Chapter 13 Solutions - Mosinee School District

Chapter 13
13
Vibrations and Waves
PROBLEM SOLUTIONS
13.1 (a) Taking to the right as positive, the spring force acting on the block at the instant of release is
F kx
s
i
130 N m 0.13 m 17 N or 17 N to the left
(b)
At this instant, the acceleration is
a
F s
m
17 N
0.60 kg
28 m s
2
or a 28 m s
2
to the left
13.2 When the object comes to equilibrium (at distance y 0 below the unstretched position of the end of the spring),
F k y mg and the force constant is
y
0 0
k
mg
y
0
4.25 kg 9.80 m s
2.62 10
2
m
2
1.59 103
N 1.59 kN
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.
1
(b) From y v t a t
2
, with v 0 0 , the time required for the ball to reach the ground is
0y
2
y
y
t
2 y 2 4.00 m
a
y
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.
13.4 (a) The spring constant is
Page 13.1

Chapter 13
k
Fs
x
mg
x
50 N
5.0 10
-2
m
1.0 103
N m
F Fs
kx
1.0 103 N m 0.11 m 1.1 102
N
(b) The graph will be a straight line passing through the origin with a slope equal to k 1.0 103
N m .
13.5 When the system is in equilibrium, the tension in the spring F k x must equal the weight of the object. Thus,
k x
mg giving
m
k x
g
47.5 N 5.00 10
2
m
9.80 m s
2
0.242 kg
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
F 0 F 2T
sin 35.0 0
x
or
T
F 375 N
2 sin 35.0 2 sin 35.0
327 N
(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
x
375 N
0.300 m
1.25 103
N m
13.7 (a) When the block comes to equilibrium, Fy
ky0 mg 0 giving
y
0
mg
k
10.0 kg 9.80 m s
475 N m
2
0.206 m
or the equilibrium position is 0.206 m below the unstretched position of the lower end of the spring.
(b) When the elevator (and everything in it) has an upward acceleration of a 2.00 m s2
second law to the block gives
, applying Newton’s
Fy
k y0
y mg ma y or y 0
F ky mg ky ma
y
Page 13.2

Chapter 13
where y = 0 at the equilibrium position of the block. Since ky0 mg 0 [see part (a)], this becomes
ky
ma and the new position of the block is
y
ma y
k
10.0 kg 2.00 m s
475 N m
2
4.21 10
2
m 4.21 cm
or 4.21 cm below the equilibrium position.
(c)
When the cable breaks, the elevator and its contents will be in free-fall with a y = g. The new “equilibrium”
position of the block is found from Fy
ky0
mg m g , which yields y 0 0
snapped, the block was at rest relative to the elevator at distance
. When the cable
y0 y 0.206 m 0.042 1 m 0.248 mbelow the new “equilibrium” position. Thus, while the elevator is
in free-fall, the block will oscillate with amplitude
0.248 m about the new “equilibrium” position, which
is the unstretched position of the spring’s lower end.
13.8 (a) When the gun is fired, the elastic potential energy initially stored in the spring is transformed into kinetic
energy of the projectile. Thus, it is necessary to have
1
2
1
kx
2
0 mv 0 or
2 2
k
mv
x
2
0
2
0
3.00 10
3
kg 45.0 m s
2
8.00 10
2
m
2
949 N m
(b)
The magnitude of the force required to compress the spring 8.00 cm and load the gun is
Fs
k x
949 N m 8.00 10
2
m 75.9 N
13.9 (a) Assume the rubber bands obey Hooke’s law. Then, the force constant of each band is
k
F s
x
15 N
1.0 10
-2
m
1.5 103
N m
Thus, when both bands are stretched 0.20 m, the total elastic potential energy is
1
PEs
kx
2
2 3
2
2 1.5 10 N m 0.20 m 60 J
(b) Conservation of mechanical energy gives KE PE s KE PE s , or
f
i
Page 13.3

Chapter 13
1
2
0 0 60 J
2 mv , so 2 60 J
v
50 10
-3
kg
49 m s
13.10 (a)
k
F
x
max
max
230 N
0.400 m
575 N m
(b)
work done PEs
kx
2 2
1
2
1 2
575 N m 0.400 46.0 J
13.11 From conservation of mechanical energy,
KE PE g PE s KE PE f
g PE s or 1
0 mghf
0 0 0 kx i
2
i
2
giving
k
2 mgh
x
2
i
f
2 0.100 kg 9.80 m s2
0.600 m
2
2.00 10
2
m
2.94 103
N m
13.12 Conservation of mechanical energy, ( KE PEg PEs ) f ( KE PEg PE s)
i
, gives
1 2 1
mv 0 0 0 0
2
2 i
2 f
kx , or
v
i
k
m
x
i
5.00 106
N m 3.16 10
2 m 2.23 m s
1000 kg
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 mv i + M(0) = (M + m)V, or the
speed of the block and embedded bullet just after collision is
V
m
M m
v
i
10.0 10
3
kg
2.01 kg
300 m s 1.49 m s
Now, we use conservation of mechanical energy from just after collision until the block comes to rest. This gives
1 2 1
0
2
2 f 2
kx M m V , or
x
f
V
M m
k
2.01 kg
1.49 m s 0.477 m
19.6 N m
Page 13.4

