Superposition and
Standing Waves
CHAPTER OUTLINE
14.1
14.2
14.3
14.4
14.5
14.6
14.7
14.8
The Principle of
Superposition
Interference of Waves
Standing Waves
Standing Waves in Strings
Standing Waves in Air
Columns
Beats: Interference in
Time
Nonsinusoidal Wave
Patterns
Context
ConnectionBuilding on
Antinodes
ANSWERS TO QUESTIONS
Q14.1
No. Waves with other waveforms are also trains of disturbance that
add together when waves from different sources move through the
same medium at the same time.
Q14.2
No. A wave is not a solid object, but a chain of disturbance. As
described by the principle of superposition, the waves move through
each other.
Q14.3
They can, wherever the two waves are nearly enough in phase that their displacements will add to
create a total displacement greater than the amplitude of either of the two original waves.
When two one-dimensional sinusoidal waves of the same amplitude interfere, this
condition is satisfied whenever the absolute value of the phase difference between the two waves is
less than 120°.
Q14.4
When the two tubes together are not an efficient transmitter of sound from source to receiver, they
are an efficient reflector. The incoming sound is reflected back to the source. The waves reflected by
the two tubes separately at the junction interfere constructively.
Q14.5
No. The total energy of the pair of waves remains the same. Energy missing from zones of
destructive interference appears in zones of constructive interference.
Q14.6
Damping, and non–linear effects in the vibration turn the energy of vibration into internal energy.
Q14.7
The air in the shower stall can vibrate in standing wave patterns to intensify those frequencies in
your voice which correspond to its free vibrations. The hard walls of the bathroom reflect sound
very well to make your voice more intense at all frequencies, and giving the room a longer
reverberation time. The reverberant sound may help you to stay on key.
Q14.8
The trombone slide and trumpet valves change the length of the air column inside the instrument,
to change its resonant frequencies.
Q14.9
In a classical guitar, vibrations of the strings are transferred to the wooden body through the bridge.
Because of its large area, the guitar body is a much more efficient radiator of sound than an
individual guitar string. Thus, energy associated with the vibration is transferred to the air relatively
rapidly by the guitar body, resulting in a more intense sound.
381
382
Q14.10
Superposition and Standing Waves
The vibration of the air must have zero amplitude at the closed end. For air in a pipe closed at one
end, the diagrams show how resonance vibrations have NA distances that are odd integer
submultiples of the NA distance in the fundamental vibration. If the pipe is open, resonance
vibrations have NA distances that are all integer submultiples of the NA distance in the
fundamental.
FIG. Q14.10
Q14.11
The bow string is pulled away from equilibrium and released, similar to the way that a guitar string
is pulled and released when it is plucked. Thus, standing waves will be excited in the bow string. If
the arrow leaves from the exact center of the string, then a series of odd harmonics will be excited.
Even harmonies will not be excited because they have a node at the point where the string exhibits
its maximum displacement.
Q14.12
What is needed is a tuning fork—or other pure-tone generator—of the desired frequency. Strike the
tuning fork and pluck the corresponding string on the piano at the same time. If they are precisely
in tune, you will hear a single pitch with no amplitude modulation. If the two pitches are a bit off,
you will hear beats. As they vibrate, retune the piano string until the beat frequency goes to zero.
Q14.13
Beats. The propellers are rotating at slightly different frequencies.
Q14.14
Walking makes the person’s hand vibrate a little. If the frequency of this motion is equal to the
natural frequency of coffee sloshing from side to side in the cup, then a large–amplitude vibration of
the coffee will build up in resonance. To get off resonance and back to the normal case of a smallamplitude disturbance producing a small–amplitude result, the person can walk faster, walk slower,
or get a larger or smaller cup. Alternatively, even at resonance he can reduce the amplitude by
adding damping, as by stirring high–fiber quick–cooking oatmeal into the hot coffee.
Q14.15
Stick a bit of chewing gum to one tine of the second fork. If the beat frequency is then faster than
4 beats per second, the second has a lower frequency than the standard fork. If the beats have
slowed down, the second fork has a higher frequency than the standard. Remove the gum, clean the
fork, add or subtract 4 Hz according to what you found, and your answer will be the frequency of
the second fork.
Q14.16
Instead of just radiating sound very softly into the surrounding air, the tuning fork makes the
chalkboard vibrate. With its large area this stiff sounding board radiates sound into the air with
higher power. So it drains away the fork’s energy of vibration faster and the fork stops vibrating
sooner. This process exemplifies conservation of energy, as the energy of vibration of the fork is
transferred through the blackboard into energy of vibration of the air.
Chapter 14
SOLUTIONS TO PROBLEMS
Section 14.1 The Principle of Superposition
P14.1
a
f
a
f
y = y1 + y 2 = 3.00 cos 4.00 x − 1.60t + 4.00 sin 5.0 x − 2.00t evaluated at the given x values.
a
a
f
f
(a)
x = 1.00 , t = 1.00
y = 3.00 cos 2.40 rad + 4.00 sin +3.00 rad = −1.65 cm
(b)
x = 1.00 , t = 0.500
y = 3.00 cos +3.20 rad + 4.00 sin +4.00 rad = −6.02 cm
(c)
x = 0.500 , t = 0
y = 3.00 cos +2.00 rad + 4.00 sin +2.50 rad = 1.15 cm
a
f
a
f
a
f
a
f
P14.2
FIG. P14.2
P14.3
(a)
a
f
a
f
y1 = f x − vt , so wave 1 travels in the +x direction
y 2 = f x + vt , so wave 2 travels in the −x direction
(b)
To cancel, y1 + y 2 = 0 :
5
=
2
for the positive root,
+5
a 3 x − 4t f + 2 a 3 x + 4t − 6 f
a3 x − 4tf = a3x + 4t − 6f
3 x − 4t = ±a3 x + 4t − 6f
2
8t = 6
2
+2
2
t = 0.750 s
(at t = 0.750 s , the waves cancel everywhere)
(c)
for the negative root,
6x = 6
(at x = 1.00 m, the waves cancel always)
x = 1.00 m
383
384
Superposition and Standing Waves
Section 14.2 Interference of Waves
P14.4
Suppose the waves are sinusoidal.
a4.00 cmf sinbkx − ω tg + a4.00 cmf sinbkx − ω t + 90.0°g
2a 4.00 cmf sinb kx − ω t + 45.0°g cos 45.0°
So the amplitude is a8.00 cmf cos 45.0° = 5.66 cm .
