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
Fs
(b)
kxi
17 N or 17 N to the left
Fs
m
17 N
0.60 kg
28 m s2
or
a
28 m s2 to the left
When the object comes to equilibrium (at distance y0 below the unstretched position of the end of the spring),
Fy
k
y0
k
13.3
0.13 m
At this instant, the acceleration is
a
13.2
130 N m
(a)
0 and the force constant is
mg
4.25 kg 9.80 m s 2
mg
y0
2.62
10
2
m
1.59
103 N
1.59 kN
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
t
1
2
v0 y t
2
ay t 2 , with v0 y
y
ay
2
4.00 m
9.80 m s2
0 , the time required for the ball to reach the ground is
0.904 s
This is one-half of the time for a complete cycle of the motion. Thus, the period is T
(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)
1.81 s .
The spring constant is
Page 13.1
Chapter 13
Fs
x
mg
x
kx
1.0
k
F
(b)
13.5
Fs
50 N
10-2 m
1.0
103 N m 0.11 m
103 N m
102 N
1.1
The graph will be a straight line passing through the origin with a slope equal to k
When the system is in equilibrium, the tension in the spring F
1.0
103 N m .
k x must equal the weight of the object. Thus,
mg giving
k x
47.5 N 5.00
k x
m
13.6
5.0
g
9.80 m
10
2
m
s2
0.242 kg
(a) The free-body diagram of the point in the center of the string is
given at the right. From this, we see that
Fx
or
(b)
T
0
F
F
2 sin 35.0
2T sin 35.0
0
375 N
2 sin 35.0
327 N
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
F
x
k
13.7
(a)
375 N
0.300 m
1.25
103 N m
When the block comes to equilibrium,
y0
mg
k
Fy
ky0
10.0 kg 9.80 m s2
475 N m
mg
0 giving
0.206 m
or the equilibrium position is 0.206 m below the unstretched position of the lower end of the spring.
(b)
2.00 m s2 , applying Newton’s
When the elevator (and everything in it) has an upward acceleration of a
second law to the block gives
Fy
k y0
y
mg
may
or
Fy
Page 13.2
ky0
mg
ky
may
Chapter 13
where y = 0 at the equilibrium position of the block. Since
ky0
0 [see part (a)], this becomes
mg
ma and the new position of the block is
ky
ma y
y
2.00 m s2
10.0 kg
k
4.21
475 N m
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 ay = g. The new “equilibrium”
position of the block is found from
Fy
ky0
mg
g , which yields y0
m
0 . When the cable
snapped, the block was at rest relative to the elevator at distance
y0
y
0.206 m
0.248 m below the new “equilibrium” position. Thus, while the elevator is
0.042 1 m
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
kx0
2
(b)
(a)
or
k
3.00
mv02
x02
10
8.00
3
kg 45.0 m s
10
2
m
2
2
k x
949 N m 8.00
10
2
m
75.9 N
Assume the rubber bands obey Hooke’s law. Then, the force constant of each band is
k
Fs
x
1.0
15 N
10-2 m
1.5
103 N m
Thus, when both bands are stretched 0.20 m, the total elastic potential energy is
PEs
(b)
949 N m
The magnitude of the force required to compress the spring 8.00 cm and load the gun is
Fs
13.9
1
mv02
2
2
1 2
kx
2
1.5
103 N m 0.20 m
Conservation of mechanical energy gives KE
PEs
Page 13.3
2
60 J
f
KE
PEs i , or
Chapter 13
1
mv2
2
13.10
13.11
0
Fmax
xmax
230 N
0.400 m
(a)
k
(b)
work done
2 60 J
60 J , so v
0
575 N m
1 2
kx
2
PEs
49 m s
10 -3 kg
50
1
575 N m 0.400
2
2
46.0 J
From conservation of mechanical energy,
KE
PEg
PEs
f
KE
PEg
PEs
i
or 0
mgh f
0
0
0
1 2
kxi
2
giving
13.12
xi2
2.00
Conservation of mechanical energy, (KE
1
2
mvi2
0
0
0
1
2
0
10
2
0.600 m
m
PEg
2
PEs ) f
(KE
2.94
103 N m
PEg
PEs )i , gives
kx2f , or
5.00 106 N m
3.16
1000 kg
k
xi
m
vi
13.13
2 0.100 kg 9.80 m s2
2 mgh f
k
10
2
m
2.23 m s
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
m
V
M
m
vi
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
0
1
2
1
2
kx2f
xf
M
V
m V 2 , or
M
m
k
1.49 m s
2.01 kg
19.6 N m
Page 13.4
0.477 m
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
(b)
When the object is distance x from the equilibrium position, the elastic potential energy is PEs
1
mv 2
2
(a)
kx 2
1 2
kx
2
3
KE
k A2 2
3k 2
x
PEs
or
2 PEs
PEs
3PEs , or
A
x
3
At maximum displacement from equilibrium, all of the energy is in the form of elastic potential energy, giving
2
kxmax
2 , and
E
2 47.0 J
2E
2
xmax
k
(b)
1
mv 2
2
or
When KE = 2PE, conservation of energy gives E
1 2
kA
2
13.15
1 2
kx
2
2
kx2 2 and
mv2 2 . At the position where KE = 2PEs, is it necessary that
the kinetic energy is KE
(c)
kA2 2 .
