Sound General Problems Solutions
67.
The unusual decrease of water corresponds to a trough in Figure 11-24. The crest or peak of the wave
is then one-half wavelength distant. The peak is 125 km away, traveling at 750 km/hr.
Dx = vt
68.
t=
®
Dx
125 km æ 60 min ö
=
ç
÷ = 10 min
v
750 km hr è 1 hr ø
Apply the conservation of mechanical energy to the car, calling condition # 1 to be before the
collision and condition # 2 to be after the collision. Assume that all of the kinetic energy of the car is
converted to potential energy stored in the bumper. We know that x1 = 0 and v2 = 0.
E1 = E2
x2 =
69.
®
1
2
m
v1 =
k
mv12 + 12 kx12 = 12 mv22 + 12 kx22
®
1
2
mv12 = 12 kx22
®
1500 kg
( 2.2 m s) = 0.11 m
550 ´ 10 3 N m
Consider the conservation of energy for the person. Call the unstretched position of the fire net the
zero location for both elastic potential energy and gravitational potential energy. The amount of
stretch of the fire net is given by x, measured positively in the downward direction. The vertical
displacement for gravitational potential energy is given by the variable y, measured positively for the
upward direction. Calculate the spring constant by conserving energy between the window height and
the lowest location of the person. The person has no kinetic energy at either location.
Etop = Ebottom
k = 2mg
(a)
(y
2
® mgytop = mgybottom + 12 kxbottom
top
- ybottom
x
2
bottom
) = 2 ( 65 kg ) 9.8 m s éë18 m - ( -1.1 m )ùû = 2.011 ´ 10
(
) (1.1 m )
2
2
4
Nm
If the person were to lie on the fire net, they would stretch the net an amount such that the
upward force of the net would be equal to their weight.
F = kx = mg
(b)
®
x=
(
)
2
mg ( 65 kg ) 9.8 m s
=
= 3.2 ´ 10 -2 m
k
2.011 ´ 10 4 N m
To find the amount of stretch given a starting height of 35 m, again use conservation of energy.
Note that ybottom = -x, and there is no kinetic energy at the top or bottom positions.
Etop = Ebottom
x2 - 2
mgytop = mgybottom + 12 kx 2
®
( 65 kg ) ( 9.8
m s2
2.011 ´ 10 4 N m
x2 - 2
®
mg
mg
x-2
ytop = 0
k
k
) x - 2 ( 65 kg ) ( 9.8 m s ) ( 35 m ) = 0
2
2.011 ´ 10 4 N m
x 2 - 0.06335 x - 2.2173 = 0
®
® x = 1.5211 m, - 1.458 m
This is a quadratic equation. The solution is the positive root, since the net must be below the
unstretched position. The result is 1.5 m .
70.
Consider energy conservation for the mass over the range of motion
from “letting go” (the highest point) to the lowest point. The mass
falls the same distance that the spring is stretched, and has no KE at
either endpoint. Call the lowest point the zero of gravitational
potential energy. The variable “x” represents the amount that the
spring is stretched from the equilibrium position.
Etop = Ebottom
1
2
2
2
2
2
mvtop
+ mgytop + 12 kxtop
= 12 mvbottom
+ mgybottom + 12 kxbottom
0 + mgH + 0 = 0 + 0 + 12 kH 2
(
®
k 2g
=
= w2
m H
®
w=
)
2g
H
2
w
1 2g
1 2 9.8 m s
f =
=
=
= 1.2 Hz
2p 2p H 2p
0.33 m
71.
(a)
From conservation of energy, the initial kinetic energy of the car will all be changed into elastic
potential energy by compressing the spring.
E1 = E2
®
1
2
mv12 + 12 kx12 = 12 mv22 + 12 kx22
®
1
2
mv12 = 12 kx22
®
( 22 m s) = 1.8392 ´ 10 4 N m » 1.8 ´ 10 4 N m
v12
= ( 950 kg )
x22
( 5.0 m )2
2
k=m
(b)
The car will be in contact with the spring for half a period, as it moves from the equilibrium
location to maximum displacement and back to equilibrium.
1
2
72.
T = 12 2p
m
( 950 kg )
=p
= 0.71 s
k
1.8392 ´ 10 4 N m
The frequency at which the water is being shaken is about 1 Hz. The sloshing coffee is in a standing
wave mode, with anti-nodes at each edge of the cup. The cup diameter is thus a half-wavelength, or
l = 16 cm. The wave speed can be calculated from the frequency and the wavelength.
v = l f = (16 cm ) (1 Hz ) = 16 cm s
73.