The sum is
P14.5
The resultant wave function has the form
FG φ IJ sinFG kx − ω t + φ IJ
H 2K H
2K
L − π 4 OP =
FφI
A = 2 A cosG J = 2a5.00f cos M
H 2K
N 2 Q
y = 2 A0 cos
(a)
(b)
P14.6
2 A0 cos
9.24 m
0
f=
ω 1 200π
=
= 600 Hz
2π
2π
FG φ IJ = A
H 2K
0
φ
so
2
= cos −1
FG 1 IJ = 60.0° = π
H 2K
3
2π
3
Thus, the phase difference is
φ = 120° =
This phase difference results if the time delay is
T
λ
1
=
=
3 3 f 3v
Time delay =
P14.7
3.00 m
= 0.500 s
3 2.00 m s
b
g
Waves reflecting from the near end travel 28.0 m (14.0 m down and 14.0 m back), while waves
reflecting from the far end travel 66.0 m. The path difference for the two waves is:
∆r = 66.0 m − 28.0 m = 38.0 m
Since
Then
or,
λ=
∆r
λ
v
f
=
a∆r f f = a38.0 mfa246 Hzf = 27.254
v
343 m s
∆r = 27.254λ
The phase difference between the two reflected waves is then
b
g
a
f
φ = 0.254 1 cycle = 0.254 2π rad = 91.3°
Chapter 14
P14.8
(a)
∆x = 9.00 + 4.00 − 3.00 = 13 − 3.00 = 0.606 m
λ=
The wavelength is
Thus,
∆x
λ
=
v 343 m s
=
= 1.14 m
f
300 Hz
0.606
= 0.530 of a wave ,
1.14
a
f
∆φ = 2π 0.530 = 3.33 rad
or
(b)
For destructive interference, we want
where ∆x is a constant in this set up.
*P14.9
(a)
385
∆x
λ
f=
= 0.500 = f
∆x
v
v
343
=
= 283 Hz
2 ∆x 2 0.606
a
f
f b
b
ga
ga f
= b 25.0 rad cmga5.00 cmf − b 40.0 rad sga 2.00 sf = 45.0 rad
φ 1 = 20.0 rad cm 5.00 cm − 32.0 rad s 2.00 s = 36.0 rad
φ1
∆φ = 9.00 radians = 516° = 156°
(b)
∆φ = 20.0 x − 32.0t − 25.0 x − 40.0t = −5.00 x + 8.00 t
At t = 2.00 s , the requirement is
a f a
f
∆φ = −5.00 x + 8.00 2.00 = 2n + 1 π for any integer n.
For x < 3.20 , −5.00 x + 16.0 is positive, so we have
a
f
−5.00 x + 16.0 = 2n + 1 π , or
x = 3.20 −
a2n + 1fπ
5.00
The smallest positive value of x occurs for n = 2 and is
x = 3.20 −
P14.10
a4 + 1fπ = 3.20 − π =
5.00
0.058 4 cm .
Suppose the man’s ears are at the same level as the lower speaker. Sound from the upper speaker is
delayed by traveling the extra distance ∆r = L2 + d 2 − L .
a
He hears a minimum when ∆r = 2n − 1
Then,
fFGH λ2 IJK with n = 1, 2, 3, K
IJ FG v IJ
KH f K
F 1IF vI
L + d = Gn − J G J + L
H 2KH f K
F 1I F vI F 1IF vI
L + d = G n − J G J + 2G n − J G J L + L
H 2K H f K H 2KH f K
FG
H
L2 + d 2 − L = n −
2
2
continued on next page
1
2
2
2
2
2
2
386
Superposition and Standing Waves
d
2
F 1I F vI
− Gn − J G J
H 2K H f K
2
2
FG
H
=2 n−
1
2
IJ FG v IJ L
KH f K
(1)
Equation 1 gives the distances from the lower speaker at which the man will hear a minimum. The
path difference ∆r starts from nearly zero when the man is very far away and increases to d when
L = 0.
(a)
The number of minima he hears is the greatest integer value for which L ≥ 0 . This is the
1 v
, or
same as the greatest integer solution to d ≥ n −
2 f
FG
H
IJ FG IJ
KH K
number of minima heard = n max = greatest integer ≤ d
(b)
(a)
.
From equation 1, the distances at which minima occur are given by
Ln =
P14.11
FG f IJ + 1
H vK 2
c
2cn −
hej
he j
d 2 − n − 12
1
2
2 v 2
f
v
f
where n = 1, 2 , K , n max .
v 344 m s
=
= 16.0 m
f 21.5 Hz
First we calculate the wavelength:
λ=
Then we note that the path difference equals
9.00 m − 1.00 m =
1
λ
2
Therefore, the receiver will record a minimum in sound intensity.
(b)
We choose the origin at the midpoint between the speakers. If the receiver is located at point
(x, y), then we must solve:
Then,
ax + 5.00f
2
+ y2 −
ax − 5.00f
ax + 5.00f
2
+ y2 =
ax − 5.00f
λ2
=λ
ax − 5.00f
Square both sides and simplify to get:
20.0 x −
Upon squaring again, this reduces to:
400 x 2 − 10.0 λ2 x +
Substituting λ = 16.0 m, and reducing,
or
4
λ4
16.0
9.00 x 2 − 16.0 y 2 = 144
y2
x2
−
=1
16.0 9.00
(When plotted this yields a curve called a hyperbola.)
