Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 21

490

14 Oscillations and Wave Motion

the phase angle φ, the maximum speed vmax , and the maximum acceleration
amax . (c) Write down the position, velocity, and acceleration in terms of time
t, then substitute with t = π/8 s and find their values.
Equilibrium position

x

x

x=0

x=0

(a)

xi

(b)

Fig. 14.33 See Exercise (12)

(14) A bullet of mass m = 10 g is fired horizontally with a speed v into a stationary
wooden block of mass M = 4 kg. The block is resting on a horizontal smooth
surface and attached to a massless spring with spring constant kH = 150 N/m,
where the other end of the spring is fixed through a wall, as shown in Fig. 14.34a.
In a very short time, the bullet penetrates the block and remains embedded
before compressing the spring, as shown in Fig. 14.34b. The maximum distance

that the block compresses the spring is 8 cm, as shown in Fig. 14.34c. (a) What
is the speed of the bullet? (b) Find the period T and frequency f of the oscillating
system.

Before collision

(a)

M

Just after collision

M+m

V

8cm

(c)

At maximum
compression

Fig. 14.34 See Exercise (14)

M+m

Stage 1

(b)

v

Stage 2

m


14.8 Exercises

491

Section 14.2 Damped Simple Harmonic Motion
(15) An object of mass m = 0.25kg oscillates in a fluid at the end of a vertical
spring of spring constant kH = 85 N/m, see Fig. 14.35. The effect of the fluid
resistance is governed by the damping constant b = 0.07kg/s. (a) Find the period
of the damped oscillation. (b) By what percentage does the amplitude of the
oscillation decrease in each cycle? (c) How long does it take for the amplitude
of the damped oscillation to drop to half of its initial value?
Fig. 14.35 See Exercise (15)

kH

m

(16) A simple pendulum has a length L and a mass m. Let the arc length s and
the angle θ measure the position of m at any time t, see Fig. 14.36. (a) When
a damped force Fd = −bvs exists, show that the equation of motion of the
pendulum is given for small angles by:
m

d2θ
mg
dθ
+
θ =0
+b
dt 2
dt
L

(b) By comparison with Eq. 14.25, show that the above differential equation
has a solution given by:
θ = θ◦ e−bt/2m cos(ωd t), ωd =

g
b2
−
L 4m2

where θ◦ is initial angular amplitude at t = 0 and ωd = 2π fd is damped angular
frequency, see Fig. 14.9b. (c) When the pendulum has L = 1 m, m = 0.1 kg,
and the angular amplitude θ becomes 0.5 θ◦ after 1 minute, find the damping
constant b and the ratio (f − fd )/f , where f is the undamped frequency.


492

14 Oscillations and Wave Motion

Fig. 14.36 See Exercise (16)

θ
L

m

s

s
O

−b s

θ

mg

−m g sinθ

Section 14.3 Sinusoidal Waves
(17) Given a sinusoidal wave represented by y = (0.2 m) sin(k x − ω t), where
k = 4 rad/m, and ω = 8 rad/s, determine the amplitude, wavelength, frequency,
and speed of this wave.
(18) A harmonic wave traveling along a string has the form y = (0.25 m) sin(3 x −
40 t), where x is in meters and t is in seconds. (a) Find the amplitude, wave
number, angular frequency, and speed of this wave. (b) Find the wavelength,
period, and frequency of this wave?

Section 14.4 The Speed of Waves on Strings

(19) A uniform string has a mass per unit length of 5 × 10−3 kg/m. The string
passes over a massless, frictionless pulley to a block of mass m = 135 kg, see
Fig. 14.37, and take g = 10 m/s2 . Find the speed of a pulse that is sent from one
end of the string toward the pulley. Does the value of the speed change when
the pulse is replaced by a sinusoidal wave?
Fig. 14.37 See Exercise (19)

At time t

τ
mg


14.8 Exercises

493

(20) Assume a transverse wave traveling on a uniform taut string of mass per unit
length μ = 4 × 10−3 kg/m. The wave has an amplitude of 5 cm, frequency of
50 Hz, and speed of 20 m/s. (a) Write an equation in SI units of the form
y = A sin(kx − ω t) for this wave. (b) Find the magnitude of the tension in the
string.