Relative to the fixed needle position, the ripples are moving with a linear velocity given by
æ rev ö æ 1 min ö æ 2p ( 0.108 m ) ö
v = ç 33
÷ø = 0.373 m s
è min ÷ø çè 60 s ÷ø çè
1 rev
This speed is the speed of the ripple waves moving past the needle. The frequency of the waves is
f=
74.
v
l
=
0.373 m s
= 220 Hz
1.70 ´ 10 -3 m
The equation of motion is x = 0.650 cos 7.40t = A cos wt.
(a)
The amplitude is A = 0.650 m
(b)
The frequency is given by w = 2p f = 7.40 rad s
(c)
The total energy is given by
Etotal = 12 kA2 = 12 mw 2 A2 =
(d)
1
2
( 2.00 kg ) ( 7.40
® f=
7.40 rad s
= 1.177 Hz » 1.18 Hz
2p rad
rad s) ( 0.650 m ) = 23.136 J » 23.1 J .
2
2
The potential energy is found by
PE = 12 kx 2 = 12 mw 2 x 2 =
1
2
( 2.00 kg ) ( 7.40
rad s) ( 0.260 m ) = 3.702 J » 3.70 J .
2
2
The kinetic energy is found by
KE = Etotal - PE = 23.136 J - 3.702 J = 19.4 J .
75.
The frequency of a simple pendulum is given by f = 1
2p
g The pendulum is accelerating vertically
.
L
which is equivalent to increasing (or decreasing) the acceleration due to gravity by the acceleration of
the pendulum.
76.
(a)
fnew =
1
2p
g+a
1 1.50 g
1
=
= 1.50
L
2p
L
2p
(b)
fnew =
1
2p
g+a
1
=
L
2p
0.5g
1
= 0.5
L
2p
g
= 1.50 f = 1.22 f
L
g
= 0.5 f = 0.71 f
L
The force of the man’s weight causes the raft to sink, and that causes the water to put a larger upward
force on the raft. This extra buoyant force is a restoring force, because it is in the opposite direction
of the force put on the raft by the man. This is analogous to pulling down on a mass-spring system
that is in equilibrium, by applying an extra force. Then when the man steps off, the restoring force
pushes upward on the raft, and thus the raft – water system acts like a spring, with a spring constant
found as follows.
k=
(a)
(
The frequency of vibration is determined by the spring constant and the mass of the raft.
fn =
(b)
)
2
F ( 75 kg ) 9.8 m s
=
= 1.8375 ´ 10 4 N m
x
4.0 ´ 10 -2 m
1
2p
k
1 1.8375 ´ 10 4 N m
=
= 1.455 Hz » 1.5 Hz
m 2p
220 kg
As explained in the text, for a vertical spring the gravitational potential energy can be ignored if
the displacement is measured from the oscillator’s equilibrium position. The total energy is
thus
Etotal = 12 kA2 =
77.
(a)
(b)
1
2
(1.8375 ´ 10
4
)(
)
2
N m 4.0 ´ 10 -2 m = 14.7J » 15 J .
The overtones are given by fn = nf1, n = 2, 3, 4 …
G:
f2 = 2 ( 392 Hz ) = 784 Hz
f3 = 3( 392 Hz ) = 1180 Hz
A:
f2 = 2 ( 440 Hz ) = 880 Hz
f3 = 3( 440 Hz ) = 1320 Hz
If the two strings have the same length, they have the same wavelength. The frequency
difference is then due to a difference in wave speed caused by different masses for the strings.
fG vG l vG
=
=
=
f A vA l vA
(c)
FT
mG L
FT
mA L
=
mA
mG
2
®
mG æ f A ö
æ 440 ö
=ç ÷ =ç
= 1.26
è 392 ÷ø
mA è fG ø
2
If the two strings have the same mass per unit length and the same tension, then the wave speed
on both strings is the same. The frequency difference is then due to a difference in wavelength.
For the fundamental, the wavelength is twice the length of the string.
fG v lG lA 2LA
=
=
=
fA v lA lG 2LG
(d)
®
LG fA 440
=
=
= 1.12
LA fG 392
If the two strings have the same length, they have the same wavelength. The frequency
difference is then due to a difference in wave speed caused by different tensions for the strings.
fG vG l vG
=
=
=
f A vA l vA
78.
(a)
FTG
m L
FTA
m L
=
FTG
FTA
2
®
FTG æ fG ö
æ 392 ö
=
=ç
= 0.794
è 440 ÷ø
FTAA çè f A ÷ø
2
Since the cord is not accelerating to the left or right, the tension in the cord must be the same
everywhere. Thus the tension is the same in the two parts of the cord. The speed difference will
then be due to the different mass densities of the two parts of the cord. Let the symbol m
represent the mass per unit length of each part of the cord.
vH
=
vL
(
(
)
m)
FT m
FT
H
L
=
mL
mH
(b)
The wavelength ratio is found as follows.
lH ( v f )H vH
=
=
=
lL ( v f )L vL
mL
mH
The two frequencies must be the same for the cord to remain continuous at the boundary. If the
two parts of the cord oscillate at different frequencies, the cord cannot stay in one piece,
because the two parts would be out of phase with each other at various times.