Chapter 13
13.14 (a) At either of the turning points, x = A, the constant total energy of the system is momentarily stored as elastic
potential energy in the spring. Thus, E kA 2 2 .
(b) When the object is distance x from the equilibrium position, the elastic potential energy is PEs
kx 2 2 and
the kinetic energy is KE mv 2 2 . At the position where KE = 2PE s , is it necessary that
1 1
2 2
1
mv kx
2
mv2 2 kx
2
or
2 2
(c) When KE = 2PE, conservation of energy gives E KE PEs 2 PEs PEs 3PE s
, or
1 1
2 2
kA2 3 kx2
x
k A 2 2
3 k 2
or
x
A
3
13.15 (a) At maximum displacement from equilibrium, all of the energy is in the form of elastic potential energy, giving
E kx , and
3
2
max 2
k
2E
x
2
max
2 47.0 J
0.240 m
2
1.63 10 N m
(b)
At the equilibrium position (x = 0), the spring is momentarily in its relaxed state and PE s = 0, so all of the
energy is in the form of kinetic energy. This gives
1
KE mv2
x 0
max E
2
47.0 J
(c)
If, at the equilibrium position, v = v max = 3.45 m/s, the mass of the block is
m
2E
v
2
max
2 47.0 J
3.45 m s
2
7.90 kg
(d) At any position, the constant total energy is, E KE PE mv2 2 kx
2
2, so at x = 0.160 m
s
v
2E
2 2 47.0 J 1.63 103
N m 0.160 m
kx
m
7.90 kg
2
2.57 m s
(e)
At x = 0.160 m, where v = 2.57 m/s, the kinetic energy is
Page 13.5

Chapter 13
1
2
1 2
7.90 kg 2.57 m s 26.1 J
KE mv
2 2
(f)
At x = 0.160 m, where KE = 26.1 J, the elastic potential energy is
PEs
E KE
47.0 J 26.1 J 20.9 J
or alternately,
1
2
1
3
2
1.63 10 N m 0.160 m 20.9 J
PEs
kx
2 2
(g)
At the first turning point (for which x < 0 since the block started from rest at x = +0.240 m and has passed
through the equilibrium at x = 0) all of the remaining energy is in the form of elastic potential energy, so
1
2
kx2
E E loss
47.0 J 14.0 J 33.0 J
and
x
2 33.0 J 2 33.0 J
k
1.63 103
N m
0.201 m
13.16 (a) F k x 83.8 N m 5.46 10
2
m 4.58 N
(b)
E PEs
kx
2 2
1 2
2
1 83.8 N m 5.46 10
2 m 0.125 J
(c) While the block was held stationary at x = 5.46 cm, F F F 0 , or the spring force was equal in
magnitude and oppositely directed to the applied force. When the applied force is suddenly removed, there is a
net force F s = 4.58 N directed toward the equilibrium position acting on the block. This gives the block an
acceleration having magnitude
x
s
a
F s
m
4.58 N
0.250 kg
18.3 m s
2
(d)
At the equilibrium position, PE s = 0, so the block has kinetic energy KE = E = 0.125 J and speed
v
2E
m
2 0.125 J
0.250 kg
1.00 m s
Page 13.6

Chapter 13
(e)
If the surface was rough, the block would spend energy overcoming a retarding friction force as it moved
toward the equilibrium position, causing it to arrive at that position with a lower speed than that computed
above. Computing a number value for this lower speed requires knowledge of the coefficient of friction
between the block and surface.
13.17 From conservation of mechanical energy,
KE PE PE KE PE PE
g s
f
g s
i
we have 1 mv 2 0 1 kx 2 0 0
1 kA 2 , or
2 2 2
k
v A x
m
2 2
(a) The speed is a maximum at the equilibrium position, x = 0.
v
max
k
A
m
2
19.6 N m 2
0.040 m 0.28 m s
0.40 kg
(b) When x = 0.015 m,
v
19.6 N m 2 2
0.040 m 0.015 m 0.26 m s
0.40 kg
(c) When x = +0.015 m,
v
19.6 N m 2 2
0.040 m 0.015 m 0.26 m s
0.40 kg
1
k
(d) If v v
2 max
, then
2 2
1 k
A x A2
m
2 m
This gives A 2 x 2 = A 2 /4, or x A 3 / 2 4.0 cm 3 / 2 3.5 cm .
1
13.18 (a) KE = 0 at x = A, so E KE PE 0 kA
2
, or the total energy is
s
2
E
1 2
1 250 N m 0.035 m 2
kA
0.15 J
2 2
Page 13.7