Section 14.5 Energy Transfer by Sinusoidal Waves on Strings
(21) A sinusoidal wave of amplitude 0.05 m is transmitted along a string that has a
linear density of 40 g/m and is under 100 N of tension. If the wave source has
a maximum power of 300 W, what is the highest frequency at which the source
can operate?
(22) A long string has a mass per unit length μ of 125 g/m and is taut under tension
τ of 32 N. A wave is supplied by a generator as shown in Fig. 14.38. This wave

travels along the string with a frequency f of 100 Hz and amplitude A of 2
cm. (a) Find the speed and the angular frequency of the wave. (b) What is the
rate of energy that must be supplied by a generator to produce this wave in the
string? (c) If the string is to transfer energy at a rate of 100 W, what must be
the required wave amplitude when all other parameters remain the same?
Fig. 14.38 See Exercise (22)

y
Vibrator

A
V

x

(23) A sinusoidal wave is traveling along a string of linear mass density μ = 75 g/m
and is described by the equation:
y = (0.25 m) sin(2 x − 40 t)
where x is in meters and t in seconds. (a) Find the speed, wavelength, and
frequency of the wave. (b) Find the power transmitted by the wave.


494

14 Oscillations and Wave Motion

Section 14.6 The Linear Wave Equation
(24) A one-dimensional wave traveling with velocity v is found to satisfy the partial
differential equation [see Eq. 14.58]:
1 ∂ 2y

∂ 2y
−
=0
∂x 2
v2 ∂ t2
Show that the following functions are the solutions to this linear wave equation:
(a) y = A sin(k x − ω t). (b) y = A cos(k x − ω t). (c) y = exp[b(x − v t)], where
b is a constant. (d) y = ln[b(x − v t)], where b is a constant. (e) Any function
y having the form y = f (x − v t).
(25) If the linear wave functions y1 = f1 (x, t) and y2 = f2 (x, t) satisfy the wave Eq.
14.58, then show that the combination y = C1 f1 (x, t)+C2 f2 (x, t) also satisfies
the same equation, where C1 and C2 are constants.

Section 14.7 Standing Waves
(26) A standing wave having a frequency of 20 Hz is established on a rope 1.5 m
long that has fixed ends. Its wavelength is observed to be twice the rope’s length.
Determine the wave’s speed.
(27) A stretched string of length 0.6 m and mass 30 g is observed to vibrate with a
fundamental frequency of 30 Hz. The amplitude of any antinodes in the standing
wave is 0.04 m. (a) What is the amplitude of a transverse wave in the string?
(b) What is the speed of a transverse wave in the string? (c) Find the magnitude
of the tension in the string.
(28) A student wants to establish a standing wave with a speed 200 m/s on a string
that is fixed at both ends and is 2.5 m long. (a) What is the minimum frequency
that should be applied? (b) Find the next three frequencies that cause standing
wave patterns on the string.
(29) Two identical waves traveling in opposite directions in a string interfere to
produce a standing wave of the form:
y = [(2 m) sin(2 x)] cos(20 t)
where x is in centimeters, t is in seconds, and the arguments of the sine and

cosine are in radians. Find the amplitude, wavelength, frequency, and speed of
the interfering waves.


14.8 Exercises

495

(30) A standing wave is produced by two identical sinusoidal waves traveling in
opposite directions in a taut string. The two waves are given by:
y1 = (2 cm) sin(2.3 x − 4 t) and y2 = (2 cm) sin(2.3 x + 4 t)
where x and y are in centimeters, t is in seconds, and the argument of the
sine is in radians. (a) Find the amplitude of the simple harmonic motion of an
element on the string located at x = 3 cm. (b) Find the position of the nodes
and antinodes on the string. (c) Find the maximum and minimum y values of
the simple harmonic motion of a string element located at any antinode.
(31) A guitar string has a length L = 64 cm and fundamental frequency
f1 = 330 Hz, see part (a) of Fig. 14.39. By pressing down with your finger on the
string, it is found that the string is shortened in a way so that it plays an F note
with a fundamental frequency f1 = 350 Hz, see part (b) of Fig. 14.39. [Assume
the speed of the wave remains constant before and after pressing] How far is
your finger from the near end of the string?
L = λ 1 /2

L′= λ ′1 /2

f1

n =1

(a)

f1′

n =1

(b)

Fig. 14.39 See Exercise (31)

(32) A violin string oscillates at a fundamental frequency of 262 Hz when unfingered. At what frequency will it vibrate if it is fingered two-fifths of the length
from its end?
(33) A string that has a length L = 1 m, mass per unit length μ = 0.1 kg/m, and
tension τ = 250 N is vibrating at its fundamental frequency. What effect on
the fundamental frequency occurs when only: (a) The length of the spring is
doubled. (b) The mass per unit length of the spring is doubled. (c) The tension
of the spring is doubled.
(34) Show that the resonance frequency fn of standing waves on a string of length L
and linear density μ, which is under a tensional force of magnitude τ , is given
√
by fn = n τ/μ/2L, where n is an integer.


