a) 1.1 × 10 2 N
(b) The graph is a straight line passing through the origin with slope equal to
k = 1.0 × 10 3 N m .
4.
(a)
8.00 s
(b)
No, the force is not of Hooke’s law form.
6.
(a)
327 N
(b)
1.25 × 10 3 N m
8.
(a)
575 N m
(b)
46.0 J
10.
2.23 m s
12.
(a)
2.61 m s
(b)
2.38 m s
14.
(a)
11 cm s
(b)
6.3 cm s
(c)
3.0 N
16.
(a) 0.15 J
(b)
0.78 m s
(c)
18 m s 2
18.
3.06 m s
20.
(a)
(b)
0.500 Hz
(c)
3.14 rad s
22.
3.95 N m
24.
2.2 Hz
26.
(a) 0.30 m, 0.24 m
(d) 6.0 s
(b)
0.30 m
(c)
1 6 Hz
28.
(a)
250 N m
(b)
T = 0.281 s, f = 3.56 Hz, ω = 22.4 rad s
(d)
5.00 cm
0.628 m s
(c) 0.313 J
(f) 0.919 cm
30.
(a)
59.6 m
(b)
37.5 s
32.
(a)
gain time
(b)
1.1 s
34.
(a)
3.65 s
(b)
6.41 s
36.
58.8 s
38.
5.67 mm
(e)
1.12 m s , 25.0 m s 2
(c)
4.24 s
469
470
CHAPTER 13
40.
0.800 m s
42.
(a)
44.
219 N
46.
1.64 m s 2
48.
7.07 m s
50.
586 m s
52.
(a)
54.
(a) 0.25 m
(d) 0.12 m s
56.
0.990 m
58.
(a)
60.
(a)
64.
32.9 ms
66.
(a)
68.
(a)
70.
1.3 cm s
0.20 Hz
(b)
0.25 Hz
(b)
0.30 m
(b)
0.47 N m
100 m s
(b)
374 J
19.8 m s
(b)
8.94 m
6.93 m s
(b)
1.14 m
28.0 J
(b)
0.446 m
0
(c)
0.23 m
Vibrations and Waves
471
Problem Solutions
13.1
(a) The force exerted on the block by the spring is
Fs = − kx = − (160 N m ) ( −0.15 m ) = + 24 N
or Fs = 24 N directed toward equilibrium position
(b) From Newton's second law, the acceleration is
a=
13.2
Fs + 24 N
m
m
=
= + 60 2 = 60 2 toward equilibrium position
m 0.40 kg
s
s
(a) The spring constant is k =
(
Fs
x
=
mg
50 N
=
= 1.0 × 10 3 N m
x
5.0 × 10 -2 m
)
F = Fs = kx = 1.0 × 10 3 N m ( 0.11 m ) = 1.1 × 10 2 N
(b) The graph will be a straight line passing through the origin with a slope equal to
k = 1.0 × 10 3 N m .
13.3
(a) Since the collision is perfectly elastic, the ball will rebound to the height of 4.00 m
before coming to rest momentarily. It will then repeat this motion over and over
again with a regular period.
(b) From ∆y = v0 y t +
ground is t =
1 2
ay t , with v0 y = 0 , the time required for the ball to reach the
2
2 ( ∆y )
ay
=
2 ( − 4.00 m )
−9.80 m s 2
= 0.904 s . This is one-half of the time for a
complete cycle of the motion. Thus, the period is T = 1.81 s .
(c)
No . The net force acting on the object is a constant given by F = − mg (except
when it is contact with the ground). This is not in the form of Hooke’s law.
472
CHAPTER 13
13.4
(a) The motion is periodic since the ball continuously repeats its back and forth motion
between the walls with no loss of energy (because of perfectly elastic collisions). The
time for the ball to travel from one wall to the other is one-half period and is given
by
T ∆x
12.0 m
=
=
= 4.00 s
2
v
3.00 m s
The period is then T = 8.00 s
(b) The motion is not simple harmonic since the force acting on the ball is not of the
form F = − kx . In fact, here F = 0 everywhere except when the ball is in contact with
a wall.
13.5
When the system is in equilibrium, the tension in the spring F = k x must equal the
weight of the object. Thus,
(
)
−2
k x ( 47.5 N ) 5.00 × 10 m
k x = mg giving m =
=
= 0.242 kg
g
9.80 m s 2
13.6
(a) The free-body diagram of the point in the center of the
string is given at the right. From this, we see that
ur 35.0°
T
ur
F
ΣFx = 0 ⇒ F − 2T sin 35.0° = 0
or T =
F
375 N
=
= 327 N
2sin 35.0° 2sin 35.0°
ur
T
35.0°
(b) Since the bow requires an applied horizontal force of 375 N to hold the string at
35.0° from the vertical, the tension in the spring must be 375 N when the spring is
stretched 30.0 cm. Thus, the spring constant is
k=
F
375 N
=
= 1.25 × 10 3 N m
x 0.300 m
Vibrations and Waves
13.7
(a) Assume the rubber bands obey Hooke’s law. Then, the force constant of each band
is
k=
Fs
15 N
=
= 1.5 × 10 3 N m
x 1.0 × 10 -2 m
Thus, when both bands are stretched 0.20 m, the total elastic potential energy is
2
1