Chapter 13
1
(b) The maximum speed occurs at the equilibrium position where PE s = 0. Thus, E mv
2
2 max
, or
v
max
2 E k
250 N m
A 0.035 m 0.78 m s
m m
0.50 kg
(c) The acceleration is a F/ m kx/
m . Thus, a = a max at x = x max = A.
a
max
k A k
m m
A
250 N m 0.035 m 18 m s
0.50 kg
2
13.19 The maximum speed occurs at the equilibrium position and is
v
max
k
m
A
Thus,
m
kA
v
2
2
max
16.0 N m 0.200 m
0.400 m s
2
2
4.00 kg ,
and
Fg
mg
4.00 kg 9.80 m s2
39.2 N
13.20
k
v A x
m
2 2
10.0 N m
50.0 10
-3
kg
2 2
0.250 m 0.125 m 3.06 m s
13.21 (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, v v 3.00 m s and the angular velocity of the tire is
t
car
v t
r
3.00 m s
0.300 m
10.0 rad s
Therefore, the period of the motion is
T
2 2
10.0 rad s
0.628 s
Page 13.8

Chapter 13
13.22 (a)
v
t
2 r
T
2 0.200 m
2.00 s
0.628 m s
(b)
f
1 1
T 2.00 s
0.500 Hz
(c)
2 2
T 2.00 s
3.14 rad s
13.23 The angle of the crank pin is t . Its x-coordinate is
x Acos
Acos
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.
13.24 The period of vibration for an object-spring system is T 2
m k . Thus, if T = 0.223 s and
m = 35.4 g = 35.4
10 3 kg, the force constant of the spring is
k
4
2m
T
2
4
2
35.4 10
3
kg
0.223 s
2
28.1 N m
13.25 The spring constant is found from
k
Fs
x
mg
x
0.010 kg 9.80 m s
3.9 10
-2
m
2
2 .5 N m
When the object attached to the spring has mass m = 25 g, the period of oscillation is
T
m 0.025 kg
2 2 0.63 s
k 2 .5 N m
13.26 The springs compress 0.80 cm when supporting an additional load of m = 320 kg. Thus, the spring constant is
k
mg
x
320 kg 9.80 m s
0.80 10
-2
m
2
3.9 105
N m
When the empty car, M = 2.0
10 3 kg, oscillates on the springs, the frequency will be
Page 13.9

Chapter 13
f
1 1 k 1 3.9 105
N m
T 2 M 2 2.0 103
kg
2.2 Hz
13.27 (a) The period of oscillation is T 2
m k where k is the spring constant and m is the mass of the object
attached to the end of the spring. Hence,
T
0.250 kg
2 1.0 s
9.5 N m
(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
k
vmax
A A m
9.5 N m
0.250 kg
4.5 10
2
m 0.28 m s
(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
k
2 2
9.5 N m 2 2
v A x
0.045 m 0.020 m 0.25 m s
m
0.250 kg
13.28 The general expression for the position as a function of time for an object undergoing simple harmonic motion with x
= 0 at t = 0 is x Asin
t . Thus, if x 5.2 cm sin 8.0
t , we have that the amplitude is
A = 5.2 cm and the angular frequency is 8.0 rad s .
(a)
The period is
T
2 2
8.0 s
-1
0.25 s
(b)
The frequency of motion is
f
1 1
T 0.25 s
4.0 s-1
4.0 Hz
(c) As discussed above, the amplitude of the motion is A 5.2 cm
(d)
Note: For this part, your calculator should be set to operate in radians mode.
Page 13.10

Chapter 13
If x = 2.6 cm, then
t
x
A
2.6 cm
5.2 cm
sin
1
sin
1
sin
1
0.50 0.52 radians
and
t
0.52 rad 0.52 rad
8.0 rad s
2.1 10
2
s 21 10
3
s 21 ms
13.29 (a) At the equilibrium position, the total energy of the system is in the form of kinetic energy and mv2 max 2 E ,
so the maximum speed is
v
max
2E
m
2 5.83 J
0.326 kg
5.98 m s
(b) The period of an object-spring system is T 2
m k , so the force constant of the spring is
k
4
2m
T
2
4
2
0.326 kg
0.250 s
2
206 N m
(c)
At the turning points, x = A, the total energy of the system is in the form of elastic potential energy, or
E kA 2 2 , giving the amplitude as
A
2E
k
2 5.83 J
206 N m
0.238 m
13.30 For a system executing simple harmonic motion, the total energy may be written as
s
1 2 1 2
2 max 2
E KE PE mv kA , where A is the amplitude and v max is the speed at the equilibrium position.
Observe from this expression, that we may write v2 2
max kA m .
1
(a) If v 2
v , then 1 2 1 2 1 2
max E mv kx mv
2 2 2 max
becomes
1 v2
max 1
m
2
1
kx mv
2 4 2 2
2
max
and gives
Page 13.11