79.
mH > m L ,
we see that
lH < lL , and so the wavelength is greater in the lighter cord .
(c)
Since
(a)
The maximum speed is given by
(
)
vmax = 2p f A = 2p ( 264 Hz ) 1.8 ´ 10 -3 m = 3.0 m s .
(b)
The maximum acceleration is given by
(
)
amax = 4p 2 f 2 A = 4p 2 ( 264 Hz ) 1.8 ´ 10 -3 m = 5.0 ´ 10 3 m s2 .
2
80.
For the pebble to lose contact with the board means that there is no normal force of the board on the
pebble. If there is no normal force on the pebble, then the only force on the pebble is the force of
gravity, and the acceleration of the pebble will be g downward, the acceleration due to gravity. This
is the maximum downward acceleration that the pebble can have. Thus if the board’s downward
acceleration exceeds g, then the pebble will lose contact. The maximum acceleration and the
amplitude are related by amax = 4p 2 f 2 A.
amax = 4p 2 f 2 A £ g
81.
®
A£
g
4p 2 f 2
£
9.8 m s2
4p 2 (1.5 Hz )
2
£ 1.1 ´ 10 -1 m
For a resonant condition, the free end of the string will be an antinode, and the fixed end of the string
will be a node. The minimum distance from a node to an antinode is l 4. Other wave patterns that fit
the boundary conditions of a node at one end and an antinode at the other end include
3l 4, 5l 4, … . See the diagrams. The general relationship is L = ( 2n - 1) l 4, n = 1, 2, 3,
Solving for the wavelength gives l = 4L , n = 1, 2, 3,
2n - 1
.
82.
2
The period of a pendulum is given by T = 2p L g , and so the length is L = T g .
4p 2
LAustin
(
)
2
T 2 gAustin ( 2.000 s ) 9.793 m s
=
=
= 0.9922 m
4p 2
4p 2
2
(a)
LParis
(
)
2
T 2 gParis ( 2.000 s ) 9.809 m s
=
=
= 0.9939 m
4p 2
4p 2
2
(b)
LParis - LAustin = 0.9939 m - 0.9922 m = 0.0016 m = 1.6 mm
83.
LMoon =
(
)
2
T 2 gMoon ( 2.00 s ) 1.62 m s
=
= 0.164 m
4p 2
4p 2
2
(c)
The spring, originally of length l0 , will be stretched downward to a new equilibrium length L when
the mass is hung on it. The amount of downward stretch L - l0 is found from setting the spring force
upward on the mass equal to the weight of the mass: k ( L - l0 ) = mg ® L = l0 + mg k. The length
of the pendulum is then L = l0 + mg k. The period of the vertical oscillations is given by
Tver = 2p m k, while the period of the pendulum oscillations is given by Tpen = 2p L g. Now
compare the periods of the two motions.
Tpen
Tver
=
2p
( l0 + mg k )
2p m k
g
=
l0 + mg k
lk
= 1+ 0 >1 ®
mg k
mg
Tpen > Tver , by a factor of 1 +
84.
l0 k
mg
Block m stays on top of block M (executing SHM relative to the ground) without slipping due to
static friction. The maximum static frictional force on m is F = m mg. This frictional force causes
fr
s
max
block m to accelerate, so mamax = ms mg ®
amax = ms g. Thus for the blocks to stay in contact
without slipping, the maximum acceleration of block M is also amax = ms g. But an object in SHM has
a maximum acceleration given by a = w 2 A =
max
k
A. Equate these two expressions for the
M total
maximum acceleration.
amax =
85.
k
A = ms g
M total
®
A=
ms g
k
( M + m) =
( 0.30 ) ( 9.8
m s2
130 N m
) ( 6.25 kg ) =
0.14 m
The speed of the pulses is found from the tension and mass per unit length of the wire.
v=
FT
=
m L
255 N
= 143.985 m s
( 0.123 kg ) (10.0 m )
The total distance traveled by the two pulses will be the length of the wire. The second pulse has a
shorter time of travel than the first pulse, by 20.0 ms.
(
L = d1 + d2 = vt1 + vt 2 = vt1 + v t1 - 2.00 ´ 10 -2
t1 =
)
L + 2.00 ´ 10 v (10.0 m ) + 2.00 ´ 10 (143.985 m s )
=
= 4.4726 ´ 10 -2 s
2v
2 (143.985 m s )
-2
-2
(
)
d1 = vt1 = (143.985 m s ) 4.4726 ´ 10 -2 s = 6.44 m
The two pulses meet 6.44 m from the end where the first pulse originated.