2
2
2
+ y2 =
1
λ
2
+ y2 +
1
λ
2
+ y2
a
= λ2 x − 5.00
f
2
+ λ2 y 2
387
Chapter 14
Section 14.3 Standing Waves
P14.12
a
f a
f a f
y = 1.50 m sin 0.400 x cos 200 t = 2 A 0 sin kx cos ω t
Therefore, k =
2π
λ
= 0.400 rad m
and ω = 2π f so
λ=
2π
= 15.7 m
0.400 rad m
f=
ω 200 rad s
=
= 31.8 Hz
2π
2π rad
The speed of waves in the medium is v = λf =
P14.13
200 rad s
λ
ω
= 500 m s
2π f = =
k 0.400 rad m
2π
The facing speakers produce a standing wave in the space between them, with the spacing between
nodes being
d NN =
λ
2
=
343 m s
v
=
= 0.214 m
2 f 2 800 s −1
e
j
If the speakers vibrate in phase, the point halfway between them is an antinode of pressure at a
distance from either speaker of
1.25 m
= 0.625 .
2
Then there is a node at
P14.14
0.625 −
0.214
= 0.518 m
2
a node at
0.518 m − 0.214 m = 0.303 m
a node at
0.303 m − 0.214 m = 0.089 1 m
a node at
0.518 m + 0.214 m = 0.732 m
a node at
0.732 m + 0.214 m = 0.947 m
and a node at
0.947 m + 0.214 m = 1.16 m from either speaker.
y = 2 A0 sin kx cos ωt
∂2y
∂x 2
= −2 A0 k 2 sin kx cos ω t
∂2 y
∂t 2
= −2 A0ω 2 sin kx cos ωt
Substitution into the wave equation gives
−2 A0 k 2 sin kx cos ω t =
This is satisfied, provided that
v=
ω
k
FG 1 IJ e−2 A ω
Hv K
2
0
2
sin kx cos ω t
j
388
P14.15
Superposition and Standing Waves
a
f
a
f
y = y + y = 3.00 sinbπ x g cosb0.600π t g + 3.00 sinbπ xg cosb0.600π t g cm
y = a6.00 cmf sinbπ x g cosb0.600π t g
(a)
We can take cosb0.600π t g = 1 to get the maximum y.
At x = 0.250 cm,
y
= a6.00 cmf sina0.250π f =
y1 = 3.00 sin π x + 0.600t cm; y 2 = 3.00 sin π x − 0.600t cm
1
2
max
a
At x = 0.500 cm ,
(c)
Now take cos 0.600π t = −1 to get y max :
(d)
a
The antinodes occur when
x=
2π
λ
= π , so
nλ
n = 1, 3 , 5 , K
4
b
g
λ = 2.00 cm
x1 =
λ
4
= 0.500 cm as in (b)
x2 =
3λ
= 1.50 cm as in (c)
4
x3 =
5λ
= 2.50 cm
4
FG
H
The resultant wave is
y = 2 A sin kx +
The nodes are located at
kx +
so
x=
φ
2
The separation of nodes is
IJ cosFGωt − φ IJ
H 2K
2K
φ
= nπ
nπ φ
−
k
2k
which means that each node is shifted
(b)
fa f
y max = 6.00 cm sin 1.50π −1 = 6.00 cm
and
(a)
f a
g
At x = 1.50 cm,
But k =
P14.16
f
y max = 6.00 cm sin 0.500π = 6.00 cm
(b)
b
f a
4.24 cm
φ
2k
to the left.
LMa f π − φ OP − LM nπ − φ OP
N k 2k Q N k 2k Q
∆x = n + 1
The nodes are still separated by half a wavelength.
∆x =
π
k
=
λ
2
Chapter 14
389
Section 14.4 Standing Waves in Strings
P14.17
L = 30.0 m ; µ = 9.00 × 10 −3 kg m; T = 20.0 N ; f1 =
where
FTI
v=G J
H µK
so
f1 =
12
= 47.1 m s
47.1
= 0.786 Hz
60.0
f 2 = 2 f1 = 1.57 Hz
f3 = 3 f1 = 2.36 Hz
P14.18
P14.19
P14.20
v
2L
f 4 = 4 f1 = 3.14 Hz
L = 120 cm , f = 120 Hz
(a)
For four segments, L = 2 λ or λ = 60.0 cm = 0.600 m
(b)
v = λ f = 72.0 m s f1 =
v
72.0
=
= 30.0 Hz
2L 2 1.20
a f
b ge
j
The tension in the string is
T = 4 kg 9.8 m s 2 = 39.2 N
Its linear density is
µ=
and the wave speed on the string is
v=
In its fundamental mode of vibration, we have
λ = 2L = 2 5 m = 10 m
Thus,
f=
(a)
m 8 × 10 −3 kg
=
= 1.6 × 10 −3 kg m
5m
L
T
µ
39.2 N
=
1.6 × 10 −3 kg m
= 156.5 m s
a f
v
λ
=
156.5 m s
= 15.7 Hz
10 m
Let n be the number of nodes in the standing wave resulting from the 25.0-kg mass. Then
n + 1 is the number of nodes for the standing wave resulting from the 16.0-kg mass. For
2L
v
, and the frequency is f = .
standing waves, λ =
n
λ
Thus,
f=
n Tn
2L µ
and also
f=
n + 1 Tn +1
2L
µ
Thus,
Tn
n +1
=
=
n
Tn +1
Therefore,
4n + 4 = 5n , or n = 4
Then,
4
f=
2 2.00 m
continued on next page
a
f
b25.0 kg gg = 5
b16.0 kg gg 4
b25.0 kg ge9.80 m s j =
2
0.002 00 kg m
350 Hz
390
Superposition and Standing Waves
(b)
*P14.21
*P14.22
f1 =
The largest mass will correspond to a standing wave of 1 loop
an = 1f so
350 Hz =
yielding
m = 400 kg
FG IJ
H K
v
T
, where v =
2L
µ
e
m 9.80 m s 2
1
2 2.00 m
a
f
j
0.002 00 kg m
12
1
.
2
(a)
If L is doubled, then f1 ∝ L−1 will be reduced by a factor
(b)
If µ is doubled, then f1 ∝ µ −1 2 will be reduced by a factor
(c)
If T is doubled, then f1 ∝ T will increase by a factor of
For the whole string vibrating, d NN = 0.64 m =
λ
2
1
2
.
2.