Chapter 13
2
3 m
2
3 m k
2
3A
x vmax
A
4 k 4 k m 4
2
or
x
A 3
2
(b)
If the elastic potential energy is PE s = E/2, we have
1 1 1
2 2 2
kx2 kA
2
or
A
2
2
x
2
and
x
A
2
13.31 (a) At t = 3.50 s,
F kx N
rad
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
k
m
5.00 N m
2.00 kg
1.58 rad s
and the period of oscillation is
2 2
T 3.97 s .
1.58 rad s
Hence the number of oscillations made in 3.50 s is
N
t
T
3.50 s
3.97 s
0.881
13.32 (a)
k
F
x
7.50 N
3.00 10
-2
m
250 N m
(b)
k
m
250 N m
0.500 kg
22.4 rad s
Page 13.12

Chapter 13
f
22.4 rad s
2 2
3.56 Hz
T
1 1
f 3.56 Hz
0.281 s
(c) At t = 0, v = 0 and x = 5.00 10 2 m, so the total energy of the oscillator is
1 1
E KE PEs
mv kx
2 2
1
2
2 2
2
0 250 N m 5.00 10
2
m 0.313 J
1
(d) When x = A, v 0 so E KE PE 0
2
s kA .
2
Thus,
A
2 E
k
2 0.313 J
250 N/m
5.00 10
2
m 5.00 cm
1
(e) At x = 0, KE mv2 2 max E , or
v
max
2 E
m
2 0.313 J
0.500 kg
1.12 m s
a
max
F
max
m
kA
m
250 N m 5.00 10
2
m
0.500 kg
25.0 m s
2
Note: To solve parts (f) and (g), your calculator should be set in radians mode.
(f)
At t = 0.500 s, Equation 13.14a gives the displacement as
250 N m
x A cos t A cos t k m 5.00 cm cos 0.500 s 0.919 cm
0.500 kg
(g)
From Equation 13.14b, the velocity at t = 0.500 s is
v A sin t A k m sin t k m
250 N m 250 N m
0.500 kg 0.500 kg
5.00 10
-2
m sin 0.500 s 1.10 m
Page 13.13

Chapter 13
and from Equation 13.14c, the acceleration at this time is
a A
2
cos t A k m cos t k m
250 N m 250 N m
5.00 10 m cos 0.500 s 4.59 m s
0.500 kg 0.500 kg
-2 2
13.33 From Equation 13.6,
k
v A x A x
m
2 2 2 2 2
Hence,
v A2 A2 cos2 t A 1 cos2
t A sin t
From Equation 13.2,
k
m
a x
2
A cos t
2A cos t
13.34 (a) The height of the tower is almost the same as the length of the pendulum. From T 2
L g , we obtain
L
9.80 m s2
15.5 s
gT2
4 4
2 2
2
59.6 m
(b)
On the Moon, where g = 1.67 m/s 2 , the period will be
T
L 59.6 m
2 2 37.5 s
g 1.67 m s2
13.35 The period of a pendulum is the time for one complete oscillation and is given by T 2
g , where l is the
length of the pendulum.
(a)
T
3.00 min 60 s
120 oscillations 1 min
1.50 s
(b)
The length of the pendulum is
g
T
2
1.50 s
9.80 m s2
0.559 m
4
2
4
2
2
Page 13.14

Chapter 13
13.36 The period in Tokyo is TT 2 LT / g T
and the period in Cambridge is TC 2 LC / g C
. We know that
T T = T C = 2.000 s, from which, we see that
L
L
g
T C
gT
g , or C C 0.994 2
C gT
LT
0.992 7
L
1.001 5
13.37 (a) The period of the pendulum is T 2
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
T
T
Moon
Earth
2 LgMoon
gEarth
9.80
2 Lg g 1.63
Earth
Moon
2 .45
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 reads 12:00 midnight, the Moon clock will have
ticked off 24.0 h/2.45 = 9.80 h and will read 9 : 48 AM .
13.38 (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 2 L / g at 20°C, and at –5.0°C it is T 2 L / g . The ratio of these
periods is T0 / T L0
/ L .
From Chapter 10, the length at –5.0°C is L L0 AlL0
T , so
L
0
L
1 1 1
1 6 1
Al( T) 0.999 4
1 24 10 C 5.0 C 20 C
1.000 6
This gives
T
L
0 0
T L
1.000 6 1.000 3
Thus in one hour (3 600 s), the chilled pendulum will gain 1.000 3 1 3 600 s 1.1 s .
13.39 (a) From T 2 L g , the length of a pendulum with period T is L = gT 2 /4 2 .
Page 13.15