0.240 m
1.63
2
103 N m
At the equilibrium position (x = 0), the spring is momentarily in its relaxed state and PEs = 0, so all of the
energy is in the form of kinetic energy. This gives
KE
(c)
1
2
mvmax
2
E
47.0 J
2 47.0 J
2E
2
vmax
3.45 m s
2
7.90 kg
At any position, the constant total energy is, E
v
(e)
0
If, at the equilibrium position, v = vmax = 3.45 m/s, the mass of the block is
m
(d)
x
kx 2
2E
m
2 47.0 J
1.63
KE
PEs
mv2 2
103 N m 0.160 m
7.90 kg
At x = 0.160 m, where v = 2.57 m/s, the kinetic energy is
Page 13.5
kx2 2 , so at x = 0.160 m
2
2.57 m s
Chapter 13
KE
(f)
1
mv2
2
1
7.90 kg 2.57 m s
2
2
26.1 J
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,
PEs
(g)
1 2
kx
2
1
1.63
2
103 N m 0.160 m
2
20.9 J
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
kx
2
E
Eloss
47.0 J
14.0 J
33.0 J
and
x
13.16
2 33.0 J
2 33.0 J
k
1.63
0.201 m
103 N m
10
2
(a)
F
k x
83.8 N m 5.46
m
4.58 N
(b)
E
PEs
1 2
kx
2
(c)
While the block was held stationary at x = 5.46 cm,
1
83.8 N m 5.46
2
10
2
m
2
Fx
0.125 J
Fs
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 Fs = 4.58 N directed toward the equilibrium position acting on the block. This gives the block an
acceleration having magnitude
a
(d)
Fs
m
4.58 N
0.250 kg
18.3 m s2
At the equilibrium position, PEs = 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,
we have
1
2
KE
PEg
mv2
0
k 2
A
m
0.40 kg
(a)
i
kA2 , or
19.6 N m
0.40 kg
0.040 m
2
0.28 m s
0.040 m
2
0.015 m
2
0.26 m s
0.040 m
2
0.015 m
2
0.26 m s
When x = +0.015 m,
19.6 N m
0.40 kg
1
2
If v
k
A2
m
vmax , then
This gives A2
13.18
1
2
0
PEs
x2
19.6 N m
v
(d)
0
PEg
When x = 0.015 m,
v
(c)
kx2
KE
f
The speed is a maximum at the equilibrium position, x = 0.
vmax
(b)
1
2
k
A2
m
v
(a)
PEs
x2 = A2/4, or x
KE = 0 at x = A, so E
E
1 2
kA
2
KE
x2
1
2
k 2
A
m
A 3 /2
PEs
4.0 cm
1
2
0
1
250 N m 0.035 m
2
2
3 /2
3.5 cm .
kA2 , or the total energy is
0.15 J
Page 13.7
Chapter 13
(b)
The maximum speed occurs at the equilibrium position where PEs = 0. Thus, E
2E
m
vmax
(c)
The acceleration is a
k
amax
13.19
k
m
A
F/m
A
2 , or
mvmax
0.78 m s
kx / m . Thus, a = amax at x = xmax = A.