PEs = 2  kx 2  = 1.5 × 10 3 N m ( 0.20 m ) = 60 J
2

(
)
(b) Conservation of mechanical energy gives ( KE + PEs ) f = ( KE + PEs ) i , or
1
mv 2 + 0 = 0 + 60 J , so v =
2
13.8
(a)
k=
2 ( 60 J )
= 49 m s
50 × 10 -3 kg
Fmax
230 N
=
= 575 N m
xmax 0.400 m
(b) work done = PEs =
13.9
1 2 1
2
kx = ( 575 N m ) ( 0.400 ) = 46.0 J
2
2
From conservation of mechanical energy,
( KE + PE
g
+ PEs
) = ( KE + PE
f
g
+ PEs
)
i
or 0 + mgh f + 0 = 0 + 0 +
1 2
kxi
2
giving
k=
13.10
473
2 mgh f
x
2
i
=
(
)
m)
2 ( 0.100 kg ) 9.80 m s 2 ( 0.600 m )
( 2.00 × 10
−2
(
Conservation of mechanical energy, KE + PEg + PEs
gives
or
= 2.94 × 10 3 N m
2
) = ( KE + PE
f
1
1
mvi2 + 0 + 0 = 0 + 0 + kx 2f ,
2
2
vi =
k
xi =
m
5.00 × 106 N m
3.16 × 10 −2 m = 2.23 m s
1000 kg
(
)
g
)
+ PEs ,
i
474
CHAPTER 13
13.11
At x = A, v = 0 and conservation of energy gives
E = KE + PEs = 0 +
2E
1 2
kA or A2 =
2
k
(a) At x = A 2 , the elastic potential energy is
2
PEs =
1  A
k
k  2E
E
k   = A2 = 
=

2  2
8
8 k 
4
From the energy conservation equation, the kinetic energy is then
KE = E − PEs = E −
E
3E
=
4
4
(b) When KE = PEs , conservation of energy yields E = KE + PEs = 2 PEs or PEs = E 2 .
Since we also have PEs = kx 2 2 , this yields
x=
13.12
2 PEs
=
k
2 ( E 2)
k
=
E
=
k
( kA 2) =
2
A
2
k
(a) From the work-energy theorem,
(
Wnc = KE + PEg + PEs
or F ⋅ x f =
) − ( KE + PE
f
g
+ PEs
)
i
1
1
mv 2f + 0 + kx 2f
2
2
This yields
vf =
2 F ⋅ x − kx 2f
m
2 ( 20.0 N )( 0.300 m ) − 19.6 N m ( 0.300 m )
= 2.61 m s
1.50 kg
2
=
Vibrations and Waves
475
(b) The work-energy theorem now contains one more nonzero term, giving
vf =
=
2 ( F − µ k mg ) ⋅ x − kx 2f
m
(
)
2
2  20.0 N − ( 0.200 ) ( 1.50 kg ) 9.80 m s 2  ( 0.300 m ) − 19.6 N m ( 0.300 m )
1.50 kg
v f = 2.38 m s
13.13
An unknown quantity of mechanical energy is converted into internal energy during the
collision. Thus, we apply conservation of momentum from just before to just after the
collision and obtain mvi + M ( 0 ) = ( M + m) V , or the speed of the block and embedded
bullet just after collision is
 10.0 × 10 −3 kg 
 m 
V=
vi = 
( 300 m s) = 1.49 m s

 M + m
2.01 kg 

Now, we use conservation of mechanical energy from just after collision until the block
1
1
comes to rest. This gives 0 + kx 2f = ( M + m) V 2 , or
2
2
xf = V
13.14
2.01 kg
M+m
= (1.49 m s )
= 0.478 m
19.6 N m
k
(
(a) In the absence of friction, KE + PEg + PEs
) = ( KE + PE
f
1
1
mv 2f + 0 + 0 = 0 + 0 + kxi2
2
2
or v f = xi
2 000 N m
k
= ( 0.30 cm )
= 11 cm s
m
1.5 kg
g
+ PEs
)
i
gives
476
CHAPTER 13
(
(b) When friction is present, Wnc = KE + PEg + PEs
) − ( KE + PE
g
f
+ PEs
1
1
 

− f ⋅ xi =  mv 2f + 0 + 0 −  0 + 0 + kxi2 
2
 

2
or
kxi2 − 2 f ⋅ xi
m
vf =
( 2 000 N m ) ( 3.0 × 10−3 m )
=
2
− 2 ( 2.0 N ) ( 3.0 × 10 −3 m )
1.5 kg
v f = 0.063 m s = 6.3 cm s
(
(c) If v f = 0 at x = 0 , then Wnc = KE + PEg + PEs
) − ( KE + PE
g
f
+ PEs
1


becomes − f ⋅ xi = ( 0 ) −  0 + 0 + kxi2 


2
−3
k xi ( 2 000 N m ) ( 3.0 × 10 m )
or f =
=
= 3.0 N
2
2
13.15
From conservation of mechanical energy,
( KE + PE
we have
g
+ PEs
) = ( KE + PE
f
g
+ PEs
)
i
(
1
1
1
mv 2 + 0 + kx 2 = 0 + 0 + kA2 , or v =
2
2
2
k
A2 − x 2
m
)
(a) The speed is a maximum at the equilibrium position, x = 0.
vmax =
k 2
A =
m
(19.6 N m) ( 0.040 m)
( 0.40 kg )
2
= 0.28 m s
(b) When x = − 0.015 m ,
v=
(19.6 N m) ( 0.040 m) − − 0.015 m
(
)
( 0.40 kg ) 
2
2
 = 0.26 m s

)
i
)
i
gives
Vibrations and Waves
(c) When x = + 0.015 m ,
v=
(d) If v =
(19.6 N m) ( 0.040 m) − + 0.015 m
(
)
( 0.40 kg ) 
2
1
vmax , then
2
This gives A2 − x 2 =
13.16
(a)
(
 = 0.26 m s