; λ = 1.28 m. The
1
speed of a pulse on the string is v = fλ = 330 1.28 m = 422 m s .
s
(a)
When the string is stopped at the fret, d NN =
λ = 0.853 m
f=
(b)
v
λ
=
λ
2
0.64 m = ;
3
2
The light touch at a point one third of the way along the
string damps out vibration in the two lowest vibration
states of the string as a whole. The whole string vibrates in
its third resonance possibility: 3d NN = 0.64 m = 3
λ = 0.427 m
f=
P14.23
FIG. P14.22(a)
422 m s
= 495 Hz .
0.853 m
v
λ
=
λ
2
;
FIG. P14.22(b)
422 m s
= 990 Hz .
0.427 m
d NN = 0.700 m
λ = 1.40 m
fλ = v = 308 m s =
T
e1.20 × 10 j a0.700f
(a)
T = 163 N
(b)
f3 = 660 Hz
−3
FIG. P14.23
Chapter 14
P14.24
a
f fv ; λ = 2L = fv
Ff I
F fI
F 392 IJ = 0.038 2 m
− G J L = L G 1 − J = a0.350 mfG 1 −
H 440 K
Hf K
H fK
λ G = 2 0.350 m =
L G − L A = LG
391
A
G
G
A
G
A
A
G
G
A
Thus, L A = LG − 0.038 2 m = 0.350 m − 0.038 2 m = 0.312 m ,
or the finger should be placed 31.2 cm from the bridge .
LA =
v
1
=
2 fA 2 fA
T
µ
dL A 1 dT
dT
;
=
2 T
4 f A Tµ L A
; dL A =
dL
0.600 cm
dT
=2 A =2
= 3.84%
T
LA
35.0 − 3.82 cm
a
*P14.25
f
In the fundamental mode, the string above the rod has only
two nodes, at A and B, with an anti-node halfway between A
and B. Thus,
λ
2
= AB =
A
L
2L
or λ =
.
cos θ
cos θ
θ
L
Since the fundamental frequency is f, the wave speed in this
segment of string is
v = λf =
Also, v =
T
µ
=
2Lf
.
cos θ
M
r
T
T
TL
where T is the tension in
=
m cos θ
m AB
2Lf
4L2 f 2
TL
TL
=
or
=
2
cos θ
m cos θ
cos θ m cos θ
r
Mg
and the mass of string above the rod is:
T cos θ
4Lf 2
r
F
θ
this part of the string. Thus,
m=
B
FIG. P14.25
[Equation 1]
Now, consider the tension in the string. The light rod would rotate about point P if the string exerted
any vertical force on it. Therefore, recalling Newton’s third law, the rod must exert only a horizontal
force on the string. Consider a free-body diagram of the string segment in contact with the end of
the rod.
Mg
∑ Fy = T sin θ − Mg = 0 ⇒ T = sin θ
Then, from Equation 1, the mass of string above the rod is
m=
FG Mg IJ cos θ =
H sinθ K 4Lf
2
Mg
4Lf 2 tan θ
.
392
P14.26
Superposition and Standing Waves
a
f db
g i db
gi
Comparing
y = 0.002 m sin π rad m x cos 100π rad s t
with
y = 2 A sin kx cos ωt
we find
k=
(a)
2π
= π m −1 , λ = 2.00 m , and ω = 2π f = 100π s −1 : f = 50.0 Hz
λ
d NN =
Then the distance between adjacent nodes is
L
and on the string are
(b)
(c)
d NN
λ
= 1.00 m
2
3.00 m
= 3 loops
1.00 m
=
e
j
For the speed we have
v = fλ = 50 s −1 2 m = 100 m s
In the simplest standing wave vibration,
d NN = 3.00 m =
and
fb =
In v 0 =
T0
µ
increases to
λb
=
100 m s
= 16.7 Hz
6.00 m
, if the tension increases to Tc = 9T0 and the string does not stretch, the speed
9T0
vc =
λc =
Then
va
λb
, λ b = 6.00 m
2
µ
=3
T0
µ
b
g
= 3 v 0 = 3 100 m s = 300 m s
v c 300 m s
=
= 6.00 m
fa
50 s −1
d NN =
λc
= 3.00 m
2
and one loop fits onto the string.
Section 14.5 Standing Waves in Air Columns
P14.27
(a)
For the fundamental mode in a closed pipe, λ = 4L , as
in the diagram.
A
λ /4
v
But v = fλ , therefore L =
4f
So, L =
(b)
343 m s
e
4 240 s −1
j
= 0.357 m
For an open pipe, λ = 2L , as in the diagram.
So, L =
343 m s
v
=
= 0.715 m
2 f 2 240 s −1
e
j
N
L
A
N
λ /2
FIG. P14.27
A
Chapter 14
P14.28
*P14.29
393
d AA = 0.320 m ; λ = 0.640 m
v
(a)
f=
(b)
λ = 0.085 0 m; d AA = 42.5 mm
λ
= 531 Hz
Assuming an air temperature of T = 37° C = 310 K , the speed of sound inside the pipe is
a
f
v = 331 m s + 0.6 m s ⋅ C° 37° C = 353 m s .
In the fundamental resonant mode, the wavelength of sound waves in a pipe closed at one end is
λ = 4L . Thus, for the whooping crane
a
f
λ = 4 5.0 ft = 2.0 × 10 1 ft
P14.30
λ=
The wavelength is
and
f=
v
λ
=
b353 m sg FG 3.281 ft IJ =
2.0 × 10 ft H 1 m K
1
57.9 Hz .
v 343 m s
=
= 1.31 m
f
261.6 s
so the length of the open pipe vibrating in its simplest (A-N-A) mode is
d A to A =
A closed pipe has
1
λ = 0.656 m
2
(N-A) for its simplest resonance,
(N-A-N-A) for the second,
P14.31
and
(N-A-N-A-N-A) for the third.
Here, the pipe length is
5d N to A =
a
f
5λ 5
= 1.31 m = 1.64 m
4
4
For a closed box, the resonant frequencies will have nodes at both sides, so the permitted
1
wavelengths will be L = nλ , n = 1, 2 , 3 , K .
2
b
i.e., L =
g
nλ nv
nv
and f =
.