k
A
m
m
250 N m
0.50 kg
0.035 m
1
2
250 N m
0.50 kg
0.035 m
18 m s2
The maximum speed occurs at the equilibrium position and is
k
A
m
vmax
Thus,
16.0 N m 0.200 m
m
kA2
2
vmax
Fg
mg
4.00 kg 9.80 m s2
x2
10.0 N m
50.0 10-3 kg
0.400 m s
2
2
4.00 kg ,
and
13.20
v
13.21
(a)
k
A2
m
39.2 N
0.250 m
2
0.125 m
2
3.06 m s
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
vt
r
vcar
3.00 m s and the angular velocity of the tire is
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)
vt
2 r
T
(b)
f
1
T
0.200 m
0.628 m s
2.00 s
1
2.00 s
2
T
(c)
13.23
2
0.500 Hz
2
2.00 s
3.14 rad s
The angle of the crank pin is
x
A cos
t . Its x-coordinate is
A cos 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
13.25
2
4
2m
4
m k . Thus, if T = 0.223 s and
10 3 kg, the force constant of the spring is
m = 35.4 g = 35.4
k
2
35.4
T2
10
0.223 s
3
kg
28.1 N m
2
The spring constant is found from
k
Fs
x
0.010 kg 9.80 m s2
mg
x
3.9
10-2 m
2 .5 N m
When the object attached to the spring has mass m = 25 g, the period of oscillation is
T
13.26
2
m
k
2
0.025 kg
2 .5 N m
0.63 s
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 s2
0.80
When the empty car, M = 2.0
10-2 m
3.9
105 N m
103 kg, oscillates on the springs, the frequency will be
Page 13.9
Chapter 13
f
13.27
(a)
1
T
1
2
k
M
3.9 10 5 N m
2.0 103 kg
1
2
The period of oscillation is T
2.2 Hz
m k where k is the spring constant and m is the mass of the object
2
attached to the end of the spring. Hence,
T
(b)
0.250 kg
9.5 N m
2
If the cart is released from rest when it is 4.5 cm from the equilibrium position, the amplitude of oscillation will
10 2 m. The maximum speed is then given by
be A = 4.5 cm = 4.5
vmax
(c)
1.0 s
A
k
m
A
4.5
10
2
m
9.5 N m
0.250 kg
0.28 m s
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
A2
m
v
13.28
9.5 N m
0.250 kg
x2
0.045 m
2
0.020 m
2
0.25 m s
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
A sin
t . Thus, if x
A = 5.2 cm and the angular frequency is
(a)
8.0
t , we have that the amplitude is
rad s .
The period is
T
(b)
5.2 cm sin 8.0
2
2
8.0 s-1
0.25 s
The frequency of motion is
f
1
T
1
0.25 s
4.0 s-1
4.0 Hz
(c)
As discussed above, the amplitude of the motion is A
(d)
Note: For this part, your calculator should be set to operate in radians mode.
Page 13.10
5.2 cm
Chapter 13
If x = 2.6 cm, then
t
x
A
1
sin
sin
2.6 cm
5.2 cm
1
sin
1
0.50
2
s
21
0.52 radians
and
0.52 rad
t
13.29
(a)
0.52 rad
8.0 rad s
2.1
10
10
3
s
21 ms
2
At the equilibrium position, the total energy of the system is in the form of kinetic energy and mvmax
2
E,
so the maximum speed is
(b)
4
2m
4
2
T2
0.326 kg
0.250 s
2
m k , so the force constant of the spring is
206 N m
2
At the turning points, x = A, the total energy of the system is in the form of elastic potential energy, or
E
kA2 2 , giving the amplitude as
2 5.83 J
2E
k
A
13.30
5.98 m s
0.326 kg
The period of an object-spring system is T
k
(c)
2 5.83 J
2E
m
vmax
206 N m
0.238 m
For a system executing simple harmonic motion, the total energy may be written as
E
KE
PEs
1
2
2
mvmax
1
2
kA2 , where A is the amplitude and vmax is the speed at the equilibrium position.
2
Observe from this expression, that we may write vmax
(a)
If v
1
2
vmax , then E
v2
1
m max
2
4
1 2
kx
2
1
2
mv2
1
2
kx2
1
2
kA2 m .