)
k
1 k 2
A2 − x 2 =
A
m
2 m
A2
3
3
, or x = A
= ( 4.0 cm )
= 3.5 cm
4
2
2
KE = 0 at x = A , so E = KE + PEs = 0 +
E=
2
1 2
kA , or the total energy is
2
1 2 1
2
kA = ( 250 N m ) ( 0.035 m ) = 0.15 J
2
2
(b) The maximum speed occurs at the equilibrium position where PEs = 0 . Thus,
1
2
E = mvmax
, or
2
vmax =
2E
250 N m
k
=A
= ( 0.035 m )
= 0.78 m s
m
m
0.50 kg
(c) The acceleration is a =
amax =
13.17
ΣF − kx
=
. Thus, a = amax at x = − xmax = − A .
m
m
 250 N m 
− k ( − A)  k 
=  A=
( 0.035 m ) = 18 m s 2

 m
m
0.50
kg


The maximum speed occurs at the equilibrium position and is
kA2 (16.0 N m ) ( 0.200 m )
k
= 4.00 kg , and
A . Thus, m = 2 =
2
vmax
m
( 0.400 m s)
2
vmax =
(
)
Fg = mg = ( 4.00 kg ) 9.80 m s 2 = 39.2 N
13.18
v=
 10.0 N m 
k
( 0.250 m ) 2 − ( 0.125 m ) 2  = 3.06 m s
A2 − x 2 = 
-3


m
 50.0 × 10 kg  
(
)
477
478
CHAPTER 13
13.19
(a) The motion is simple harmonic because the tire is rotating with constant velocity
and you are looking at the uniform circular motion of the “bump” projected on a
plane perpendicular to the tire.
(b) Note that the tangential speed of a point on the rim of a rolling tire is the same as
the translational speed of the axle. Thus, vt = vcar = 3.00 m s and the angular
velocity of the tire is
ω=
vt 3.00 m s
=
= 10.0 rad s
0.300 m
r
Therefore, the period of the motion is
T=
13.20
13.21
2π
ω
=
2π
= 0.628 s
10.0 rad s
(a)
vt =
2π r 2π ( 0.200 m )
=
= 0.628 m s
2.00 s
T
(b)
f=
1
1
=
= 0.500 Hz
T 2.00 s
(c)
ω=
2π
2π
=
= 3.14 rad s
T
2.00 s
The angle of the crank pin is θ = ω t . Its
x-coordinate is x = A cos θ = A cos ω t where
A is the distance from the center of the
wheel to the crank pin. This is of the
correct form to describe simple harmonic
motion. Hence, one must conclude that
the motion is indeed simple harmonic.
wt
A
x=0
x=0
13.22
x(t)
x(t)
The period of oscillations of a mass-spring system is given by T = 2π m k and the
frequency is
f=
1
1
=
T 2π
k
m
(
Thus, k = 4π 2 f 2 m = 4π 2 5.00 s -1
) ( 4.00 × 10
2
−3
)
kg = 3.95 N m
Vibrations and Waves
13.23
479
The spring constant is found from
(
)
2
Fs mg ( 0.010 kg ) 9.80 m s
k= =
=
= 2.5 N m
x
x
3.9 × 10 -2 m
When the object attached to the spring has mass m = 25 g , the period of oscillation is
0.025 kg
m
= 2π
= 0.63 s
2.5 N m
k
T = 2π
13.24
The springs compress 0.80 cm when supporting an additional load of m = 320 kg . Thus,
the spring constant is
(
)
2
mg ( 320 kg ) 9.80 m s
k=
=
= 3.9 × 10 5 N m
0.80 × 10 -2 m
x
When the empty car, M = 2.0 × 10 3 kg , oscillates on the springs, the frequency will be
f =
13.25
1
1
=
T 2π
1
k
=
M 2π
3.9 × 10 5 N m
= 2.2 Hz
2.0 × 10 3 kg
(a) The period of oscillation is T = 2π m k where k is the spring constant and m is the
mass of the object attached to the end of the spring. Hence,
0.250 kg
= 1.0 s
9.5 N m
T = 2π
(b) If the cart is released from rest when it is 4.5 cm from the equilibrium position, the
amplitude of oscillation will be A = 4.5 cm = 4.5 × 10 −2 m . The maximum speed is
then given by
vmax = Aω = A
(
k
= 4.5 × 10 −2 m
m
)
9.5 N m
= 0.28 m s
0.250 kg
(c) When the cart is 14 cm from the left end of the track, it has a displacement of
x = 14 cm − 12 cm = 2.0 cm from the equilibrium position. The speed of the cart at
this distance from equilibrium is
v=
(
)
k
A2 − x 2 =
m
9.5 N m
( 0.045 m ) 2 − ( 0.020 m ) 2  = 0.25 m s

0.250 kg 
480
CHAPTER 13
13.26
(a) At t = 0 , x = ( 0.30 m ) cos ( 0 ) = 0.30 m , and at t = 0.60 s ,
 π


x = ( 0.30 m ) cos   rad s ( 0.60 s )  = ( 0.30 m ) cos ( 0.20 π rad ) = 0.24 m

 3

(b)
A = xmax = ( 0.30 m )(1) = 0.30 m
(c)
π 
x = ( 0.30 m ) cos  t is of the form x = A cos (ω t ) with an angular frequency of
3 
ω=
π
3
rad s . Thus, f =
(d) The period is T =
13.27
ω π 3 1
Hz
=
=
2π 2π
6
1
= 6.0 s
f
(a) At t = 3.50 s ,
N
rad 



