Chapter 18
Superposition and Standing Waves.
Solutions of Selected Problems
18.1
Problem 18.8 (In the text book)
Two loudspeakers are placed on a wall 2.00 m apart. A listener stands 3.00 m from the
wall directly in front of one of the speakers. A single oscillator is driving the speakers at a
frequency of 300 Hz.
(a) What is the phase difference between the two waves when they reach the observer?
(b) What If? What is the frequency closest to 300 Hz to which the oscillator may be
adjusted such that the observer hears minimal sound?
Solution
(a) The listener is 3 m away from one speaker and 32 + 22 = 13 m away from the other
speaker. The path difference x is then:
x = 13 3 = 0.606 m
and the wavelength is:
=
343
v
=
= 1.14 m
f
300
and the phase difference is:
= 2
x
0.606
= 2
= 3.34 rad
1.14
CHAPTER 18. SUPERPOSITION AND STANDING WAVES. SOLUTIONS OF
SELECTED PROBLEMS
(b) Minimum sound means destructive interference, and for that we must have:
1
1v
x = =
2
2f
Physics 111:Introductory Physics II, Chapter 18
or
f=
winter 2005
v
343
=
= 283 Hz
2x
2 0.606
Ahmed H. Hussein
18.2. PROBLEM 18.17 (IN THE TEXT BOOK)
18.2
Problem 18.17 (In the text book)
Two sinusoidal waves combining in a medium are described by the wave functions
y1 = (3.0cm) sin (x + 0.60t)
and
y2 = (3.0cm) sin (x 0.60t)
where x is in centimeters and t is in seconds. Determine the maximum transverse position
of an element of the medium at
(a) x = 0.250 cm,
(b) x = 0.500 cm, and
(c) x = 1.50 cm.
(d) Find the three smallest values of x corresponding to antinodes.
Solution
The combined wave function is:
y = y1 + y2 = 3.00 sin[(x + 0.600t)] + 3.00 sin[(x 0.600t)]
Using:
sin(a b) = sin(a) cos(b) cos(a) sin(b)
we get for the combined wave:
y = 3.00[sin(x) cos(0.600t) + cos(x) sin(0.600t)] + 3.00[sin(x) cos(0.600t) cos(x) sin(0.600t)]
= 6.00 sin(x) cos(0.600t)
(a) To find the maximum transverse position of an element of the medium; we start by
setting cos(0.600t) = +1, we then get at x = 0.25 cm:
ymax = 6.00(cm) sin(0.250) = 4.24 cm
(b) At x = 0.500 cm:
ymax = 6.00 sin(0.500) = 6.00 cm
Physics 111:Introductory Physics II, Chapter 18
winter 2005
Ahmed H. Hussein
CHAPTER 18. SUPERPOSITION AND STANDING WAVES. SOLUTIONS OF
SELECTED PROBLEMS
(c) Now at x = 1.5 cm, we take cos(0.600t) = 1:
ymax = 6.00 sin(1.5) = 6.00 cm
(d) The antinodes occur when:
x=
n
4
n = 1, 3, 5,
From the combined wave function above, we find k = , we then get:
k=
2
=
or
= 2.00 cm
The smallest values of x corresponding to antinodes are given by n = 1, 3, 5, so:
= 0.500 cm
4
3
=
= 1.50 cm
4
5
=
= 2.50 cm
4
x1 =
x2
x3
Physics 111:Introductory Physics II, Chapter 18
winter 2005
Ahmed H. Hussein
18.3. PROBLEM 18.31 (IN THE TEXT BOOK)
18.3
Problem 18.31 (In the text book)
A standing-wave pattern is observed in a thin wire with a length of 3.00 m. The equation
of the wave is
y = (0.002m) sin(x) cos(100t)
where x is in meters and t is in seconds.
(a) How many loops does this pattern exhibit?
(b) What is the fundamental frequency of vibration of the wire?
(c) What If? If the original frequency is held constant and the tension in the wire is increased
by a factor of 9, how many loops are present in the new pattern?