2
becomes
mvmax
1 2
mvmax
2
and gives
Page 13.11
Chapter 13
3 m 2
v
4 k max
x2
3 m
4 k
3 A2
4
k 2
A
m
or
A 3
2
x
(b)
If the elastic potential energy is PEs = E/2, we have
1 2
kx
2
13.31
(a)
or
A2
2
x2
and
or F
kx
5.00
N
(3.00 m) cos
m
1.58
rad
(3.50 s)
s
11.0 N directed to the left
The angular frequency is
k
m
5.00 N m
2.00 kg
1.58 rad s
and the period of oscillation is
2
T
2
1.58 rad s
3.97 s .
Hence the number of oscillations made in 3.50 s is
t
T
N
13.32
(a)
(b)
x
A
2
At t = 3.50 s,
F
(b)
1 1 2
kA
2 2
k
F
x
k
m
3.50 s
3.97 s
0.881
7.50 N
3.00 10-2 m
250 N m
250 N m
0.500 kg
22.4 rad s
Page 13.12
11.0 N ,
Chapter 13
f
T
(c)
22.4 rad s
2
2
1
f
1
3.56 Hz
KE
10 2 m, so the total energy of the oscillator is
1
mv 2
2
PEs
1 2
kx
2
1
250 N m 5.00
2
0
(d)
0.281 s
At t = 0, v = 0 and x = 5.00
E
3.56 Hz
When x = A, v
0 so E
KE
2
10
m
2
PEs
0
5.00
10
0.313 J
1 2
kA .
2
Thus,
A
(e)
2 0.313 J
2E
k
250 N/m
1
2
At x = 0, KE
vmax
2E
m
amax
Fmax
m
2
mvmax
2
m
5.00 cm
E , or
2 0.313 J
0.500 kg
kA
m
1.12 m s
250 N m 5.00
10
2
m
0.500 kg
25.0 m s2
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
x
(g)
A cos
t
A cos t k m
5.00 cm cos
0.500 s
250 N m
0.500 kg
From Equation 13.14b, the velocity at t = 0.500 s is
v
A sin
5.00
t
10 -2 m
A k m sin t k m
250 N m
sin
0.500 kg
0.500 s
Page 13.13
250 N m
0.500 kg
1.10 m
0.919 cm
Chapter 13
and from Equation 13.14c, the acceleration at this time is
a
2
A
cos
5.00
13.33
t
A k m cos t k m
250 N m
cos
0.500 kg
10-2 m
250 N m
0.500 kg
0.500 s
4.59 m s 2
From Equation 13.6,
k
A2
m
v
x2
2
A2
x2
Hence,
A2
v
A2 cos2
t
A 1
cos2
t
A sin
t
From Equation 13.2,
k
x
m
a
13.34
(a)
2
t
A cos
t
9.80 m s2
g T2
4 2
4
15.5 s
2
L g , we obtain
2
59.6 m
2
On the Moon, where g = 1.67 m/s2, the period will be
T
13.35
A cos
The height of the tower is almost the same as the length of the pendulum. From T
L
(b)
2
2
L
g
2
59.6 m
1.67 m s2
37.5 s
The period of a pendulum is the time for one complete oscillation and is given by T
length of the pendulum.
3.00 min
60 s
120 oscillations 1 min
(a)
T
(b)
The length of the pendulum is
T2
g
4 2
9.80 m
1.50 s
s2
1.50 s
4
2
2
0.559 m
Page 13.14
2
g , where ℓ is the
Chapter 13
13.36
The period in Tokyo is TT
LT / gT and the period in Cambridge is TC
2
2
LC / gC . We know that
TT = TC = 2.000 s, from which, we see that
LT
gT
13.37
(a)
LC
g
, or C
gT
gC
LC
LT
0.994 2
0.992 7
The period of the pendulum is T
1.001 5
L g . Thus, on the Moon where the free-fall acceleration is smaller,
2
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
TEarth
2
L gMoon
2
L gEarth
gEarth
gMoon
9.80
1.63
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
periods is T0 / T
L / g at 20°C, and at –5.0°C it is T
2
L0
1
1
L / g . The ratio of these
L0 / L .