F = − kx = −  5.00
 ( 3.00 m ) cos  1.58
 ( 3.50 s )  = −11.0 N ,

m
s


or F = 11.0 N directed to the left
(b) The angular frequency is ω =
5.00 N m
= 1.58 rad s and the period of
2.00 kg
2π
= 3.97 s . Hence the number of oscillations made in
ω 1.58 rad s
∆t 3.50 s
3.50 s is N =
=
= 0.881
T 3.97 s
oscillation is T =
13.28
(a)
k=
(b) ω =
and
2π
k
=
m
=
7.50 N
F
=
= 250 N m
x 3.00 × 10 -2 m
k
=
m
250 N m
ω 22.4 rad s
= 22.4 rad s , f =
=
= 3.56 Hz ,
0.500 kg
2π
2π
T=
1
1
=
= 0.281 s
f 3.56 Hz
Vibrations and Waves
(c) At t = 0 , v = 0 and x = 5.00 × 10 −2 m , so the total energy of the oscillator is
E = KE + PEs =
=0+
1
1
mv 2 + kx 2
2
2
(
1
( 250 N m ) 5.00 × 10 −2 m
2
)
2
= 0.313 J
(d) When x = A , v = 0 so E = KE + PEs = 0 +
Thus, A =
2 ( 0.313 J )
2E
=
= 5.00 × 10 −2 m = 5.00 cm
k
250 N/m
(e) At x = 0 , KE =
or vmax =
amax =
(f)
1 2
kA .
2
1
2
mvmax
= E,
2
2E
=
m
2 ( 0.313 J )
= 1.12 m s
0.500 kg
(
)
−2
Fmax k A ( 250 N m ) 5.00 × 10 m
=
=
= 25.0 m s 2
0.500 kg
m
m
At t = 0.500 s ,
x = A cos (ω t ) = ( 5.00 cm ) cos ( 22.4 rad s) ( 0.500 s)  = 0.919 cm
13.29
From Equation 13.6, v = ±
(
)
(
k
A2 − x 2 = ± ω 2 A2 − x 2
m
)
Hence, v = ± ω A2 − A2 cos 2 (ω t ) = ± ω A 1 − cos 2 (ω t ) = ± ω A sin (ω t )
From Equation 13.2, a = −
13.30
k
x = − ω 2  A cos (ω t )  = − ω 2 A cos (ω t )
m
(a) The height of the tower is almost the same as the length of the pendulum. From
T = 2π L g , we obtain
L=
(
)
9.80 m s 2 (15.5 s)
gT2
=
= 59.6 m
4π 2
4π 2
2
481
482
CHAPTER 13
(b) On the Moon, where g = 1.67 m s 2 , the period will be
T = 2π
13.31
59.6 m
L
= 2π
= 37.5 s
1.67 m s 2
g
The period of a simple pendulum is T = 2π L g where L is its length. The number of
complete oscillations per second (that is, the frequency) for this pendulum is then given
by
9.80 m s 2
= 0.352 s -1
2.00 m
g
1
=
L 2π
1
1
f= =
T 2π
Hence, the number of oscillations in a time ∆t = 5.00 min = 300 s is
(
)
N = f ( ∆t ) = 0.352 s -1 ( 300 s ) = 105.7 or 105 complete oscillations
13.32
(a) The lower temperature will cause the pendulum to contract. The shorter length will
produce a smaller period, so the clock will run faster or gain time .
(b) The period of the pendulum is T0 = 2π
and at –5.0°C it is T = 2π
L0
at 20°C,
g
T
L
. The ratio of these periods is 0 =
g
T
L0
.
L
From Chapter 10, the length at –5.0°C is L = L0 + α Al L0 ( ∆T ) , so
L0
1
1
1
=
=
=
= 1.000 6
−
1
L 1 + α Al ( ∆T ) 1 +  24 × 10 −6 ( °C )  [ −5.0°C − 20°C] 0.999 4