Chapter 13
On Earth, with T = 1.0 s,
L
9.80 m s2
1.0 s
4
2
2
0.25 m 25 cm
If T = 1.0 s on Mars,
L
2 2
3.7 m s 1.0 s
4
2
0.094 m 9.4 cm
(b) The period of an object on a spring is T 2
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
m
kT2
10 N m 1.0 s
4
2
4
2
2
0.25 kg
13.40 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 hanging from a vertical string in the elevator and
appears to be at rest to the elevator passengers. These passengers believe the tension in the string is the negative
of the object’s weight, or
T m g app
where
app
g is the apparent free-fall acceleration in the elevator.
An observer located outside the elevator applies Newton’s second law to this object by writing
F T mg ma e
where
e
T m ae
g mg app
, which gives
app e
a is the acceleration of the elevator and all its contents. Thus,
g g a .
(a) If we choose downward as the positive direction, then a 5.00 m s2
in this case and
g 9.80 5.00 m s2 14.8 m s2
app
(downward). The period of the pendulum is
e
T
L
5.00 m
2 2 3.65 s
g 14.8 m s2
app
(b) Again choosing downward as positive, a 5.00 m s2
and g 9.80 5.00 m s2 4.80 m s2
(downward) in this case. The period is now given by
e
app
T
L
5.00 m
2 2 6.41 s
g 4.80 m s2
app
Page 13.16

Chapter 13
(c) If a 5.00 m s2
horizontally, the vector sum gapp g a e is
e
as shown in the sketch at the right. The magnitude is
g app
2 2
2 2 2
5.00 m s 9.80 m s 11.0 m s
and the period of the pendulum is
T
L
5.00 m
2 2 4.24 s
g 11.0 m s2
app
13.41 (a) The distance from the bottom of a trough to the top of a crest
is twice the amplitude of the wave. Thus, 2A = 8.26 cm and
A
4.13 cm
(b)
The horizontal distance from a crest to a trough is a half
wavelength. Hence,
2
5.20 cm
and
10.4 cm
Figure P13.41
(c)
The period is
T
1 1
f 18.0 s
1
5.56 10
2
s
(d) The wave speed is v f 10.4 cm 18.0 s
1
187 cm s 1.87 m s
13.42 (a) The amplitude is the magnitude of the maximum
displacement from equilibrium (at x = 0). Thus,
A
2.00 cm
(b)
The period is the time for one full cycle of the motion.
Therefore,
T
4.00 s
(c) The period may be written as T 2
frequency is
, so the angular
Figure P13.42
2 2
T 4.00 s 2
rad s
Page 13.17

Chapter 13
1 1
(d) The total energy may be expressed as E mv2 kA
2. Thus, vmax A k / m , and since km,
this becomes vmax
A and yields
2 max 2
vmax A rad s 2.00 cm cm s
2
(e) The spring exerts maximum force, F k x , when the object is at maximum distance from equilibrium, i.e.,
at
x A 2.00 cm . Thus, the maximum acceleration of the object is
2
max kA
2 2
F
amax A rad s 2.00 cm 4.93 cm s
m m
2
(f)
The general equation for position as a function of time for an object undergoing simple harmonic motion with t
= 0 when x = 0 is x = A sin ( t). For this oscillator, this becomes
x 2.00 cm sin t 2
13.43 (a) The period and the frequency are reciprocals of each other. Therefore,
T
1 1 1
f 101.9 MHz 101.9 10 s
6 1
9.81 10
9
s 9.81 ns
(b)
v
f
3.00 108
m s
101.9 10 s
6 1
2.94 m
13.44 (a) The frequency of a transverse wave is the number of crests that pass a given point each second. Thus, if 5.00
crests pass in 14.0 seconds, the frequency is
f
5.00
14.0 s
0.357 s
1
0.357 Hz
(b)
The wavelength of the wave is the distance between successive maxima or successive minima. Thus,
2.76 m and the wave speed is
v f
2.76 m 0.357 s
1
0.985 m s
13.45 The speed of the wave is
Page 13.18