From Chapter 10, the length at –5.0°C is L
L0
L
2
Al (
Al L0
T , so
1
T)
1
24
10
6
C
1
5.0 C
20 C
1
0.999 4
1.000 6
This gives
T0
T
L0
L
1.000 6
1.000 3
Thus in one hour (3 600 s), the chilled pendulum will gain 1.000 3
13.39
(a)
From T
2
1 3 600 s
L g , the length of a pendulum with period T is L = gT2/4 2.
Page 13.15
1.1 s .
Chapter 13
On Earth, with T = 1.0 s,
9.80 m s2
L
4
2
1.0 s
2
0.25 m
25 cm
0.094 m
9.4 cm
If T = 1.0 s on Mars,
3.7 m s2
L
(b)
4
1.0 s
2
2
The period of an object on a spring is T
m k , which is independent of the local free-fall acceleration.
2
Thus, the same mass will work on Earth and on Mars. This mass is
13.40
10 N m 1.0 s
k T2
4 2
m
4
2
2
0.25 kg
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
m gapp where g app is the apparent free-fall acceleration in the elevator.
of the object’s weight, or T
An observer located outside the elevator applies Newton’s second law to this object by writing
F
T
mg
mae where a e is the acceleration of the elevator and all its contents. Thus,
g
ae .
m ae
(a)
If we choose downward as the positive direction, then ae
gapp
T
(b)
g
m gapp , which gives gapp
T
9.80
2
5.00
L
gapp
m s2
2
14.8 m s2 (downward). The period of the pendulum is
5.00 m
14.8 m s2
Again choosing downward as positive, ae
3.65 s
5.00 m s2 and gapp
(downward) in this case. The period is now given by
T
2
L
gapp
2
5.00 m s2 in this case and
5.00 m
4.80 m s2
6.41 s
Page 13.16
9.80
5.00
m s2
4.80 m s2
Chapter 13
(c)
5.00 m s2 horizontally, the vector sum gapp
If ae
g
ae is
as shown in the sketch at the right. The magnitude is
2
5.00 m s2
gapp
9.80 m s2
2
11.0 m s2
and the period of the pendulum is
T
13.41
(a)
2
L
gapp
5.00 m
11.0 m s2
2
4.24 s
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
(b)
4.13 cm
The horizontal distance from a crest to a trough is a half
wavelength. Hence,
Figure P13.41
2
(c)
10.4 cm
The period is
T
13.42
5.20 cm and
1
f
1
18.0 s
5.56
1
10
2
s
(d)
The wave speed is v
(a)
The amplitude is the magnitude of the maximum
f
10.4 cm 18.0 s
1
187 cm s
1.87 m s
displacement from equilibrium (at x = 0). Thus,
A
(b)
2.00 cm
The period is the time for one full cycle of the motion.
Therefore, T
4.00 s
Figure P13.42
(c)
The period may be written as T
2
, so the angular
frequency is
2
T
2
4.00 s
2
rad s
Page 13.17
Chapter 13
(d)
this becomes vmax
vmax
(e)
A
1
2
2
mvmax
kA2 . Thus, vmax
A k / m , and since
k m,
A and yields
rad s
2
2.00 cm
The spring exerts maximum force, F
at x
cm s
k x , when the object is at maximum distance from equilibrium, i.e.,
2.00 cm . Thus, the maximum acceleration of the object is
A
Fmax
amax
(f)
1
2
The total energy may be expressed as E
m
2
kA
m
2A
2
rad s
4.93 cm s2
2.00 cm
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
13.43
(a)
1
f
v
f
(b)
(a)
2
t
The period and the frequency are reciprocals of each other. Therefore,
T
13.44
2.00 cm sin
1
101.9 MHz
1
101.9
3.00 108 m s
101.9 106 s 1
106 s
1
9.81
10
9
s
9.81 ns
2.94 m
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
(b)
5.00
14.0 s
0.357 s
1
0.357 Hz
The wavelength of the wave is the distance between successive maxima or successive minima. Thus,
2.76 m and the wave speed is
v
13.45
f
2.76 m 0.357 s
1
0.985 m s
The speed of the wave is
Page 13.18
Chapter 13
x
t
v
425 cm
10.0 s
42 .5 cm s
and the frequency is
40.0 vib