T0
L0
=
= 1.000 6 = 1.000 3 . Thus in one hour (3 600 s), the chilled
T
L
pendulum will gain ( 1.000 3 − 1)( 3 600 s ) = 1.1 s .
This gives
Vibrations and Waves
13.33
483
(a) The period of the pendulum is T = 2π L g . Thus, on the Moon where the free-fall
acceleration is smaller, the period will be longer and the clock will run slow .
(b) The ratio of the pendulum’s period on the Moon to that on Earth is
TMoon 2π L g Moon
=
=
TEarth 2π L g Earth
gEarth
=
g Moon
9.80
= 2.45
1.63
Hence, the pendulum of the clock on Earth makes 2.45 “ticks” while the clock on the
Moon is making 1.00 “tick”. After the Earth clock has ticked off 24.0 h and again
24.0 h
reads 12:00 midnight, the Moon clock will have ticked off
= 9.79 h and will
2.45
read 9 : 47 AM .
13.34
The apparent free-fall acceleration is the vector sum of the actual
free-fall acceleration and the negative of the elevator’s acceleration.
To see this, consider an object that is suspended by a string in the
elevator and that appears to be at rest to the elevator passengers.
These passengers believe the tension in the string is the negative of
G
G
G
the object’s weight, or T = − m g app where g app is the apparent free-fall
acceleration in the elevator.
ur
T
ur
mg
An observer located outside the elevator applies Newton’s second law to this object by
G G
G
G
G
writing ΣF = T + m g = ma e where a e is the acceleration of the elevator and all its
G
G
G
G
G
G G
contents. Thus, T = ma e − m g = − m g app , which gives g app = g − a e .
G
(a) If we choose downward as the positive direction, then a e = − 5.00 m s 2 in this case
G
and g app = ( 9.80 + 5.00 ) m s 2 = +14.8 m s 2 (downward). The period of the
pendulum is
T = 2π
L
5.00 m
= 2π
= 3.65 s
g app
14.8 m s 2
G
(b) Again choosing downward as positive, a e = 5.00 m s 2 and
G
g app = ( 9.80 − 5.00 ) m s 2 = + 4.80 m s 2 (downward) in this case. The period is now
given by
T = 2π
L
= 2π
g app
5.00 m
= 6.41 s
4.80 m s 2
484
CHAPTER 13
G
G
G G
(c) If a e = 5.00 m s 2 horizontally, the vector sum g app = g − a e
ur
– ae
is as shown in the sketch at the right. The magnitude is
g app =
( 5.00 m s ) + ( 9.80 m s )
2 2
2 2
= 11.0 m s ,
2
ur
g
ur
gapp
and the period of the pendulum is
T = 2π
13.35
L
5.00 m
= 2π
= 4.24 s
g app
11.0 m s 2
(a) From T = 2π L g , the length of a pendulum with period T is L =
( 9.80 m s ) (1.0 s)
2
On Earth, with T = 1.0 s , L =
4π 2
( 3.7 m s ) (1.0 s)
If T = 1.0 s on Mars, L =
2
4π 2
gT2
.
4π 2
2
= 0.25 m = 25 cm
2
= 0.094 m = 9.4 cm
(b) The period of an object on a spring is T = 2π m k , which is independent of the
local free-fall acceleration. Thus, the same mass will work on Earth and on Mars.
This mass is
k T 2 (10 N m ) (1.0 s )
=
= 0.25 kg
m=
4π 2
4π 2
2
13.36
The frequency of the wave (that is, the number of crests passing the cork each second) is
f = 2.00 s -1 and the wavelength (distance between successive crests) is λ = 8.50 cm .
Thus, the wave speed is
(
)
v = λ f = ( 8.50 cm ) 2.00 s -1 = 17.0 cm s = 0.170 m s
and the time required for the ripples to travel 10.0 m over the surface of the water is
∆t =
∆x
10.0 m
=
= 58.8 s
v
0.170 m s
Vibrations and Waves
13.37
485
(a) The amplitude, A, is the maximum displacement from equilibrium. Thus, from
1
Figure P13.37, A = (18.0 cm ) = 9.00 cm
2
(b) The wavelength, λ , is the distance between successive crests (or successive
troughs). From Figure P13.37, λ = 2 (10.0 cm ) = 20.0 cm
(c) The period is T =
1
1
=
= 4.00 × 10 −2 s = 40.0 ms
f 25.0 Hz
(d) The speed of the wave is v = λ f = ( 0.200 m )( 25.0 Hz ) = 5.00 m s
13.38
From v = λ f , the wavelength (and size of smallest detectable insect) is
λ=
13.39
(a)
T=
(b) λ =
13.40
340 m s
v
=
= 5.67 × 10 -3 m = 5.67 mm
f 60.0 × 10 3 Hz
1
1
=
= 1.14 × 10 −8 s = 11.4 ns
f 88.0 × 10 6 Hz
v 3.00 × 108 m s
=
= 3.41 m
f
88.0 × 10 6 Hz
The distance between successive maxima in a transverse wave is the wavelength of that
wave. Hence, λ = 1.20 m . The frequency (number of crests passing a fixed point each
second) is
f=
8
= 0.667 s -1 = 0.667 Hz
12.0 s
(
)
Therefore, the wave speed is v = λ f = (1.20 m ) 0.667 s -1 = 0.800 m s
13.41
The speed of the wave is
and the frequency is
Thus, λ =
f=
v=
∆x 425 cm
=
= 42.5 cm s
10.0 s
∆t
40.0 vib
= 1.33 Hz
30.0 s
v 42.5 cm s
=
= 31.9 cm
f
1.33 Hz
486
CHAPTER 13
13.42
(a) When the boat is at rest in the water, the speed of the wave relative to the boat is the
same as the speed of the wave relative to the water, v = 4.0 m s . The frequency
detected in this case is
f=
v
λ
=
4.0 m s
= 0.20 Hz
20 m
G
G
G
(b) Taking westward as positive, v boat , water = v boat , wave + v wave , water gives
G
G
G
v boat , wave = v boat , water − v wave , water = +1.0 m s − ( − 4.0 m s ) = +5.0 m s
f=
Thus,
13.43
vboat , wave
λ
=
5.0 m s
= 0.25 Hz
20 m
The down and back distance is 4.00 m + 4.00 m = 8.00 m .
The speed is then v =
Now, µ =
dtotal 4 ( 8.00 m )
=
= 40.0 m s = F µ
0.800 s
t
m 0.200 kg
=
= 5.00 × 10 −2 kg m , so
L
4.00 m
(
)
F = µv 2 = 5.00 × 10 −2 kg m ( 40.0 m s ) = 80.0 N
13.44
The speed of the wave is v =
the rope is µ =
2
∆x 20.0 m
=
= 25.0 m s , and the mass per unit length of
∆t 0.800 s
m
= 0.350 kg m . Thus, from v = F µ , we obtain
L
F = v 2 µ = ( 25.0 m s ) ( 0.350 kg m ) = 219 N
2
13.45
(a) The mass per unit length is µ =
m 0.0600 kg
=
= 0.0120 kg m
L
5.00 m
From v = F µ , the required tension in the string is
F = v 2 µ = ( 50.0 m s ) ( 0.0120 kg m ) = 30.0 N ,
2
(b) v =
F
µ
=
8.00 N
= 25.8 m s
0.0120 kg m
Vibrations and Waves
13.46
The mass per unit length of the wire is
µ=
m 4.00 × 10 -3 kg
=
= 2.50 × 10 -3 kg m ,
L
1.60 m
and the speed of the pulse is v =
L
1.60 m
=
= 44.3 m s .
∆t 0.0361 s
Thus, the tension in the wire is
(
)
F = v 2 µ = ( 44.3 m s ) 2.50 × 10 -3 kg m = 4.91 N
2
But, the tension in the wire is the weight of a 3.00-kg object on the Moon. Hence, the
local free-fall acceleration is
g=
13.47
F
4.91 N
=
= 1.64 m s 2
m 3.00 kg
The period of the pendulum is T = 2π
(
)
L
, so the length of the string is
g
9.80 m s 2 ( 2.00 s )
gT 2
L=
=
= 0.993 m
4π 2
4π 2
2
Then mass per unit length of the string is then
µ=
kg
m 0.060 0 kg
=
= 0.060 4
L
0.993 m
m
When the pendulum is vertical and stationary, the tension in the string is
(
)
F = Mball g = ( 5.00 kg ) 9.80 m s 2 = 49.0 N ,
and the speed of transverse waves in it is
v=
F
µ
=
49.0 N
= 28.5 m s
0.060 4 kg m
487
488
CHAPTER 13
13.48
If µ1 = m1 L is the mass per unit length for the first string, then µ 2 = m2 L = m1 2L = µ1 2
is that of the second string. Thus, with F2 = F1 = F , the speed of waves in the second
string is
F
v2 =
13.49
µ2
=
 F
= 2
 = 2 v1 = 2 ( 5.00 m s ) = 7.07 m s
µ1
 µ1 
2F
(
)
(a) The tension in the string is F = mg = ( 3.0 kg ) 9.80 m s 2 = 29 N . Then, from
v = F µ , the mass per unit length is
µ=
F
29 N
=
2 = 0.051 kg m
2
v
( 24 m s)
(b) When m = 2 .00 kg , the tension is
(
)
F = mg = ( 2.0 kg ) 9.80 m s 2 = 20 N
and the speed of transverse waves in the string is
v=
13.50
F
µ
=
20 N
= 20 m s
0.051 kg m
If the tension in the wire is F, the tensile stress is Stress = F A , so the speed of transverse
waves in the wire may be written as
v=
F
µ
=
A ⋅ Stress
=
mL
Stress
m ( A ⋅ L)
But, A ⋅ L = V =volume , so m ( A ⋅ L ) = ρ = density . Thus, v =
Stress
ρ
When the stress is at its maximum, the speed of waves in the wire is
vmax =
( Stress)max
ρ
=
2.70 × 10 9 Pa
= 586 m s
7.86 × 10 3 kg m 3
.
Vibrations and Waves
13.51
489
From v = F µ , the tension in the string is F = v 2 µ . Thus, the ratio of the new tension to
the original is
2
2
 30.0 m s 
v 
F2 v22
= 2 , giving F2 =  2  F1 = 
( 6.00 N ) = 13.5 N
F1 v1
 v1 
 20.0 m s 
13.52
(a) If the end is fixed, there is inversion of the pulse upon reflection. Thus, when they
meet, they cancel and the amplitude is zero .
(b) If the end is free there is no inversion on reflection. When they meet the amplitude
is A′ = 2 A = 2 ( 0.15 m ) = 0.30 m .
13.53
(a)
Constructive interference produces the maximum amplitude
Amax
′ = A1 + A2 = 0.50 m
(b)
Destructive interference produces the minimum amplitude
Amin
′ = A1 − A2 = 0.10 m
13.54
We are given that x = A cos(ω t ) = ( 0.25 m ) cos( 0.4π t ) .
(a) By inspection, the amplitude is seen to be A = 0.25 m
(b) The angular frequency is ω = 0.4π rad s . But ω = k m , so the spring constant is
k = m ω 2 = ( 0.30 kg )( 0.4π rad s ) = 0.47 N m
2
(c) At t = 0.30 s , x = ( 0.25 m ) cos ( 0.4π rad s ) ( 0.30 s)  = 0.23 m
(d) From conservation of mechanical energy, the speed at displacement x is given by
v = ω A2 − x 2 . Thus, at t = 0.30 s , when x = 0.23 m , the speed is
v = ( 0.4 π rad s)
( 0.25 m ) 2 − ( 0.23 m ) 2
= 0.12 m s
490
CHAPTER 13
13.55
The maximum acceleration of the oscillating system is
ur
n
amax = ω 2 A = ( 2 π f ) A
2
B
The friction force exerted between the two blocks must be
capable of accelerating block B at this rate. When block B is
on the verge of slipping, f s = ( f s ) max = µs n = µ s mg = mamax and
we must have
ur
fs
ur
ur
Fg = mg
amax = ( 2 π f ) A = µ s g
2
Thus,
13.56
( 0.600 ) ( 9.80 m s 2 )
= 6.62 × 10 −2
A=
2 =
2
2
2
1.50
Hz
π
π
f