Chapter 13
v
x
t
425 cm 42 .5 cm s
10.0 s
and the frequency is
f
40.0 vib
30.0 s
1.33 Hz
Thus,
v
f
42.5 cm s
1.33 Hz
31.9 cm
13.46 From v = f, the wavelength (and size of smallest detectable insect) is
v
f
340 m s
60.0 103
Hz
5.67 10
-3
m 5.67 mm
13.47 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
8.50 cm . Thus, the wave speed is
v 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
v
10.0 m
0.170 m s
58.8 s
13.48 (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
4.0 m s
20 m
0.20 Hz
(b) Taking eastward as positive, vwave, boat vwave, water v boat,
water
(see the discussion of relative velocity in
Chapter 3 of the textbook) gives
v wave, boat 4.0 m s 1.0 m s 5.0 m s and vboat, wave v wave, boat 5.0 m s
Thus,
Page 13.19

Chapter 13
f
v boat wave
, 5.0 m s
20 m
0.25 Hz
13.49 The down and back distance is 4.00 m + 4.00 m = 8.00 m.
The speed is then
v
d total
t
4 8.00 m
0.800 s
40.0 m s
F
Now,
m
L
0.200 kg
4.00 m
5.00 10
2
kg m
so
F v
5.00 10 kg m 40.0 m s 80.0 N
2 2
2
13.50 The speed of the wave is
v
x
t
20.0 m
0.800 s
25.0 m s
and the mass per unit length of the rope is m/ L 0.350 kg m . Thus, from v F , we obtain
2
2
25.0 m s 0.350 kg m 219 N
F v
13.51 (a) The speed of transverse waves in the cord is v F , where mL is the mass per unit length. With
the tension in the cord being F = 12.0 N, the wave speed is
v
F F FL
m L m
12.0 N 6.30 m
0.150 kg
22.4 m s
(b)
The time to travel the length of the cord is
t
L
v
6.30 m
22.4 m s
0.281 s
13.52 (a) In making 6 round trips, the pulse travels the length of the line 12 times for a total distance of 144 m. The
speed of the pulse is then
Page 13.20

Chapter 13
v
x
t
12L
t
12 12.0 m
2.96 s
48.6 m s
(b) The speed of transverse waves in the line is v F , so the tension in the line is
2
m
2
0.375 kg
F v v
L 12.0 m
2
48.6 m s 73.8 N
13.53 (a) The mass per unit length is
m
L
0.0600 kg
5.00 m
0.0120 kg m
From v F , the required tension in the string is
2 2
F v 50.0 m s 0.0120 kg m 30.0 N ,
(b)
v
F
8.00 N
0.0120 kg m
25.8 m s
13.54 The mass per unit length of the wire is
m
L
4.00 10
-3
kg
1.60 m
2 .50 10
-3
kg m
and the speed of the pulse is
v
L
t
1.60 m
0.0361 s
44.3 m s
Thus, the tension in the wire is
F v
2 2
-3
44.3 m s 2.50 10 kg m 4.91 N
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
F
m
4.91 N
3.00 kg
1.64 m s
2
13.55 The period of the pendulum is T 2 L / g , so the length of the string is
Page 13.21

Chapter 13
L
9.80 m s2
2.00 s
gT
2
4 4
2 2
2
0.993 m
Then mass per unit length of the string is then
m
L
0.060 0 kg kg
0.060 4
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 s2
49.0 N
and the speed of transverse waves in it is
v
F
49.0 N
0.060 4 kg m
28.5 m s
13.56 If
1 m1
L is the mass per unit length for the first string, then
2 m2 L m1 2L 1 2 is that of the
second string. Thus, with F2 F1
F , the speed of waves in the second string is
F 2 F F
v 2 2 v 2 5.00 m s 7.07 m s
2 1
2 1 1
13.57 (a) The tension in the string is F mg 3.0 kg 9.80 m s2
29 N . Then, from v F , the mass per
unit length is
F
v2
29 N
24 m s
2
0.051 kg m
(b)
When m = 2.00 kg, the tension is
F mg
2.0 kg 9.80 m s2
20 N
and the speed of transverse waves in the string is
v
F
20 N
0.051 kg m
20 m s
13.58 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
Page 13.22

Chapter 13
v
F A Stress Stress
m/ L m /( A L)
But,
Stress
A L V =volume , so m A L density . Thus, v
.
When the stress is at its maximum, the speed of waves in the wire is
v
( Stress) 9
max 2.70 10 Pa
max 3 3
7.86 10 kg/m
586 m s
13.59 (a) The speed of transverse waves in the line is v F , with mL being the mass per unit length.
Therefore,
v
F F FL
m L m
12.5 N 38.0 m
2.65 kg
13.4 m s
(b)
The worker could throw an object, such as a snowball, at one end of the line to set up a pulse, and use a
stopwatch to measure the time it takes a pulse to travel the length of the line. From this measurement, the wo
rker would have an estimate of the wave speed, which in turn can be used to estimate the tension.
13.60 (a) In making n round trips along the length of the line, the total distance traveled by the pulse is
x n 2L 2nL . The wave speed is then
v
x
t
2nL
t
(b) From v F as the speed of transverse waves in the line, the tension is
F
v
2
2 2 2 2 2
M 2nL M 4n L 4n ML
L t L t2 t2
13.61 (a) Constructive interference produces the maximum amplitude
Amax A1 A 2 0.50 m
(b)
Destructive interference produces the minimum amplitude
Page 13.23