30.0 s
f
1.33 Hz
Thus,
v
f
13.46
42.5 cm s
1.33 Hz
From v = f, the wavelength (and size of smallest detectable insect) is
v
f
13.47
31.9 cm
340 m s
60.0 103 Hz
5.67
10-3 m
5.67 mm
The frequency of the wave (that is, the number of crests passing the cork each second) is f
wavelength (distance between successive crests) is
v
f
8.50 cm 2.00 s-1
2.00 s-1 and the
8.50 cm . Thus, the wave speed is
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
x
v
t
13.48
(a)
10.0 m
0.170 m s
58.8 s
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
f
(b)
v
4.0 m s
20 m
4.0 m s . The frequency detected in this case is
0.20 Hz
Taking eastward as positive, vwave,boat
vwave, water
vboat , water (see the discussion of relative velocity in
Chapter 3 of the textbook) gives
vwave, boat
4.0 m s
1.0 m s
5.0 m s
Thus,
Page 13.19
and
vboat, wave
vwave, boat
5.0 m s
Chapter 13
vboat , wave
f
13.49
5.0 m s
20 m
0.25 Hz
The down and back distance is 4.00 m + 4.00 m = 8.00 m.
The speed is then
dtotal
t
v
4 8.00 m
40.0 m s
0.800 s
F
Now,
m
L
0.200 kg
4.00 m
5.00
10
2
kg m
so
F
13.50
v2
5.00
10
2
kg m 40.0 m s
80.0 N
The speed of the wave is
x
t
v
20.0 m
0.800 s
25.0 m s
and the mass per unit length of the rope is
F
13.51
2
(a)
v2
25.0 m s
2
m/L
0.350 kg m
The speed of transverse waves in the cord is v
0.350 kg m . Thus, from v
F
, we obtain
219 N
F
, where
m L is the mass per unit length. With
the tension in the cord being F = 12.0 N, the wave speed is
F
v
(b)
(a)
FL
m
12.0 N 6.30 m
0.150 kg
22.4 m s
The time to travel the length of the cord is
t
13.52
F
m L
L
v
6.30 m
22.4 m s
0.281 s
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
x
t
v
(b)
(a)
v2
From v
m 2
v
L
0.375 kg
12.0 m
F
v
0.0600 kg
5.00 m
F
v2
F
13.54
48.6 m s
2.96 s
, so the tension in the line is
F
48.6 m s
2
73.8 N
The mass per unit length is
m
L
(b)
12 12.0 m
The speed of transverse waves in the line is v
F
13.53
12 L
t
0.0120 kg m
, the required tension in the string is
2
50.0 m s
0.0120 kg m
8.00 N
0.0120 kg m
30.0 N ,
25.8 m s
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
v2
44.3 m s
2
2 .50
10-3 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
13.55
F
m
4.91 N
3.00 kg
The period of the pendulum is T
1.64 m s2
2
L / g , so the length of the string is
Page 13.21
Chapter 13
L
9.80 m s2
gT 2
4 2
4
2
2.00 s
0.993 m
2
Then mass per unit length of the string is then
m
L
0.060 0 kg
0.993 m
0.060 4
kg
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
13.56
If
1
F
49.0 N
0.060 4 kg m
m1 L is the mass per unit length for the first string, then
second string. Thus, with F2
F
v2
13.57
(a)
28.5 m s
2
m2 L
m1 2L
1
2 is that of the
F , the speed of waves in the second string is
F1
2F
2
F
2
1
2 v1
1
The tension in the string is F
2 5.00 m s
3.0 kg 9.80 m s2
mg
7.07 m s
29 N . Then, from v
F
, the mass per
unit length is
F
v2
(b)
29 N
24 m s
2
0.051 kg m
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
13.58
F
20 N
0.051 kg m
20 m s
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
F
v
But, A L
A Stress
m/L
Stress
m /( A L )
density . Thus, v
V =volume , so m A L
Stress
.
When the stress is at its maximum, the speed of waves in the wire is
13.59
(a)
2.70 10 9 Pa
7.86 103 kg/m 3
(Stress)max
vmax
The speed of transverse waves in the line is v
586 m s
F
, with
m L being the mass per unit length.