496

14 Oscillations and Wave Motion

(35) Show by direct substitution that the standing wave given by Eq. 14.62,
y = (2 A sin k x) cos ω t
is a solution of the general linear wave Eq. 14.58:

∂ 2y
1 ∂ 2y
− 2 2 =0
2
∂x
v ∂t
(36) End A of a string is attached to a vibrator that vibrates with a constant frequency
f, while the other end B passes over a pulley to a block of mass m, see Fig. 14.40.
The separation L between points A and B is 2.5 m and the linear mass density
of the string is 0.1 kg/m. When the mass m of the block is either 16 or 25 kg,
standing waves are observed; however, standing waves are not observed for
masses between these two values. Take g = 10 m/s2 . (Hint: The greater the
tension in the string, the smaller the number of nodes in the standing wave)
(a) What is the frequency of the vibrator? (b) Find the largest m at which a
standing wave could be observed.

L

Vibrator A

B

m
Fig. 14.40 See Exercise (36)

(37) Two identical sinusoidal waves traveling in opposite directions on a string of
length L = 3 m interfere to produce a standing wave pattern of the form:
y = [(0.2 m) sin(2π x)] cos(20π t)
where x is in meters, t in seconds, and the arguments of the sine and cosine
are in radians. (a) How many loops are there in this pattern? (b) What is the

fundamental frequency of vibration of the string?
(38) Two strings 1 and 2, each of length L = 0.5 m, but different mass densities
μ1 and μ2 , are joined together with a knot and then stretched between two


14.8 Exercises

497

fixed walls as shown in Fig. 14.41. For a particular frequency, a standing wave
is established with a node at the knot, as shown in the figure. (a) What is
the relation between the two mass densities? (b) Answer part (a) when the
frequency is changed so that the next harmonic in each string is established.
Knot

μ1

Srting 1

μ2

Srting 2

Fig. 14.41 See Exercise (38)

(39) The strings 1 and 2 of exercise 38 have L1 = 0.64 m, μ1 = 1.8 g/m, L2 = 0.8 m,
and μ2 = 7.2 g/m, respectively, and both are held at a uniform tension τ =
115.2 N. Find the smallest number of loops in each string and the corresponding
standing wave frequency.
(40) In the case of the smallest number of loops in exercise 39, determine the total

number of nodes and the position of the nodes measured from the left end of
string 1.


Sound Waves

15

Sound waves are the most common examples of longitudinal waves. The speed of
sound waves in a particular medium depends on the properties of that medium and
the temperature. As discussed in Chap. 14, sound waves travel through air when air
elements vibrate to produce changes in density and pressure along the direction of
the wave’s motion.
Sound waves can be classified into three frequency ranges:
(1) Audible waves: within the range of human ear sensitivity and can be generated
by a variety of ways such as human vocal cords, etc.
(2) Infrasonic waves: below the audible range but perhaps within the range of
elephant-ear sensitivity.
(3) Ultrasonic waves: above the audible range and lie partly within the range of
dog-ear sensitivity.

15.1

Speed of Sound Waves

The motion of a one-dimensional, longitudinal pulse through a long tube containing
undisturbed gas is shown in Fig. 15.1. When the piston is suddenly pushed to the
right, the compressed gas (or the change in pressure) travels as a pulse from one
region to another toward the right along the pipe with a speed v.
The speed of sound waves depends on the compressibility and density of the

√
medium. We can apply equation v = τ/μ, which gives the speed of a transverse
wave along a stretched string, to the speed of longitudinal sound waves in fluids or

H. A. Radi and J. O. Rasmussen, Principles of Physics,
Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_15,
© Springer-Verlag Berlin Heidelberg 2013

499


500

15 Sound Waves

Fig. 15.1 Motion of a
longitudinal sound pulse in a
gas-filled tube

Undisturbed gas
Compressed gas

Compressed gas

Compressed gas


16

Superposition of Sound Waves

In this chapter, we explore the phenomenon that occurs when combining two or more
waves at one point in the same medium. This phenomenon is known as interference.
We first combine waves having the same frequencies. Then we combine waves that
have slightly different frequencies. In both cases we only consider waves with small
amplitudes so that we can use the superposition principle.

16.1

Superposition and Interference

To analyze complex combinations of traveling waves where each wave has a small
amplitude, we use the superposition principle:

The Superposition Principle
If y1 and y2 are two traveling waves produced separately by two sources, then
the resultant wave y at any point is the algebraic sum y1 + y2 when the two
sources act together.