)
( )
 (
µs g
Since the spring is “light”, we neglect any small amount of energy lost in the collision
with the spring, and apply conservation of mechanical energy from when the block first
starts until it comes to rest again. This gives
( KE + PE
Thus, xmax =
13.57
m = 6.62 cm
g
+ PEs
) = ( KE + PE
f
2mghi
=
k
g
)
+ PEs , or 0 + 0 +
i
(
1 2
kxmax = 0 + 0 + mghi
2
)
2 ( 0.500 kg ) 9.80 m s 2 ( 2.00 m )
20.0 N m
= 0.990 m
Choosing PEg = 0 at the initial height of the block, conservation of mechanical energy
(
gives KE + PEg + PEs
) = ( KE + PE
f
g
)
+ PEs , or
i
1
1
mv 2 + mg ( − x ) + kx 2 = 0 ,
2
2
where v is the speed of the block after falling distance x.
(a) When v = 0 , the non-zero solution to the energy equation from above gives
1 2
kxmax = mgxmax ,
2
(
)
2
2 mg 2 ( 3.00 kg ) 9.80 m s
or k =
=
= 588 N m
0.100 m
xmax
Vibrations and Waves
491
(b) When x = 5.00 cm = 0.050 0 m , the energy equation gives
v = 2 gx −
kx 2
, or
m
v = 2 ( 9.80 m s
13.58
2
) ( 0.050 0 m )
( 588
−
N m )( 0.050 0 m )
3.00 kg
2
= 0.700 m s
(a) We apply conservation of mechanical energy from just after the collision until the
block comes to rest.
( KE + PEs ) f = ( KE + PEs )i
gives 0 +
1 2 1
kx f = MV 2 + 0
2
2
or the speed of the block just after the collision is
V=
k x 2f
M
=
( 900
N m )( 0.050 0 m )
1.00 kg
2
= 1.50 m s
Now, we apply conservation of momentum from just before impact to immediately
after the collision. This gives
m ( vbullet ) i + 0 = m ( vbullet ) f + MV
 M
or ( vbullet ) f = ( vbullet ) i −   V
 m
 1.00 kg 
= 400 m s − 
(1.5 m s) = 100 m s
-3
 5.00 × 10 kg 
(b) The mechanical energy converted into internal energy during the collision is
∆E = KEi − ΣKE f =
1
1
1
2
2
m ( vbullet ) i − m ( vbullet ) f − MV 2
2
2
2
or
∆E =
(
)
1
2
2
1
2
5.00 × 10 −3 kg ( 400 m s ) − (100 m s )  − (1.00 kg )(1.50 m s)


2
2
∆E = 374 J
492
CHAPTER 13
13.59
Choose PEg = 0 when the blocks start from rest. Then, using conservation of mechanical
energy from when the blocks are released until the spring returns to its unstretched
(
length gives KE + PEg + PEs
) = ( KE + PE
f
g
)
+ PEs , or
i
1
1
m1 + m2 ) v 2f + ( m1 g x sin 40° − m2 g x ) + 0 = 0 + 0 + k x 2
(
2
2
(
)
1
( 25 + 30 ) kg  v 2f + ( 25 kg ) 9.80 m s 2 ( 0.200 m ) sin 40°