=
2
2f
2L
Therefore, with L = 0.860 m and L ′ = 2.10 m, the resonant frequencies are
a
fn = n 206 Hz
a
f
for L = 0.860 m for each n from 1 to 9
and fn′ = n 84.5 Hz
f
for L ′ = 2.10 m for each n from 2 to 23.
394
P14.32
Superposition and Standing Waves
The air in the auditory canal, about 3 cm long, can vibrate with a node at the closed end and
antinode at the open end,
with
d N to A = 3 cm =
so
λ = 0.12 m
and
f=
v
λ
=
λ
4
343 m s
≈ 3 kHz
0.12 m
A small-amplitude external excitation at this frequency can, over time, feed energy
into a larger-amplitude resonance vibration of the air in the canal, making it audible.
P14.33
For both open and closed pipes, resonant frequencies are equally spaced as numbers. The set of
resonant frequencies then must be 650 Hz, 550 Hz, 450 Hz, 350 Hz, 250 Hz, 150 Hz, 50 Hz. These are
odd-integer multipliers of the fundamental frequency of 50.0 Hz . Then the pipe length is
d NA =
P14.34
λ
4
=
v 340 m s
=
= 1.70 m .
4f
4 50 s
b g
λ=
The wavelength of sound is
The distance between water levels at resonance is d =
and
P14.35
t=
λ
2
= d AA =
Since λ =
L
or
n
v
f
With v = 343 m s and
L=
v
f
v
2f
∴ Rt = π r 2 d =
π r 2v
2 Rf
2f
.
nλ
2
L=n
π r 2v
for n = 1, 2 , 3 , K
FG v IJ
H2fK
for n = 1, 2 , 3 , K
f = 680 Hz,
L=n
F 343 m s I = na0.252 mf
GH 2a680 Hzf JK
for n = 1, 2 , 3 , K
a
f
Possible lengths for resonance are: L = 0.252 m, 0.504 m, 0.757 m, K , n 0.252 m .
P14.36
The length corresponding to the fundamental satisfies f =
343
v
v
: L1 =
=
= 0.167 m .
4 f 4 512
4L
a f
Since L > 20.0 cm, the next two modes will be observed, corresponding to f =
or L 2 =
3v
5v
= 0.502 m and L3 =
= 0.837 m .
4f
4f
3v
5v
and f =
.
4L 2
4L3
Chapter 14
P14.37
f = 384 Hz
For resonance in a narrow tube open at one end,
f =n
(a)
b
395
g
v
n = 1, 3 , 5 , K .
4L
warm
air
22.8 cm
Assuming n = 1 and n = 3 ,
68.3 cm
3v
v
384 =
and 384 =
.
4 0.228
4 0.683
a
a
f
f
In either case, v = 350 m s .
*P14.38
b
e
For the next resonance n = 5 , and L =
(a)
For the fundamental mode of an open tube,
L=
(b)
a
λ
2
=
g
j
5 v 5 350 m s
=
= 1.14 m .
4f
4 384 s −1
(b)
FIG. P14.37
343 m s
v
=
= 0.195 m .
2 f 2 880 s −1
e
j
f
v = 331 m s + 0.6 m s⋅° C −5° C = 328 m s
We ignore the thermal expansion of the metal.
f=
v
λ
=
328 m s
v
=
= 842 Hz
2L 2 0.195 m
a
f
The flute is flat by a semitone.
Section 14.6 Beats: Interference in Time
P14.39
f ∝v∝ T
f new = 110
540
= 104.4 Hz
600
∆f = 5.64 beats s
P14.40
(a)
The string could be tuned to either 521 Hz or 525 Hz from this evidence.
(b)
Tightening the string raises the wave speed and frequency. If the frequency were originally
521 Hz, the beats would slow down.
Instead, the frequency must have started at 525 Hz to become 526 Hz .
continued on next page
396
Superposition and Standing Waves
(c)
v
From f =
λ
=
T µ
2L
=
1 T
2L µ
FG IJ
H K
f
f2
T2
=
and T2 = 2
f1
f1
T1
2
T1 =
FG 523 Hz IJ
H 526 Hz K
2
T1 = 0.989T1 .
The fractional change that should be made in the tension is then
fractional change =
T1 − T2
= 1 − 0.989 = 0.011 4 = 1.14% lower.
T1
The tension should be reduced by 1.14% .
P14.41
For an echo f ′ = f
bv + v g the beat frequency is f
bv − v g
s
s
b
= f′− f .
Solving for fb .
gives fb = f
b2 v g
s
bv − v g when approaching wall.
2a1.33 f
= a 256f
a343 − 1.33f = 1.99 Hz beat frequency
s
(a)
fb
(b)
When he is moving away from the wall, v s changes sign. Solving for v s gives
vs =
a fa f
a fa f
5 343
fb v
=
= 3.38 m s .
2 f − fb
2 256 − 5
Section 14.7 Nonsinusoidal Wave Patterns
P14.42
We evaluate
s = 100 sin θ + 157 sin 2θ + 62.9 sin 3θ + 105 sin 4θ
+51.9 sin 5θ + 29.5 sin 6θ + 25.3 sin 7θ
where s represents particle displacement in nanometers
and θ represents the phase of the wave in radians. As θ
advances by 2π , time advances by (1/523) s. Here is the
result:
FIG. P14.42
Chapter 14
*P14.43
397
We list the frequencies of the harmonics of each note in Hz:
Note
A
C#
E
1
440.00
554.37
659.26
Harmonic
3
1 320.0
1 663.1
1 977.8
2
880.00
1 108.7
1 318.5
4
1 760.0
2 217.5
2 637.0
5
2 200.0
2 771.9
3 296.3
The second harmonic of E is close the the third harmonic of A, and the fourth
harmonic of C# is close to the fifth harmonic of A.
Section 14.8 Context Connection
Building on Antinodes
P14.44
v=
9.15 m
= 3.66 m s
2.50 s
(a)
The wave speed is
(b)
From the figure, there are antinodes at both ends of the pond, so the distance between
adjacent antinodes
λ
is
d AA =
and the wavelength is
λ = 18.3 m
The frequency is then
f=
v
λ
2
=
= 9.15 m ,
3.66 m s
= 0.200 Hz
18.3 m
We have assumed the wave speed is the same for all wavelengths.