This principle is extremely important in all types of wave motion and applies not
only to sound waves, but to string waves, light waves, and, in fact, to wave motion
of any sort.
The general term interference is applied to the effect produced by two (or more)
traveling waves when they are simultaneously passing through a given region. When
the resultant wave has larger amplitude than that of either individual wave, we refer
to their superposition as constructive interference. However, when the resultant

H. A. Radi and J. O. Rasmussen, Principles of Physics,
Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_16,
© Springer-Verlag Berlin Heidelberg 2013

531


532

16 Superposition of Sound Waves

wave has smaller amplitude than that of either individual wave, we refer to their
superposition as destructive interference.

Superposition of Sinusoidal Waves
Let us apply the principle of superposition to two sinusoidal waves traveling to
the right in a homogeneous medium and having a different phase φ but the same
frequency f, wavelength λ, and amplitude A. Accordingly, we write their individual
waves as follows:
y1 = A sin(kx − ω t),

(16.1)

y2 = A sin(kx − ω t + φ)

where, as usual, k = 2π/λ, ω = 2π f , and φ is the phase constant. The superposition
of y1 and y2 gives the following resultant:
y = y1 + y2 = A [sin(kx − ω t) + sin(kx − ω t + φ)]

(16.2)

To simplify the previous expression, we use the trigonometric identity:
sin a + sin b = 2 cos

a +b
a−b
sin
2
2

(16.3)

If we substitute in this identity with a = kx−ω t and b = kx−ω t +φ, then a−b = −φ
and a + b = 2kx − 2ω t + φ. Accordingly, we will find that the resultant wave y is
reduced to:
y = 2A cos

φ
2

sin kx − ω t +

φ
2

(16.4)

The resultant wave y has the following important characteristics:
(1) It is a sinusoidal wave and has the same frequency f and wavelength λ as any
one of the contributing waves y1 and y2 ,
(2) It has an amplitude of 2A cos(φ/2),
(3) It has a phase of φ/2.
Now, let us consider the following three cases:

(a) If φ = 0, then cos(φ/2) = + 1 and the amplitude of y is +2 A, i.e. twice the
amplitude of either one of the individual waves. In this case, the waves are
said to be in phase and thus interfere constructively, see Fig. 16.1a. In general,


16.1 Superposition and Interference

533

constructive interference occurs if the phase φ is an even multiple of π, i.e.
φ = 0, 2π, 4π, . . . rad, then cos(φ/2) = ±1.
(b) If φ is an odd multiple of π, i.e. φ = π, 3π, 5π, . . . rad, then cos(φ/2) = 0,
and the crests of one wave occur at the same positions as the troughs of the
second wave to produce a resultant amplitude of zero. In this case, the waves
are canceling each other out and are said to be out of phase and thus interfere
destructively, see Fig. 16.1b.
(c) If φ has an arbitrary value other than an odd or even multiple of π, then the
resultant wave has an amplitude between 0 and 2 A, see Fig. 16.1c.

y

y

y1

and

y2

y

are identical

x

0

y1

y2

y

x

0

φ = 0°

φ =180°

(a)

(b)

y

y
y1

y2

x

0
φ = 60 °
(c)

Fig. 16.1 Two identical waves, y1 (blue) and y2 (green), traveling in the same direction are added to
each other at time t = 0 to give a resultant wave y (red). (a) When y1 and y2 are in phase (φ = 0), they
undergo constructive interference with a resultant wave y = y1 + y2 that has double the amplitude of
either one of y1 or y2 . (b) When y1 and y2 are out of phase (φ = π rad = 180◦ ), they undergo destructive
interference with a resultant wave y = y1 + y2 = 0, i.e. they cancel each other out. (c) When the phase
is different from 0 or π rad, the resultant wave y falls somewhere between part (a) and part (b)

16.2

Spatial Interference of Sound Waves

Figure 16.2 depicts an acoustic interferometer device used to demonstrate sound
interference. Sound energy from the source S is divided into two equal parts at the


534

16 Superposition of Sound Waves

T-shaped junction of the tube. This means that the sound wave that reached the
receiver R traveled along either path A or path B. The distance along any path is
called the path length L. The upper path length LA is adjusted by a U-shaped tube,

while the lower path length LB is kept fixed.

Sliding
tube

Path A

LA

S

R

LB
Path B
Fig. 16.2 A device used to demonstrate the interference of sound waves. Sound energy from the speaker
(S) is divided into two parts at the T-shaped junction of the tube. Before reaching the receiver (R), half of
the wave energy propagates through path A of length LA , while the other half propagates through path B
of length LB . The upper path length LA can be varied by sliding the U-tube up or down

Constructive interference occurs when the difference in the path length
L = |LA − LB | is given by:
λ
L = |LA − LB | = (2n) , n = 0, 1, 2, . . .
2

Constructive
interference

(16.5)

Therefore, the two waves reaching the receiver at any time are in phase (φ = 0,
2π, 4π, . . . rad), as shown in Fig. 16.1a, and hence a maximum sound intensity is
detected at the receiver R.
Destructive interference occurs when the difference in the path length
L = |LA − LB | is given by:
λ
L = |LA − LB | = (2n + 1) , n = 0, 1, 2, . . .
2

Destructive
interference

(16.6)

Therefore, the two waves reaching the receiver at any time are completely out of phase
(φ = π, 3π, 5π, . . . rad), as shown in Fig. 16.1b, and hence no sound is detected at
the receiver R.