Chapter 13
Amin A1 A 2 0.10 m
13.62 We are given that x Acos( t) (0.25 m)cos(0.4 t ) .
(a) By inspection, the amplitude is seen to be A 0.25 m
(b) The angular frequency is 0.4 rad s . But km, so the spring constant is
k m
2
(0.30 kg)(0.4 rad s)
2
0.47 N m
(c)
Note: Your calculator must be in radians mode for part (c).
At t = 0.30 s, x 0.25 m cos 0.4 rad s 0.30 s 0.23 m
(d) From conservation of mechanical energy, the speed at displacement x is given by v A2 x
2
. Thus, at t
= 0.30 s, when x = 0.23 m, the speed is
v
0.4 rad s (0.25 m)
2
(0.23 m)
2
0.12 m s
13.63 (a) The period of a vibrating object-spring system is T 2 2
m k , so the spring constant is
k
4
2m
T
2
4
2
2.00 kg
0.600 s
2
219 N m
(b)
If T = 1.05 s for mass m 2 , this mass is
m
kT (219 N/m)(1.05 s)
4 4
2 2
2 2 2
6.12 kg
13.64 (a) The period is the reciprocal of the frequency, or
T
1 1
f 196 s
1
5.10 10
3
s 5.10 ms
(b)
v sound
f
343 m s
196 s
1
1.75 m
13.65 (a) The period of a simple pendulum is T 2
g , so the period of the first system is
Page 13.24

Chapter 13
T
0.700 m
2 2 1.68 s
g 9.80 m s
1 2
(b) The period of mass-spring system is T 2
m k , so if the period of the second system is T 2 = T 1 , then
2 m k 2
l g and the spring constant is
k
mg
1.20 kg 9.80 m s 2
0.700 m
16.8 N m
13.66 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 g PE s KE PE f
g PE s , or 1
0 0 kx
2 0 0
i
2 max
mgh
i
Thus,
x
max
2mgh i
k
2 0.500 kg 9.80 m s2
2.00 m
20.0 N m
0.990 m
13.67 Choosing PE g = 0 at the initial height of the block, conservation of mechanical energy gives
KE PE PE KE PE PE , or
g s
f
g s
i
1 1
2 2
mv2 mg x kx
2
0 ,
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
2 3.00 kg 9.80 m s
2
1
max
2
max
2 kx mgx or 2 mg
k
xmax
0.100 m
588 N m
(b)
When x = 5.00 cm = 0.050 0 m, the energy equation gives
v
2gx
kx2
m ,
or
Page 13.25

Chapter 13
v
2
588 N m 0.050 0 m
2 9.80 m s2
0.050 0 m 0.700 m s
3.00 kg
13.68 (a) We apply conservation of mechanical energy from just after the collision until the block comes to rest.
KE PE s KE PE f
s
gives 1 2 1
0 k x 2 0
i
f MV
2 2
or the speed of the block just after the collision is
V
kx2
f
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 v 0 m v MV
bullet
i
bullet
f
or
M
vbullet
v
f bullet V
i
m
1.00 kg
400 m s 1.5 m s 100 m s
5.00 10
-3
kg
(b)
The mechanical energy converted into internal energy during the collision is
1 2 1 2 1
i f bullet i
bullet f
E KE KE 2 m v 2 m v 2
MV
2
or
E
1 1
2 2
3
2 2 2
5.00 10 kg 400 m s 100 m s 1.00 kg 1.50 m s
E
374 J
13.69 Choose PE g = 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 PE PE KE PE PE , or
g s
f
g s
i
Page 13.26

Chapter 13
1 1
m m v m g x sin 40 m g x 0 0 0 k x
2 2
2 2
1 2 f 1 2
1
2
25 30 kg v2 25 kg 9.80 m s2
f
0.200 m sin 40
30 kg 9.80 m s2
0.200 m 200 N m 0.200 m
1
2
2
yielding
v f
1.1 m s
13.70 (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 1 mv
2 1 kx 2 , or
2 2 i
v
x
i
k
m
9.80 N m
0.200 m 19.8 m s
1.00 10
3
kg
1
(b) From y v t a t
2
, the time required for the pellet to drop 1.00 m to the floor, starting with v 0 0 ,
is
0y
2
y
y
t
2 y 2 1.00 m
a
y
9.80 m s
2
0.452 s
The range (horizontal distance traveled during the flight) is then
x v0xt
19.8 m s 0.452 s 8.94 m
Page 13.27