P14.45
The wave speed is
v = gd =
e9.80 m s ja36.1 mf = 18.8 m s
2
The bay has one end open and one closed. Its simplest resonance is with a node of horizontal
velocity, which is also an antinode of vertical displacement, at the head of the bay and an antinode
of velocity, which is a node of displacement, at the mouth. The vibration of the water in the bay is
like that in one half of the pond shown in Figure P14.44.
Then,
d NA = 210 × 10 3 m =
λ
4
3
and
λ = 840 × 10 m
Therefore, the period is
T=
1 λ 840 × 10 3 m
= =
= 4.47 × 10 4 s = 12 h 24 min
18.8 m s
f v
This agrees precisely with the period of the lunar excitation , so we identify the extra-high tides as
amplified by resonance.
398
Superposition and Standing Waves
Additional Problems
P14.46
The distance between adjacent nodes is one-quarter of the circumference.
d NN = d AA =
v
so λ = 10.0 cm and f =
λ
=
λ
2
20.0 cm
= 5.00 cm
4
=
900 m s
= 9 000 Hz = 9.00 kHz .
0.100 m
The singer must match this frequency quite precisely for some interval of time to feed enough
energy into the glass to crack it.
P14.47
f = 87.0 Hz
speed of sound in air: v a = 340 m s
ja
e
f
v = fλ b = 87.0 s −1 0.400 m
λb = l
(a)
v = 34.8 m s
λ a = 4L
va = λ a f
(b)
*P14.48
UV
W
A
L=
340 m s
va
=
= 0.977 m
4 f 4 87.0 s −1
e
j
A
A
d NA
N
N
I
d NA
N
A
N
d NA
A
II
N
(a)
d NA
A
N
d NA
III
(b)
FIG. P14.48
(a)
µ=
v=
5.5 × 10 −3 kg
= 6.40 × 10 −3 kg m
0.86 m
T
µ
=
1.30 kg ⋅ m s 2
6.40 × 10 −3 kg m
(b)
In state I, d NA = 0.860 m =
(c)
λ = 3.44 m
f=
In state II, d NA =
a
f
v
λ
=
λ
4
14.3 m s
= 4.14 Hz
3.44 m
a
f
1
0.86 m = 0.287 m
3
λ = 4 0.287 m = 1.15 m
continued on next page
= 14.3 m s
f=
v
λ
=
14.3 m s
= 12.4 Hz
1.15 m
FIG. P14.47
Chapter 14
In state III, d NA =
f=
P14.49
v
λ
=
a
399
f
1
0.86 m = 0.172 m
5
14.3 m s
= 20.7 Hz
4 0.172 m
a
f
Moving away from station, frequency is depressed:
343
343 − −v
f ′ = 180 − 2.00 = 178 Hz :
178 = 180
Solving for v gives
v=
Therefore,
v = 3.85 m s away from station
a f
a2.00fa343f
178
Moving toward the station, the frequency is enhanced:
*P14.50
343
343 − v
f ′ = 180 + 2.00 = 182 Hz :
182 = 180
Solving for v gives
4=
Therefore,
v = 3.77 m s toward the station
(a)
a2.00fa343f
182
Use the Doppler formula
f′= f
bv ± v g .
bv m v g
0
s
With f1′ = frequency of the speaker in front of student and
f 2′ = frequency of the speaker behind the student.
m s + 1.50 m sg
= 458 Hz
f b343b343
m s − 0g
b343 m s − 1.50 m sg = 454 Hz
f ′ = a 456 Hz f
b343 m s + 0g
a
f1′ = 456 Hz
2
Therefore, fb = f1′ − f 2′ = 3.99 Hz .
(b)
The waves broadcast by both speakers have λ =
between them has d AA =
time ∆t =
λ
2
v 343 m s
=
= 0.752 m . The standing wave
f
456 s
= 0.376 m . The student walks from one maximum to the next in
0.376 m
1
= 0.251 s , so the frequency at which she hears maxima is f = = 3.99 Hz .
T
1.50 m s
400
P14.51
Superposition and Standing Waves
Call L the depth of the well and v the speed of sound.
f λ4 = a2n − 1f 4vf = a2n4− 511fb.5343s m sg
e
j
a2n + 1fb343 m sg
λ
v
= a 2n + 1f
=
L = 2an + 1f − 1
4
4f
4e60.0 s j
a2n − 1fb343 m sg = a2n + 1fb343 m sg
4e51.5 s j
4e60.0 s j
a
Then for some integer n
1
L = 2n − 1
−1
1
2
and for the next resonance
2
Thus,
−1
and we require an integer solution to
The equation gives n =
−1
−1
2n + 1 2n − 1
=
60.0
51.5
111.5
= 6.56 , so the best fitting integer is n = 7 .
17
Then
L=
and
L=
af b
g = 21.6 m
4e51.5 s j
2a7 f + 1 b343 m sg
= 21.4 m
4e60.0 s j
2 7 − 1 343 m s
−1
−1
suggest the best value for the depth of the well is 21.5 m .
P14.52
v=
a48.0fa2.00f = 141 m s
4.80 × 10 −3
d NN = 1.00 m ; λ = 2.00 m ; f =
λa =
P14.53
(a)
v
λ
= 70.7 Hz
v a 343 m s
=
= 4.85 m
f
70.7 Hz
Since the first node is at the weld, the wavelength in the thin wire is 2L or 80.0 cm. The
frequency and tension are the same in both sections, so
f=
(b)
1 T
1
=
2L µ 2 0.400
a
f
4.60
= 59.9 Hz .
2.00 × 10 −3
As the thick wire is twice the diameter, the linear density is 4 times that of the thin wire.
µ ′ = 8.00 g m
1
2f
thin wire.
so L ′ =
T
µ′
L′ =
LM 1 OP
N a2fa59.9f Q
4.60
= 20.0 cm half the length of the
8.00 × 10 −3
Chapter 14
P14.54
The second standing wave mode of the air in the pipe reads ANAN, with d NA =
so
λ = 2.33 m
and
f=
v
λ
=
λ
4
=
1.75 m
3
343 m s
= 147 Hz
2.33 m
For the string, λ and v are different but f is the same.