16.2 Spatial Interference of Sound Waves

535

Because a path difference of a complete wave length λ corresponds to a phase
L to the phase angle φ by the

angle of 2π rad, one can relate path difference
relation:
L=

φ
λ
2π

(16.7)

Example 16.1

Two identical speakers, S1 and S2 , are placed horizontally at a distance d = 2 m
apart. Each emits sound waves of wavelength λ = 80 cm driven by the same oscillator, see Fig. 16.3. A listener is originally located at point O, which is midway
between the two speakers. The listener walks to point P, which is a distance x
from O, and reaches the first minimum in sound intensity. Find x.

.

S2

.

x

O

S1

P
d/2

d/2

Fig. 16.3

Solution: If L1 and L2 are the distances from S1 and S2 to point P, respectively,
then according to Fig. 16.3 we have:
L1 =

d
d
− x, L2 = + x
2
2

From these two relations and Eq. 16.6, the condition for the first destructive interference at point P leads to the following:
|L2 −L1 | =

λ
2

⇒

d
d
+x −
−x
2
2

=

λ

2

⇒

x=

λ 80 cm
=
= 20 cm
4
4

Example 16.2

Two identical speakers, S1 and S2 , are placed vertically at a distance d = 2 m apart
and emit sound waves driven by the same oscillator, see Fig. 16.4. A listener is


536

16 Superposition of Sound Waves

originally located at point O, which is a distance R = 5 m from the center of the
line connecting the two speakers. The listener walks to point P, which is a distance
y = 0.5 m above O, and thus reaches the first minimum in sound intensity. Find
the wavelength λ of the sound wave.

S1

L1

d/2-y

.
.
P

d/2
R
d/2

y

O

d/2+y

L2
S2

Fig. 16.4

Solution: The first minimum in sound intensity occurs when the two waves reaching the listener at point P are 180◦ out of phase. In other words, when their path
difference equals λ/2. As per Fig. 16.4, we first calculate the path lengths L1 and
L2 as follows:
L1 =

R2 + (d/2 − y)2 =

(5 m)2 + [(2 m)/2 − 0.5 m]2 = 5.0249 m

R2 + (d/2 + y)2 =

(5 m)2 + [(2 m)/2 + 0.5 m]2 = 5.2202 m

and
L2 =

Thus, from Eq. 16.6, the first destructive interference at point P leads to the following:
|L2 − L1 | =

λ
2

⇒

|5.2202 m − 5.0249 m| = λ/2

Therefore:
λ = 0.3906 m = 39.06 cm

⇒

0.1953 m = λ/2


16.3 Standing Sound Waves

16.3

537

Standing Sound Waves

Assume we have two identical sound sources that face each other as shown in Fig. 16.5
and driven by the same oscillator. In this case, they produce two identical traveling
waves each with a speed v. These waves would be moving in opposite directions in the
same medium. Of course, these two waves combine according to the superposition
principle.

Fig. 16.5 Two identical sound sources emitting traveling waves towards each other, each with a
speed v. When the two waves overlap, they produce standing waves (not shown in the figure)

To analyze this situation, we assume that the two sound sources generate sound
waves that have the same frequency f, wavelength λ, and amplitude A but differ
by traveling in opposite directions. Therefore, we can write these two waves in the
following form:
y1 = A sin(kx − ω t),
y2 = A sin(kx + ω t)

(16.8)

where y1 represents a wave traveling in the positive x-direction and y2 represents a
wave traveling in the negative x-direction. The superposition of y1 and y2 gives the
following resultant:
y = y1 + y2 = A [sin(kx − ω t) + sin(kx + ω t)]

(16.9)

To simplify this expression, we use the trigonometric identity:

sin(a ± b) = sin a cos b ± cos a sin b

(16.10)

If we substitute in this identity with a = k x and b = ω t, then the resultant wave y
reduces to:
y = (2 A sin kx) cos ω t

(16.11)