Solution
The general wave function of a standing wave in a wire is:
y = 2A sin(kx) cos(t)
comparing this general function with the given function we get:
k==
or
= 2.00 m
and
= 100 = 2f
or
f = 50.0 Hz
(a) The distance between adjacent nodes dN N = /2 and the number of loops N on the wire
with length L = 3.00 m is:
N=
L
dN N
L
2L
2 3.00
=
=
= 3 loops
2
2.00
and the speed of the wave in the wire is:
v = f = 50.0 2.00 = 100 m/s
(b) The standing wave that has the fundamental frequency fits one-half of a wavelength in
the entire length of the wire, so its wavelength is:
= 2L = 6.00 m
This wave travels the wire with the same speed as the wave that fits 3 loops in the length
of the wire, so fundamental frequency is then:
f=
Physics 111:Introductory Physics II, Chapter 18
v
100 m/s
=
= 16.7 Hz
6.00 m
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Ahmed H. Hussein
CHAPTER 18. SUPERPOSITION AND STANDING WAVES. SOLUTIONS OF
SELECTED PROBLEMS
(c) Let the initial tension be T1 , the velocity v1 = 100 m/s, the new tension T2 = 9T1 and
the new velocity v2 , then
s
s
s
T2
9T1
T1
v2 =
=
=3
= 3v1 = 3 100 = 300 m/s
The new wavelength is:
2 =
v2
300
=
= 6.00 m
f
50
and the number of loops in this case:
1
dN N = 2 = 3.00 m
2
Physics 111:Introductory Physics II, Chapter 18
and
winter 2005
N=
L
1
3.00
= 1 loop
3.00
Ahmed H. Hussein
18.4. PROBLEM 18.53 (IN THE TEXT BOOK)
18.4
Problem 18.53 (In the text book)
A student holds a tuning fork oscillating at 256 Hz. He walks toward a wall at a constant
speed of 1.33 m/s.
(a) What beat frequency does he observe between the tuning fork and its echo?
(b) How fast must he walk away from the wall to observe a beat frequency of 5.00 Hz?
Solution
The student and the tuning fork are moving toward the wall with a speed of vs = 1.33 m/s.
The frequency at the wall f 0 is:
v
f0 =
f
v vs
Now waves with frequency f 0 are reflected from the wall and heard by the student who is
walking, with a speed of 1.33 m/s toward the wall which is the source of the the reflected
waves. The frequency of the reflected waves as heard by the student f 00 is:
v + vs 0 v + vs
v
v + vs
f 00 =
f =
f=
f
v
v
v vs
v vs
The student, actually, hears the wave produced by tunning fork and the wave reflected
00 from
the wall. The two waves combine and produce beats. The beats frequency fb = f f is
then:
v + vs
v + vs
v + vs v + vs
2vs
00
f f =f
1 =f
=f
fb = f f =
v vs
v vs
v vs
v vs
(a) The last formula is for the case when the student is walking toward the wall, so fb is
then:
2 1.33
= 1.99 Hz
fb = 256
343 1.33
(b) When student is walking away from the wall, vs changes sign and ,
v vs
f 00 =
f
v + vs
In this case fb = f f 00 ,
v vs
v + vs v + vs
2vs
00
f =f
=f
fb = f f = f
v + vs
v + vs
v + vs
from that we find vs
fb v + fb vs = 2vs f
Physics 111:Introductory Physics II, Chapter 18
or
vs =
fb v
5 343
=
= 3.38 m/s
2f fb
2 256 5
winter 2005
Ahmed H. Hussein
CHAPTER 18. SUPERPOSITION AND STANDING WAVES. SOLUTIONS OF
SELECTED PROBLEMS
18.5
Problem 18.63 (In the text book)
Two wires are welded together end to end. The wires are made of the same material, but the
diameter of one is twice that of the other. They are subjected to a tension of 4.60 N . The
thin wire has a length of 40.0 cm and a linear mass density of 2.00 g/m. The combination
is fixed at both ends and vibrated in such a way that two antinodes are present, with the
node between them being right at the weld.
(a) What is the frequency of vibration?
(b) How long is the thick wire?
Solution
(a) The frequency of the standing wave and the tension are the same along the entire length
of the wire. Since there is a node at the weld, then there is a half-wave fitted on the
length of the thin wire. So, the wavelength = 2L1 = 80.0 cm, where the L1 is the
length of the thin wire. The frequency of the wave is then:
s
r
1
T
4.60
1
= 59.9 Hz
f=
=
2L 1
80.0 2.00 103
where 1 is the linear density of the this wire.
(b) Since the thick wire is twice the diameter of thin wire then its linear density 2 is four
times that of the thin wire, i.e. 2 = 41 = 8.00 103 kg/m. The length of the thick
wire L2 is then:
s
r
T
1
4.60
1
L2 =
=
= 20.0 cm
2f 2
2 59.9 8.00 103
Physics 111:Introductory Physics II, Chapter 18
winter 2005
Ahmed H. Hussein