2
1
2
− ( 30 kg ) 9.80 m s 2 ( 0.200 m ) = ( 200 N m ) ( 0.200 m )
2
(
v f = 1.1 m s
yielding
13.60
)
(a) When the gun is fired, the energy initially stored as elastic potential energy in the
spring is transformed into kinetic energy of the bullet. Assuming no loss of energy,
we have 12 mv 2 = 12 kxi2 , or
9.80 N m
k
= ( 0.200 m )
= 19.8 m s
1.00 × 10 −3 kg
m
v = xi
(b) From ∆y = v0 y t + 12 ay t 2 , the time required for the pellet to drop 1.00 m to the floor,
starting with v0 y = 0 , is
t=
2 ( ∆y )
ay
=
2 ( −1.00 m )
= 0.452 s
−9.80 m s 2
The range (horizontal distance traveled during the flight) is then
∆x = v0 x t = (19.8 m s ) ( 0.452 s ) = 8.94 m
13.61
(a) ω =
k
=
m
500 N m
= 15.8 rad s
2.00 kg
Vibrations and Waves
493
(b) Apply Newton’s second law to the block while the elevator is accelerating:
ΣFy = Fs − mg = may
With Fs = kx and ay = g 3 , this gives kx = m ( g + g 3) , or
(
)
2
4 mg 4 ( 2.00 kg ) 9.80 m s
x=
=
= 5.23 × 10 −2 m = 5.23 cm
3k
3 ( 500 N m )
13.62
(a) When the block is given some small upward displacement, the net restoring force
exerted on it by the rubber bands is
Fnet = ΣFy = −2 F sin θ , where tan θ =
y
L
For small displacements, the angle θ will be very small. Then sin θ ≈ tan θ =
y
, and
L
the net restoring force is
 y
 2F
Fnet = −2 F   = − 
y
 L
 L 
(b) The net restoring force found in part (a) is in the form of Hooke’s law F = − ky , with
2F
k=
. Thus, the motion will be simple harmonic, and the angular frequency is
L
ω=
k
=
m
2F
mL
494
CHAPTER 13
13.63
The free-body diagram at the right shows the forces
acting on the balloon when it is displaced distance
s = Lθ along the circular arc it follows. The net force
tangential to this path is
ur
B
+y
q
s = Lq
Fnet = ΣFx = − B sin θ + mg sin θ = − ( B − mg ) sin θ
For small angles, sin θ ≈ θ =
s
L
equilibrium
position
q
ur
mg
+x
L
ur
T
q
Also, mg = ( ρHeV ) g
and the buoyant force is B = ( ρairV ) g . Thus, the net
restoring force
 ( ρair − ρHe ) Vg 
acting on the balloon is Fnet ≈ − 
s
L


Observe that this is in the form of Hooke’s law, F = − k s ,
with k = ( ρair − ρHe ) Vg L
Thus, the motion will be simple harmonic and the period is given by
T=
m
1 2π
=
= 2π
= 2π
f
k
ω
 ρHe  L
ρHeV
= 2π 
 ρair − ρHe  g
( ρair − ρHe ) Vg L
 0.180
 ( 3.00 m )
This yields T = 2 π 
= 1.40 s
 1.29 − 0.180  9.80 m s 2
(
13.64
)
Observe in the sketch at the right that
2d + L 2 = D , or
d=
D − L 2 2.00 m − 1.50 m
=
= 0.250 m
2
2
Thus,
D
L
–
4
q
A
L
–
2
d
 d 
 0.250 m 
= cos −1 
= 70.5°

 0.750 m 
 L 4
θ = cos −1 
m
q
B
d
m
495
Vibrations and Waves
Now, consider a free body diagram of point A:
ΣFx = 0 ⇒ F = T2 cos ( 70.5°)
and
ΣFy = 0 ⇒ T2 sin ( 70.5° ) = mg = 19.6 N
ur
T2
y
q
A
x
ur
ur
T1 = mg
Hence, the tension in the section between A and B is
F=
ur
F
19.6 N
= 6.93 N
tan ( 70.5°)
The mass per unit length of the string is
µ=
10.0 × 10 −3 kg
= 3.33 × 10 -3 kg m
3.00 m
so the speed of transverse waves in the string between points A and B is
v=
F
µ
=
6.93 N
= 45.6 m s
3.33 × 10 -3 kg m
The time for the pulse to travel from A to B is
t=
13.65
L2
1.50 m
=
= 3.29 × 10 −2 s = 32.9 ms
v
45.6 m s
Newton's law of gravitation is
GMm
4

F = − 2 , where M = ρ  π r 3 


r
3
4

Thus, F= -  πρGm r
3

which is of Hooke’s law form, F = -k r , with
k=
4
πρGm
3
m
M
r
F
RE
496
CHAPTER 13
13.66
(a) Apply the work-energy theorem from the instant
the firefighter starts from rest until just before contact
with the platform.
(
Wnc = KE + PEg
) − ( KE + PE )
g
f
i
m
gives
h
1

− f ⋅ h =  mv 2 + 0 − ( 0 + mgh ) , or v =
2

v=
f

2 g −  h

m
M
s

300 N 
2  9.80 m s 2 −
( 5.00 m ) = 6.93 m s
60 kg 

(b) Next, apply conservation of momentum to find the speed V of the firefighter and
platform immediately after the perfectly inelastic collision. This gives
( m + M ) V = mv + M ( 0 )
 m 
 60.0 
or V = 
v=
( 6.93 m s) = 5.20 m s

 m + M
 60.0+20.0 
Finally, apply the work-energy theorem from just after the collision until the
firefighter comes to rest.
(
Wnc = KE + PEg + PEs
) − ( KE + PE
f
g
+ PEs
)
i
gives
1  1