538

16 Superposition of Sound Waves

The resultant y represented by Eq. 16.11 gives a special kind of simple harmonic
motion in which every element of the medium oscillates in simple harmonic motion
with the same angular frequency ω (through the factor cos ω t) and an amplitude
(given by the factor 2 A sin kx) that varies with position x. This wave is called a
standing wave because there is no motion of the disturbance along the x-direction.
A standing wave is distinguished by stationary positions with zero amplitudes
called nodes (see Fig. 16.6). This happens when x satisfies the condition sin kx = 0,
that is, when:
kx = 0, π, 2π, 3π, . . .
When using k = 2π/λ, these values give x = 0,
x = 0,

λ
3λ
, λ,

, . . . , that is:
2
2

3λ
λ
λ
, λ,
, . . . = n , (n = 0, 1, 2, . . .)
2
2
2

(Nodes)

(16.12)

In addition, a standing wave is distinguished by elements with greatest possible
displacements called antinodes (see Fig. 16.6). This happens when x satisfies the
condition sin kx = ±1, that is, when:
kx =

π 3π 5π
,
,
, ...
2 2
2

λ 3λ 5λ

Also, using k = 2π/λ, these values give x = ,
,
, . . . , that is:
4 4
4
x=

λ 3λ 5λ
λ
,
,
, . . . = (n + 21 ) , (n = 0, 1, 2, . . .)
4 4
4
2

(Antinodes)

(16.13)

Equations 16.12 and 16.13 indicate the following general features of nodes and
antinodes (see Fig. 16.6):
Soptlight
(1) The distance between adjacent nodes is λ/2.
(2) The distance between adjacent antinodes is λ/2.
(3) The distance between a node and an adjacent antinode is λ/4.


16.3 Standing Sound Waves

539

y
A

When

t= 0

When

t = p /2

When

t= p

o

2 A sin k x
A

A

A

.........x
N

N

N

Antinode=A
Node=N

N

N

λ
Fig. 16.6 The time dependence of the vertical displacement (from equilibrium) of any individual element
in a standing wave y is governed by cos ω t. Each element vibrates within the confines of the envelope
2 A sin kx. The nodes (N) are points of zero displacement, and the antinodes (A) are points of maximum
displacement

In Fig. 16.7a, at t = 0 (ω t = 0), the two oppositely traveling waves are in phase,
producing a wave pattern in which each element of the medium is experiencing its
maximum displacement from equilibrium. In Fig. 16.7b, at t = T /4 (ω t = π/2), the
traveling waves have moved one quarter of a wavelength (one to the right and the
other to the left). At this time, each element of the medium is passing through the
equilibrium position in its simple harmonic motion. The result is zero displacement
for each element at all values of x. In Fig. 16.7c, at t = T /2 (ω t = π ), the traveling
waves are again in phase, producing a wave pattern that is inverted relative to the
t = 0 pattern. The patterns at t = 3T /4 and t = T are similar to t = T /4 and t = 0,
respectively.
y

y

1

y

y

y

1

.

A

.

A

(a) t = 0

y

2

. . . . . x
N
. N N .A N
A

N

1

y

. . . . .
N

N

N

N

(b) t = T / 4

N

2

x

.

A

y

.

y

A

. . . . .
. N N A. N N
A

N

2

x

(c) t = T / 2

Fig. 16.7 Standing-wave patterns resulting from two oppositely traveling identical waves y1 and y2 at
different phases. The displacement is zero for each node (N), and is maximum for each antinode (A)


540

16 Superposition of Sound Waves

Example 16.3

Two opposing speakers are shown in Fig. 16.8. A standing wave is produced
from two sound waves traveling in opposite directions; each can be described as
follows:
y1 = (5 cm) sin(4x − 2 t),

y2 = (5 cm) sin(4x + 2 t).
where x and y, are in centimeters and t is in seconds. (a) What is the amplitude of
the simple harmonic motion of a medium element lying between the two speakers
at x = 2.5 cm? (b) Find the amplitude of the nodes and antinodes. (c) What is the
maximum amplitude of an element at an antinode?
Fig. 16.8

Standing wave

Speaker

0

Speaker

x

Solution: (a) Using the general form of a standing wave given by Eq. 16.11, we
find A = 5 cm, k = 4 rad/cm, and ω = 2 rad/s. Thus:
y = (2 A sin kx) cos ω t = [(10 cm) sin(4x)] cos(2t)
The amplitude of the simple harmonic motion of an element lying between the two
speakers at x = 2.5 cm is the absolute value of the coefficient of cos(2t) evaluated
at this point. Thus:
Amplitude = |(10 cm) sin(4x)|x = 2.5 |
= |(10 cm) sin(10 rad)| = |−5.4 cm| = 5.4 cm
(b) With k = 2π/λ = 4 rad/cm, we have λ = π/2 cm. Then, from Eq. 16.12 we
find that the nodes are located at:
x=n