Therefore,
F
v
(b)
F
m L
FL
m
12.5 N 38.0 m
2.65 kg
13.4 m s
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
v
(b)
From v
F
13.61
(a)
x
t
2 nL
t
F
v2
as the speed of transverse waves in the line, the tension is
M
L
2nL
t
2
M
L
4n2 L2
t2
4n2 ML2
t2
Constructive interference produces the maximum amplitude
Amax
(b)
2nL . The wave speed is then
A1
A2
0.50 m
Destructive interference produces the minimum amplitude
Page 13.23
Chapter 13
Amin
13.62
A1
A2
We are given that x
0.10 m
(0.25 m) cos(0.4 t ) .
A cos( t )
(a)
By inspection, the amplitude is seen to be A
(b)
The angular frequency is
k
(c)
2
m
rad s)2
k m , so the spring constant is
0.47 N m
Note: Your calculator must be in radians mode for part (c).
At t = 0.30 s, x
(d)
rad s . But
0.4
(0.30 kg)(0.4
0.25 m
0.25 m cos
0.4
rad s 0.30 s
0.23 m
From conservation of mechanical energy, the speed at displacement x is given by v
A2
x2 . Thus, at t
= 0.30 s, when x = 0.23 m, the speed is
v
13.63
(a)
0.4
2m
4
(a)
13.65
(a)
0.12 m s
2
2.00 kg
0.600 s
2
2
m k , so the spring constant is
219 N m
2
If T = 1.05 s for mass m2, this mass is
kT 2
4 2
(219 N/m)(1.05 s)2
4 2
6.12 kg
The period is the reciprocal of the frequency, or
T
(b)
4
T2
m2
13.64
(0.23 m)2
The period of a vibrating object-spring system is T
k
(b)
(0.25 m)2
rad s
1
f
vsound
f
1
196 s
1
343 m s
196 s 1
5.10
10
3
s
5.10 ms
1.75 m
The period of a simple pendulum is T
2
g , so the period of the first system is
Page 13.24
Chapter 13
T1
(b)
0.700 m
9.80 m s2
2
g
1.68 s
The period of mass-spring system is T
2
mk
k
13.66
2
m k , so if the period of the second system is T2 = T1, then
2
l g and the spring constant is
2
1.20 kg 9.80 m s2
mg
16.8 N m
0.700 m
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
PEg
PEs
KE
f
PEg
PEs
i
, or 0
0
1 2
kxmax
2
0
0
mghi
Thus,
13.67
2 0.500 kg 9.80 m s2
2mghi
k
xmax
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
1
mv2
2
f
mg
KE
x
PEg
1 2
kx
2
PEs , or
i
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
1 2
kxmax
2
(b)
mgxmax
or
k
2 mg
xmax
2 3.00 kg 9.80 m s2
When x = 5.00 cm = 0.050 0 m, the energy equation gives
v
2 gx
kx 2
,
m
or
Page 13.25
0.100 m
588 N m
Chapter 13
v
13.68
(a)
2 9.80 m
s2
588 N m 0.050 0 m
0.050 0 m
2
0.700 m s
3.00 kg
We apply conservation of mechanical energy from just after the collision until the block comes to rest.