λ
2
so
= d NN =
λ = 0.400 m
a
fa
f
T
v = λf = 0.400 m 147 Hz = 58.8 m s =
2
e
T = µv = 9.00 × 10
P14.55
(a)
0.400 m
2
−3
f=
n T
2L µ
so
f′ L
L 1
=
=
=
f L ′ 2L 2
jb
kg m 58.8 m s
g
µ
2
= 31.1 N
The frequency should be halved to get the same number of antinodes for twice the
length.
(b)
n′
T
=
T′
n
so
FG IJ = LM n OP
H K Nn + 1 Q
L n OP T
T′ = M
Nn + 1 Q
T ′ F nf ′L ′ I
=G
J
T H n ′fL K
T′
n
=
T
n′
2
2
2
The tension must be
(c)
f ′ n ′L T ′
=
f nL ′ T
FG IJ
H K
T′
3
=
T
2⋅2
P14.56
(a)
2
so
2
T′ 9
=
to get twice as many antinodes.
T 16
For the block:
∑ Fx = T − Mg sin 30.0° = 0
so T = Mg sin 30.0° =
(b)
(c)
1
Mg .
2
The length of the section of string parallel to the incline is
h
= 2 h . The total length of the string is then 3h .
sin 30.0°
The mass per unit length of the string is
continued on next page
µ=
m
3h
FIG. P14.56
401
402
Superposition and Standing Waves
v=
FG Mg IJ FG 3h IJ =
H 2 KH m K
3 Mgh
2m
The speed of waves in the string is
(e)
In the fundamental mode, the segment of length h vibrates as one loop. The distance
between adjacent nodes is then d NN =
λ
(g)
f=
v
λ
1 3 Mgh
=
2h
2m
=
3 Mg
8mh
The period of the standing wave of 3 nodes (or two loops) is
e
j
fb = 1.02 f − f = 2.00 × 10 −2 f =
e2.00 × 10 j
−2
1 λ
2m
= =h
=
3 Mgh
f v
2mh
3 Mg
3 Mg
8mh
We look for a solution of the form
a
f
a
f
b
g
= A sina 2.00 x − 10.0tf cos φ + A cosa 2.00 x − 10.0t f sin φ
5.00 sin 2.00 x − 10.0t + 10.0 cos 2.00 x − 10.0t = A sin 2.00 x − 10.0t + φ
This will be true if both
5.00 = A cos φ and 10.0 = A sin φ ,
requiring
a5.00f + a10.0f
2
2
= A2
A = 11.2 and φ = 63.4°
a
f
The resultant wave 11.2 sin 2.00 x − 10.0t + 63.4°
P14.58
FG λ IJ and
H 2K
λ= h .
T=
(h)
=
When the vertical segment of string vibrates with 2 loops (i.e., 3 nodes), then h = 2
the wavelength is
(f)
µ
= h , so the wavelength is λ = 2h.
2
The frequency is
P14.57
T
(d)
is sinusoidal.
0.010 0 kg
For the wire, µ =
= 5.00 × 10 −3 kg m :
2.00 m
T
v=
µ
e200 kg ⋅ m s j
2
=
5.00 × 10 −3 kg m
v = 200 m s
If it vibrates in its simplest state, d NN = 2.00 m =
(a)
λ
2
:
f=
v
λ
=
b200 m sg = 50.0 Hz
4.00 m
The tuning fork can have frequencies 45.0 Hz or 55.0 Hz .
continued on next page
Chapter 14
b
g
If f = 45.0 Hz , v = fλ = 45.0 s 4.00 m = 180 m s .
(b)
b
Then, T = v 2 µ = 180 m s
g e5.00 × 10
2
or if f = 55.0 Hz , T = v 2 µ = f 2
P14.59
j
λ µ = b55.0 sg a 4.00 mf e5.00 × 10
−3
kg m = 162 N
2
2
2
−3
j
kg m = 242 N .
Let θ represent the angle each slanted rope
makes with the vertical.
(a)
In the diagram, observe that:
sin θ =
r
g
1.00 m 2
=
1.50 m 3
or θ = 41.8° .
Considering the mass,
∑ Fy = 0 : 2T cos θ = mg
b12.0 kg ge9.80 m s j =
or T =
FIG. P14.59
2
2 cos 41.8°
(b)
P14.60
d AA =
λ
2
78.9 N
T
78.9 N
= 281 m s.
0.001 00 kg m
The speed of transverse waves in the string is
v=
For the standing wave pattern shown (3 loops),
d=
3
λ
2
or
λ=
2 2.00 m
= 1.33 m .
3
Thus, the required frequency is
f=
µ
a
v
λ
=
=
f
281 m s
= 211 Hz .
1.33 m
= 7.05 × 10 −3 m is the distance between antinodes.
Then λ = 14.1 × 10 −3 m
and f =
v
λ
=
3.70 × 10 3 m s
14.1 × 10 −3 m
= 2.62 × 10 5 Hz .
FIG. P14.60
18
The crystal can be tuned to vibrate at 2 Hz , so that binary counters can
derive from it a signal at precisely 1 Hz.