λ

π
= n cm, (n = 0, 1, 2, . . .)
2
4

From Eq. 16.13, we find that the antinodes are located at:
x = (n + 21 )

λ
π
= (n + 21 ) , (n = 0, 1, 2, . . .)
2
4

(c) The maximum amplitude of antinodes will be 2 A = 10 cm


16.3 Standing Sound Waves

541

Example 16.4

Two sinusoidal sound waves, equal in amplitude and traveling in opposite directions along the x-axis, are superimposed on each other. The resultant wave is of
the form:
y = (2 m) sin

π
π
x cos

t
L
T

where x is in meters and t in seconds and the arguments of the sine and cosine functions are in radians. (a) What are the mathematical formulas of the two sinusoidal
sound waves that are superimposed to give this resultant? (b) Find the values of
the wavelength and the frequency of the two sinusoidal waves when L = 2 m and
T = 1 s. (c) What are the velocities of the two sinusoidal waves?
Solution: (a) Using the general form of the standing waves given by Eq. 16.11,
we find A = 1 m, k = π/L rad/m, and ω = π/T rad/s. Using Eq. 16.8, we find the
two sinusoidal waves as follows:
π
x−
L
π
y2 = (1 m) sin
x+
L
y1 = (1 m) sin

π
t ,
T
π
t
T

(b) Using k = 2π/λ, and ω = 2π f when L = 2 m and T = 1 s, we have:
k=

π
π
2π
= =
λ
L
2m

ω = 2π f =

π
π
=
T
1s

⇒

⇒

λ=4m

f = 0.5 s−1 = 0.5 Hz

(c) Using v = ω/k, we find the speed of each of the sinusoidal waves as follows:
v=

2π f
ω
=

= λf = (4 m)(0.5 s−1 ) = 2 m/s
k
2π/λ

The velocity of y1 is v1 = +2 m/s (in the direction of increasing x) and the velocity
of y2 is v2 = −2 m/s (in the direction of decreasing x).

16.4

Standing Sound Waves in Air Columns

In Chap. 14, we saw how a standing wave can be generated either on a stretched string
with fixed ends or when one end is fixed and the other is left free to move. We learned


542

16 Superposition of Sound Waves

that this happens when the wavelengths of the waves suitably match the length of the
string, in which case the superposition of the traveling and reflecting waves produce a
standing wave pattern. For such a match, the wavelength corresponds to the resonant
frequency of the string.
We can set up standing sound waves in air-filled pipes in a way similar to that for
strings. Here is how we can compare the two:
1. The closed end of a pipe is similar to the fixed end of a string in that it must
be a displacement node. This is because the pipe’s wall at this end does not
allow longitudinal motion of the air and acts like a pressure antinode (point of
maximum pressure variation).
2. The open end of a pipe acts like the end of a string that is free to move, so there

must be a displacement antinode there1 . This is because the pipe’s open end
allows longitudinal motion of the air and acts like a pressure node (point of no
pressure variation, since the end must remain at atmospheric pressure).
It is interesting to know how sound waves reflect from the open end of a pipe.
To get insight into this, we start with the fact that sound waves are in fact pressure
waves. Next, we know that any compression region must be contained inside the pipe
(between its two ends). Furthermore, any compression region that exists at an open
end is free to expand into the atmosphere. This change in behavior of the air inside
and outside the pipe is sufficient to allow some reflection.
With the boundary conditions of nodes and antinodes at the ends of air columns,
we must set the normal modes of oscillations as we did in the case of stretched
strings.

Air Columns of Two Open Ends
First, we consider a pipe of length L that is open at both ends. By representing the
horizontal displacement of air elements on the vertical axis and applying the boundary
condition that meets the case of two open ends, see Fig. 16.9, the normal modes of
oscillations can be explained by considering the following first three patterns:
(1) The first normal mode (the first harmonic, or the fundamental):
The simplest pattern is shown in Fig. 16.9a. There are two imposed antinodes
1

The antinode of an open end of a pipe is located slightly beyond the end because sound compression reaching an open end does not reflect until it passes the end. Therefore, the effective length
of the air column is little greater than the true length L of the pipe.