Chapter 13
13.71 The free-body diagram at the right shows the forces acting on the balloon
when it is displaced distance s = L along the circular arc it follows. The
net force tangential to this path is
F F Bsin mg sin B mg sin
net
x
For small angles, sin s/
L
Also,
mg
He
V g
and the buoyant force is
acting on the balloon is
B
air
V g . Thus, the net restoring force
F
net
air
L
He
Vg
s
Observe that this is in the form of Hooke’s law, F k s , with k air He
Vg L
Thus, the motion will be simple harmonic and the period is given by
T
1 2 m
HeV
He L
2 2 2
f k Vg L
g
air
He
air
He
This yields
T
0.180 3.00 m
2 1.40 s
1.29 0.180 9.80 m s2
13.72 (a) When the block is given some small upward displacement, the net restoring force exerted on it by the rubber
bands is
F F 2 F sin , where tan
net
y
y
L
For small displacements, the angle will be very small. Then sin tan
is
y
, and the net restoring force
L
Page 13.28

Chapter 13
y 2 F
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 k = 2F/L. Thus, the motion
will be simple harmonic, and the angular frequency is
k
m
2 F
m L
13.73 Newton’s law of gravitation is
GMm
4
F , where M r
r2
3
3
Thus,
F=
4
3
Gm r
which is of Hooke’s law form, F
k r , with
4
k = 3
Gm
13.74 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
k
4.7 10
4
N m
inner, max spring, max 3
v v A m
0.20 cm 0.25 cm s
0.30 10 kg
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, we may write
v
r
outer
outer
v
r
inner
inner
The maximum speed of the outer tip is then
Page 13.29

Chapter 13
v
r
outer
outer, max vinner, max
rinner
15.0 mm 0.25 cm s 1.3 cm s
3.00 mm
13.75 (a)
k
m
500 N m
2.00 kg
15.8 rad s
(b)
Apply Newton’s second law to the block while the elevator is accelerating:
Fy Fs mg ma y
With F kx and a g 3 , this gives kx m g g 3 , or
s
y
x
4 mg 4 2.00 kg 9.80 m s
3k
3 500 N m
2
5.23 10
2
m 5.23 cm
13.76 (a) Note that as the spring passes through the vertical position, the object is moving in a circular arc of radius
L y f . Also, observe that the y-coordinate of the object at this point must be negative y f 0 so the spring is
stretched and exerting an upward tension force of magnitude greater than the object’s weight. This is necessary
so the object experiences a net force toward the pivot to supply the needed centripetal acceleration in this
position. This is summarized by Newton’s second law applied to the object at this point, stating
mv2
Fy
ky f mg
L y
f
(b) Conservation of energy requires that E KEi PEg, i PEs, i KEf PEg, f PE s,
f
, or
1
2
1
E 0 mgL 0 mv mgy
2
f ky f , reducing to 2mg L y
2 2
f mv ky f
2 2
(c) From the result of part (a), observe that mv2 ( L y )( ky mg ) . Substituting this into the result from
part (b) gives 2 mg( L y ) ( L y )( ky mg)
ky
2
. After expanding and regrouping terms, this
f f f f
becomes (2 k) y2
(3 mg kL) y ( 3 mgL ) 0 , which is a quadratic equation ay2 by c 0 , with
f
f
f
f
f
f
a 2k
2 1 250 N m 2.50 103
N m
b 3mg kL 3 5.00 kg 9.80 m s2 1 250 N m 1.50 m 1.73 103
N
and
Page 13.30

Chapter 13
c 3mgL
3 5.00 kg 9.80 m s2
1.50 m 221 N m
Applying the quadratic formula, keeping only the negative solution [see the discussion in part (a)] gives
y
f
b b2
4ac
2
3 3 3
1.73 10 1.73 10 4 2.50 10 221
2a
2 2.50 10
3
or
y f
0.110 m
(d)
Because the length of this pendulum varies and is longer throughout its motion than a simple pendulum of
length L, its period will be longer than that of a simple pendulum.
13.77 The maximum acceleration of the oscillating system is
a A f A
2
max
2
2
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 f n mg ma and we must have
s s max s s
max
2
amax 2 f A
sg
Thus,
A
s g
0.600 9.80 m s
2 2
2 f 2 1.50 Hz
2
6.62 10
2
m 6.62 cm
Page 13.31