ANSWERS TO EVEN PROBLEMS
P14.2
see the solution
P14.6
0.500 s
P14.4
5.66 cm
P14.8
(a) 3.33 rad; (b) 283 Hz
403
404
Superposition and Standing Waves
(a) number of minima heard = n max =
P14.36
0.502 m, 0.837 m
greatest integer ≤ d
P14.38
(a) 0.195 m; (b) 842 Hz
P14.40
(a) 521 Hz or 525 Hz; (b) 526 Hz;
(c) reduced by 1.14%
n = 1, 2 , K , n max
P14.42
see the solution
P14.12
15.7 m, 31.8 Hz, 500 m/s
P14.44
(a) 3.66 m/s; (b) 0.200 Hz
P14.14
see the solution
P14.46
9.00 kHz
P14.16
(a) see the solution; (b) see the solution
P14.48
(a) 14.3 m/s; (b) 0.860 m , 0.287 m, 0.172 m;
(c) 4.14 Hz , 12.4 Hz , 20.7 Hz
P14.18
(a) 0.600 m; (b) 30.0 Hz
P14.50
(a) 3.99 Hz ; (b) 3.99 Hz
P14.20
(a) 350 Hz; (b) 400 kg
P14.52
4.85 m
P14.22
(a) 495 Hz; (b) 990 Hz
P14.54
31.1 N
P14.24
31.2 cm from the bridge, 3.84%
P14.10
(b) Ln =
d2
FG f IJ + 1 ;
H vK 2
− cn − h e j
where
2cn − he j
2
1 2 v
2
f
1
2
v
f
P14.26
(a) 3 loops; (b) 16.7 Hz; (c) 1 loop
P14.28
(a) 531 Hz; (b) 42.5 mm
3 Mgh
1
m
Mg ; (b) 3h; (c)
; (d)
;
2
3h
2m
3 Mg
2mh
; (f)
(e)
; (g) h;
8mh
3 Mg
P14.30
0.656 m, 1.64 m
(h) 2.00 × 10 −2
P14.32
around 3 kHz, A small-amplitude external
excitation at this frequency can, over time,
feed energy into a larger-amplitude
resonance vibration of the air in the canal,
making it audible.
P14.34
P14.56
(a)
e
j
3 Mg
8mh
P14.58
(a) 45.0 or 55.0 Hz; (b) 162 or 242 N
P14.60
2.62 × 10 5 Hz
π r 2v
2 Rf
CONTEXT 3 CONCLUSION SOLUTIONS TO PROBLEMS
CC3.1
Let point 1 be r = 10 km from the epicenter and point 2 be at 20 km. The intensity is proportional to
1
C
according to I = , where C is some constant. Intensity is defined as the energy a wave carries
r
r
each second through a unit area of wavefront, so it is proportional to the amplitude squared
according to I = DA 2 , where D is another constant. Then the factors of change are related by
continued on next page
Chapter 14
405
I 2 DA 2 2 r1C
=
=
I 1 DA1 2 r2C
A2
r
= 1
A1
r2
A 2 = A1
CC3.2
r1
10 km
= 5.0 cm
= 3.5 cm
r2
20 km
As in Equation 13.23, the rate of energy transfer in a seismic wave is proportional to the speed and to
the amplitude squared. We write P = FvA 2 , where F is some constant. If no wave energy is reflected
2
2
or turns into internal energy, Fv bedrock A bedrock
= Fv mudfill Amudfill
FG
H
v mudfill
A bedrock
=
v bedrock
Amudfill
IJ = FG A
K H 5A
2
bedrock
bedrock
IJ
K
2
=
1
25
The speed decreases by a factor of 25.
CC3.3
METHOD ONE
From the graph, we have for the speed of S waves vS =
395 km
= 3.95 km s , and for the speed of P
100 s
400 km
= 8.33 km s . From the data of station 1 we can find a value for the time the
48 s
200 km
quake started: 15 h:46 min:06 s −
= 15 h:45 min:66 s − 24 s = 15 h:45 min:42 s . Similarly
8.33 km s
160 km
from the data of the other stations, the quake began at 15:46:01 −
= 15:45:41.8 or
8.33 km s
105 s
15:45:54 −
= 15:45:41.4 . For the most probable value for the actual time we take the average,
8.33
15:45:41.7 ± 0.3 s .Then the S-wave arrival time should be
waves v P =
15:45:41.7 +
200 km
= 15:46:32 for station 1,
3.95 km s
160 s
= 15:46:22 for station 2,
3.95
105 s
15:45:41.7 +
= 15:46:08 for station 3,
3.95
15:45:41.7 +
all with uncertainties of ±1 s
METHOD TWO
With no significant loss of precision, we can use the graph of travel times to read the S wave arrival
times almost directly.
For station #1, locate 200 km on the horizontal axis. Vertically above it, read the size of the
space between the P and S lines as 27 s. Add this S wave delay time to the P wave arrival time,
15:46:06, to obtain 15:46:33 as the S wave arrival time at station #1.
Similarly for station #2, the S wave should arrive at 21 s + 15:46:01 = 15:46:22.
For station #3, the graph shows that at range 105 km an S wave arrives 14 s after a P wave,
placing it at 15:45:54 + 14 = 15:46:08.
406
Superposition and Standing Waves
Note: The theory developed in the context conclusion can be applied to the following topical problem.
CC3.4
An earthquake or a landslide can produce an ocean wave carrying great energy, called a tsunami,
when its wavelength is large compared to the ocean depth d, the speed of a water wave is given
approximately by gd .
(a)
Explain why the amplitude of the wave increases as the wave approaches shore. What can
you consider to be constant in the motion of one wave crest?
(b)
Assume that an earthquake, all along a plate boundary running north and south, produces a
straight tsunami wave crest moving everywhere to the west. If the wave has amplitude
1.8 m when its speed is 200 m s , what will be its amplitude where the water is 9.00 m deep?
(c)
Explain why the amplitude at the shore should be expected to be still greater, but cannot be
meaningfully predicted by your model.
ANSWERS TO EVEN CONTEXT 3 CONCLUSION PROBLEMS
CC3.2
The speed decreases by a factor is 25.
CC3.4
(a) The energy a wave crest carries is
constant in the absence of absorption.
Then the rate at which energy passes a
stationary point, which is the power of the
wave, is constant. The power is
proportional to the square of the
amplitude end to the wave speed. The
speed decreases as the wave moves into
shallower water near shore, so the
amplitude must increase.
(b) For the wave described, with a single
direction of energy transport, the intensity
is the same at the deep-water location 1
and at the place 2 with depth 9 m. To
express the constant intensity we write
A12 v1 = A 22 v 2 = A 22 gd 2
a1.8 mf
2
200 m s = A 22
e9.8 m s j9 m
2
= A 22 9.39 m s
A2
F 200 m s I
= 1.8G
H 9.39 m s JK
12
= 8.31 m
(c) As the water depth goes to zero, our
model would predict zero speed and
infinite amplitude. In fact the amplitude
must be finite as the wave comes ashore. As
the speed decreases the wavelength also
decreases. When it becomes comparable to
the water depth, or smaller, our formula
gd for wave speed no longer applies.