KE
PEs
KE
f
PEs
1 2
kxf
2
gives 0
i
1
MV 2
2
0
or the speed of the block just after the collision is
V
k x 2f
900 N m 0.050 0 m
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
0
i
m vbullet
MV
f
or
vbullet
vbullet
f
M
V
m
i
1.00 kg
5.00 10-3 kg
400 m s
(b)
1.5 m s
100 m s
The mechanical energy converted into internal energy during the collision is
E
KEi
E
1
5.00
2
E
374 J
1
m vbullet
2
KE f
1
m vbullet
2
2
i
1
MV 2
2
2
f
or
13.69
10
3
kg
400 m s
2
100 m s
2
1
1.00 kg 1.50 m s
2
2
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
f
KE
PEg
PEs
i
, or
Page 13.26
Chapter 13
1
m1
2
1
2
m2 v2f
25
m1g x sin 40
30 kg v 2f
13.70
(a)
0
25 kg 9.80 m s 2
30 kg 9.80 m s2
yielding v f
m2 g x
0
0
0.200 m sin 40
1
200 N m 0.200 m
2
0.200 m
2
1.1 m s
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
v
(b)
1
k x2
2
xi
From
y
k
m
0.200 m
v0 y t
1
2
9.80 N m
1.00 10 3 kg
2
y
ay
v0 x t
1
2
kxi2 , or
ay t 2 , the time required for the pellet to drop 1.00 m to the floor, starting with v0 y
2
1.00 m
9.80 m s2
0.452 s
The range (horizontal distance traveled during the flight) is then
x
mv2
19.8 m s
is
t
1
2
19.8 m s 0.452 s
8.94 m
Page 13.27
0,
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
Fnet
Fx
B sin
For small angles, sin
Also, mg
HeV
mg sin
B
mg sin
s/ L
g
and the buoyant force is B
airV
g . Thus, the net restoring force
acting on the balloon is
air
Fnet
He
Vg
s
L
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
f
2
2
m
k
HeV
2
air
He
Vg L
He
2
air
He
L
g
This yields
T
13.72
(a)
0.180
1.29 0.180
2
3.00 m
1.40 s
9.80 m s2
When the block is given some small upward displacement, the net restoring force exerted on it by the rubber
bands is
Fnet
Fy
2 F sin , where tan
For small displacements, the angle
y
L
will be very small. Then sin
is
Page 13.28
tan
y
, and the net restoring force
L
Chapter 13
Fnet
(b)
y
L
2F
2F
y
L
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
13.73
2F
mL
Newton’s law of gravitation is
GMm
, where M
r2
F
4 3
r
3
Thus,
4
3
F=
Gm r
which is of Hooke’s law form, F
k=
13.74
4
3
k r , with
Gm
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
k
m
0.20 cm
4.7 10 4 N m
0.30 10 3 kg
0.25 cm s
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
vouter
router
vinner
rinner
The maximum speed of the outer tip is then
Page 13.29
Chapter 13
router
rinner
vouter , max
13.75
k
m
(a)
(b)
500 N m
2.00 kg
With Fs
Fs
mg
1.3 cm s
15.8 rad s
may
kx and ay
g 3 , this gives kx
4 2.00 kg 9.80 m s2
4 mg
3k
x
(a)
0.25 cm s
Apply Newton’s second law to the block while the elevator is accelerating:
Fy
13.76
15.0 mm
3.00 mm
vinner , max
5.23
3 500 N m
g 3 , or
m g
10
2
m
5.23 cm
Note that as the spring passes through the vertical position, the object is moving in a circular arc of radius
L
yf. 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
L yf
Fy
(b)
mg
Conservation of energy requires that E
E
(c)
ky f
0
mgL
1
mv2
2
0
KEi
PEg,i
(L
part (b) gives 2mg(L
mg)
becomes (2k ) y2f
a
2k
b
3mg
(3mg
(L
kL)y f
2 1 250 N m
kL
KE f
1 2
ky f , reducing to 2mg L
2
mgy f
From the result of part (a), observe that mv2
yf )
PEs,i
y f )(ky f
( 3mgL)
2.50
y f )(ky f
PEg, f
yf
PEs, f , or
mv2
ky2f
mg) . Substituting this into the result from
ky2f . After expanding and regrouping terms, this
0 , which is a quadratic equation ay2f
by f
103 N m
3 5.00 kg 9.80 m s2
1 250 N m 1.50 m
and
Page 13.30
1.73
103 N
c
0 , with
Chapter 13
c
3 5.00 kg 9.80 m s2 1.50 m
3mgL
221 N m
Applying the quadratic formula, keeping only the negative solution [see the discussion in part (a)] gives
yf
or y f
(d)
b2
2a
b
1.73
4ac
103
1.73
103
2 2.50
2
4 2.50
221
103
0.110 m
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
103
The maximum acceleration of the oscillating system is
2
amax
A
2
f
2
A
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,
fs
fs
max
sn
s mg
amax
2 f
2
mamax and we must have
A
s
g
Thus,
A
0.600 9.80 m s2
sg
2
f
2
2
1.50 Hz
2
6.62
Page 13.31
10
2
m
6.62 cm