− f ⋅ s =  0 + 0 + ks2  −  ( m + M ) V 2 + ( m + M ) gs + 0




2
2
or s2 −
2
 m + M 2
V =0
( m + M ) g − f  s − 
 k 
k
Using the given data, we obtain s2 − ( 0.387 m ) s − 0.865 m 2 = 0 , and the quadratic
formula gives a positive solution of s = 1.14 m
Vibrations and Waves
13.67
497
(a) Using conservation of mechanical energy, ( KE + PEs ) f = ( KE + PEs ) i , from the
moment of release to the instant of separation gives
1
1
m1 + m2 ) v 2 + 0 = 0 + kA2
(
2
2
or v = A
k
= ( 0.20 m )
m1 + m2
100 N m
= 0.50 m s
( 9.0 + 7.0) kg
(b) After the two blocks separate, m1 oscillates with new amplitude A′ found by
applying ( KE + PEs ) f = ( KE + PEs ) i to the m1 + spring system from the moment of
separation until the spring is fully stretched the first time.
0+
1
1
kA′ 2 = m1v 2
2
2
or A′ = v
9.0 kg
m1
= ( 0.50 m s )
= 0.15 m
k
100 N m
The period of this oscillation is T = 2 π
9.0 kg
m1
= 2π
= 1.9 s ,
k
100 N m
T
= 0.47 s after separation.
4
During this time, m2 has moved distance x = vt from the point of separation. Thus,
the distance separating the two blocks at this instant is
so the spring is fully stretched for the first time at t =
D = vt − A′ = ( 0.50 m s ) ( 0.47 s ) − 0.15 m = 0.086 m = 8.6 cm
13.68
(a) Apply the work-energy theorem from the instant before the block contacts the
spring until the instant the block leaves the spring.
Wnc = KE f − KEi =
=
(
1
m v 2f − vi2
2
)
2
2
1
8.00 kg ) ( 3.00 m s ) − ( 4.00 m s )  = −28.0 J
(


2
or the mechanical energy lost is Wnc = 28.0 J
498
CHAPTER 13
(b) The energy spent overcoming the friction force while the block is in contact with the
spring is f ⋅ s = Wnc , where s = 2 xmax with xmax being the maximum distance the
spring was compressed. Hence,
xmax =
=
13.69
Wnc
2f
=
Wnc
2 µk n
=
Wnc
2 µ k mg
28.0 J
= 0.446 m
2 ( 0.400 ) ( 8.00 kg ) 9.80 m s 2
(
)
(a) As seen in the sketch at the right, the length of the chord
∆x followed by the longitudinal wave is shorter than
the arc length ∆s traveled by the transverse wave. Also,
the longitudinal wave travels faster than does the
transverse wave. Thus, the longitudinal wave will
A
RE
q
RE
Dx
Ds
B
arrive first.
(b) With θ = 60.0° = π 3 radians , the arc length, ∆s , traveled by the transverse wave is
π

∆s = REθ = RE  rad
3

∆s RE (π 3)
=
vt
vt
From the sketch, observe that the chord labeled ∆x is one side of an isosceles
triangle. Thus, ∆x = RE and the travel time for the longitudinal wave is
and the time for this wave to travel from A to B is:
∆tA =
∆tt =
∆x RE
=
vA
vA
The difference in the arrival times of the two waves is then

 π
1
π
1
∆t = ∆tt − ∆tA = RE 
−  = 6.37 × 10 3 km 
−
 3vt vA 
 3 ( 4.50 km s ) 7.80 km
(
 1 min 
or ∆t = ( 666 s ) 
= 11.1 min
 60 s 
)


s 
499
Vibrations and Waves
13.70
The inner tip of the wing is attached to the end of the spring and always moves with the
same speed as the end of the vibrating spring. Thus, its maximum speed is
vinner , max = vspring , max = A
4.7 × 10 −4 N m
k
= ( 0.20 cm )
= 0.25 cm s
0.30 × 10 −3 kg
m
Treating the wing as a rigid bar, all points in the wing have the same angular velocity at
any instant in time. As the wing rocks on the fulcrum, the inner tip and outer tips follow
circular paths of different radii. Since the angular velocities of the tips are always equal,
v
v
we may write ω = outer = inner . The maximum speed of the outer tip is then
router
rinner
r 
 15.0 mm 
vouter , max =  outer  vinner , max = 
( 0.25 cm s) = 1.3 cm s
 3.00 mm 
 rinner 
13.71
(a) If the surface is frictionless, the total mechanical energy of the system is conserved.
Thus, taking the initial position at the point when the block starts from rest and the
final position at the equilibrium position, we have
1
1
mv 2f + 0 = 0 + kxi2
2
2
KE f + PE f = KEi + PEi ⇒
or v f = xi
(
k
= 2.0 × 10 −2 m
m
)
1.0 × 10 3 N m
= 0.50 m s
1.6 kg
(b) When a constant friction force retards the motion of the block, the work-energy
theorem gives
Wnc = ( KE + PE) f − ( KE + PE) i or − f ⋅ s =
1
1
1
1
mv 2f + kx 2f − mvi2 − kxi2
2
2
2
2
where s is the total distance the block moves between the initial and final states. If
the block is initially at rest at xi = 2.0 cm , and the final state is when the block
makes its first pass through the equilibrium position, then s = xi and we have
(
)
− ( 4.0 N ) ⋅ 2.0 × 10 −2 m =
or v f =
(
)(
1
1
1.6 kg ) v 2f + 0 − 0 − 1.0 × 10 3 N m 2.0 × 10 −2 m
(
2
2
0.20 J − 0.080 J
= 0.39 m s
0.80 kg
)
2
500
CHAPTER 13
(c) When a friction force is present, the system is continuously spending energy as the
friction force does negative work on it. The block will cease to move when all of the
original energy (initially stored as elastic potential energy) has been used doing
work to overcome friction. That is, until
(
)(
1.0 × 10 3 N m 2.0 × 10 −2 m
kx 2
1
f ⋅ s = kxi2 or s = i =
2f
2 ( 4.0 N )
2
)
2
= 5.0 × 10 −2 m = 5.0 cm