16.4 Standing Sound Waves in Air Columns

543

at the two ends and only one node in the middle of the pipe. Also, there is only
half a wavelength in the length L. Thus, this standing wave pattern has:
λ1 = 2L and f1 =

v
v
=
.
λ1
2L

(2) The second normal mode (the second harmonic):
The second pattern is shown in Fig. 16.9b. This pattern has three antinodes and
two nodes. This standing wave pattern has:
v
v
= = 2 f1
λ2
L

λ2 = L and f2 =

(3) The third normal mode (the third harmonic):
The third pattern is shown in Fig. 16.9c. This pattern has four antinodes and
three nodes. This standing wave pattern has:
v
3v
=
= 3 f1
λ3

2L

λ3 = 2L/3 and f3 =

L

λ1 = 2 L

(a)
A

A

N

λ1 =

f1=

n=1

First
harmonic

n=2

Second
harmonic

n=3

Third
harmonic

2L

λ2= L

(b)
A

A

N

A

N

f2=

λ2 =

L= 2 f 1

λ3 = 2 L 3

(c)
A

N

A

N

A

N

A

f3=

λ3 =3

2L= 3 f 1

Fig. 16.9 The first three standing wave patterns (a), (b), and (c) of a longitudinal sound wave established
in an organ pipe that is open to the atmosphere at both ends. The horizontal motion of air elements in
the pipe is displayed vertically by using a red color. The difference between successive harmonics is the
fundamental frequency f1 , and each harmonic is an integer multiple of the fundamental frequency f1

Generally, the relation between the wavelength λn of the various normal modes
and the length L of a pipe of two open ends is:
λn =

2L
, (n = 1, 2, 3, . . .)
n

(Pipe, two open ends)

(16.14)


544

16 Superposition of Sound Waves

Also, according to the relation f = v/λ, where the speed v of the sound wave
is the same for all frequencies, the resonance frequencies fn associated with these
modes are (see Fig. 16.9):
fn =

v
v
= n , (n = 1, 2, 3, . . .)
λn
2L

(Pipe, two open ends)

(16.15)

The expressions of λn and fn are the same as for the string, except that v is the speed of
waves on the strings as in Eq. 14.66, whereas v in Eq. 16.15 is the speed of sound in air.
The relation between the resonance frequencies and the fundamental frequency is:
fn = n f1 , (n = 1, 2, 3, . . .)

(Pipe, two open ends)

(16.16)

Air Columns of One Closed End
Second, we consider a pipe of length L that is open at one end and closed at the
other. By applying the boundary condition that meets this case, the normal modes of
oscillations can be explained by considering the following first three patterns:
(1) The first normal mode (the first harmonic, or the fundamental):
Fig. 16.10a shows the simplest pattern. The standing wave extends from an
antinode at the open end to the adjacent node at the closed end. The fundamental standing wave pattern has:
λ1 = 4 L and f1 =

v
v
.
=
λ1
4L

(2) The third normal mode (the third harmonic):
The next pattern is shown in Fig. 16.10b. This pattern has two antinodes and two
nodes. Thus, this standing wave pattern has:
λ3 = 4L/3 and f3 =

v
3v
=
= 3 f1
λ3

4L

(3) The fifth normal mode (the fifth harmonic):
The next pattern is shown in Fig. 16.10c. This pattern has four antinodes and
four nodes. Thus:
λ5 = 4L/5 and f5 =

v
5v
=
= 5 f1
λ5
4L


16.4 Standing Sound Waves in Air Columns
L

545

λ1 = 4 L

(a)
A

f1=

N

λ1 =

4L

n=1

First
harmonic

n=3

Third
harmonic

n=5

Fifth
harmonic

λ 3= 4L/ 3

(b)
A

N

N

A

f3=

λ3 = 3

4 L= 3 f 1

λ5= 4L 5

(c)
A

N

A

N

A

N

f5=

λ5 = 5

4 L= 5 f 1

Fig. 16.10 The first three standing wave patterns (a), (b), and (c) of a longitudinal sound wave established
in an organ pipe that is open to the atmosphere at only one end. The horizontal motion of air elements
in the pipe is displayed vertically by using a red color. The harmonic frequencies are the odd-integer
multiples of f1 , and the successive difference is 2 f1

Generally, λn and fn of the various normal modes for a pipe of length L with only
one end open are given as (see Fig. 16.10):
λn =

fn =

4L
, (n = 1, 3, 5, . . .)
n

(Pipe, one open end)

v
v
= n , (n = 1, 3, 5, . . .)
λn
4L

fn = nf1 , (n = 1, 3, 5, . . .)

(Pipe, one open end)

(Pipe, one open end)

(16.17)

(16.18)

(16.19)

Figure 16.11 shows a simple apparatus for demonstrating the resonance of sound
waves in air columns. A tube that is open from both ends is immersed into a container
filled with water, and a tuning fork of unknown frequency f and wavelength λ is placed
at its top. The sound waves generated by the fork are reinforced when the length L
corresponds to one of the resonance frequencies of the tube. Thus:
λ=

v
v
4Ln
, f = =n
, (n = 1, 3, 5, . . .)
n
λ
4Ln

(16.20)