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Chapter 16
TRAVELING WAVES
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms as it is used in physics, and use the term in an
operational definition:
frequency
wavelength
amplitude
longitudinal wave
transverse wave
phase
intensity
reflection
refraction
superposition principle
interference
diffraction
standing wave
Fourier's theorem
dispersion
Wave Forms
Sketch a longitudinal wave and a transverse wave.
Wave Problems
Solve wave problems involving the relationships that exist between the different
characteristics of waves.
Superposition and Fourier's Theorem
Use the superposition principle and Fourier's theorem to explain the wave form of a
complex wave.
Standing Waves
Use the superposition principle to explain the formation of standing waves in different
situations.
Inverse Square Law
Use the inverse square law to calculate the intensity of a wave emanating from a point
source.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 5, Energy,
Chapter 13, Elastic Properties of Materials, and Chapter 15, Simple Harmonic Motion.
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Chapter 16
TRAVELING WAVES
16.1 Introduction
Have you ever taken the loose end of a rope tied to a fence and given it a shake? Have
you ever dropped a stone into a quiet lake? Have you ever shaken the dirt out of a rug?
Have you ever given a towel a flip? Each of these is a way of getting energy to travel
from one place to another in the form of undulatory, or wave, motion.
Many of your interactions with your environment take place in conjunction with
wave motion. Your senses of hearing and sight involve the detection of pressure and
electromagnetic traveling waves. In this chapter you are introduced to the quantitative
treatment of traveling waves and their associated phenomena.
16.2 Waves
We have previously considered energy transport as it occurs in direct contact
interactions with matter. The baseball possesses kinetic energy and makes direct contact
with the bat. The gas molecules carry their kinetic energy into a direct contact
interaction with the piston of a heat engine. In this chapter we return to the interactionat-a-distance model and consider wave motion. Energy from a stone dropped into the
lake is transported to a floating stick by the water waves generated by the stone. The
speaker system of a stereo transfers energy to your ears through the sound waves it
generates with its vibrations. The energy we receive from the sun is transported in the
form of electromagnetic waves.
A water wave is characterized by its crests and troughs moving past the floating
stick. Every form of wave can be characterized by a change in a physical variable that is
propagated through space. The wave model calls for energy propagation as a result of a
succession of oscillations (crests and troughs) at neighboring points in space. The wave
model requires no net transfer of matter for its energy transportation.
For example, in sound, the physical variable undergoing oscillation is the pressure.
Sound is referred to as a matter wave, because the vibratory motion of matter, i.e.,
molecules, is involved in its propagation. Light is referred to as an electromagnetic
wave, not a matter wave. The physical variables oscillating in an electromagnetic wave
are the electric and magnetic fields. Electromagnetic waves, unlike matter waves, can be
propagated through a vacuum.
What are the characteristics of the wave model that distinguish it from the particle,
or matter, transport model? The answer lies in the results of the superposition principle
when applied to the wave model. You have confronted such results when you hear
sound around the corner from the source of the sound. The color of an oil film on water
is another example of the results of superimposed waves. The formal designations of
these unique superposition results are interference and diffraction.
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16.3 Longitudinal and Transverse Waves
The two different types of waves are illustrated in Figure 16.1. The longitudinal wave
consists of oscillations parallel to the propagation direction. In a child's toy called a
Slinky the wave vibrating parallel to its axis is longitudinal. A transverse wave consists of
oscillations perpendicular to the direction of propagation. The wave on a vibrating
violin string is an example of a transverse wave. Sound and light are understood as
longitudinal and transverse waves respectively.
16.4 Wave Propagation
Consider what happens if you pick up a long rope and start shaking the end of the rope
by moving your hand up and down. The shape of the rope for the first few positions of
your hand is shown inFigure 16.2, which shows you moving your hand back and forth
with an amplitude of motion A and a frequency of one complete cycle every 4t seconds.
Notice that the waves traveling down the rope have a characteristic wavelength λ.
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After you have been shaking the rope for several complete cycles of oscillation, the
shape of the rope at some instant in time, t 0, may be given by Figure 16.3a, and the
displacement of the rope at some position, x0, may be given by Figure 16.3b.
Note that the amplitude or displacement is repeating itself after each wavelength of
distance. The wavelength λ of a wave is defined to be the period of the wave in space.
Likewise, if you imagine watching one point of the rope you will observe that this point
reaches maximum amplitude periodically in time. This time period is the time required
for the wave to make one complete oscillation. Thus we have the following relation
between period (time) and frequency of a wave:
f = 1/period (seconds) = f (hertz)
(16.1)
We now combine these two periodic aspects of the wave motion to give us a useful
wave equation. The distance that the traveling wave moves during one time period is
one wavelength. The speed of the wave is given by the distance traveled divided by the
time or
υwave = λ/period = λ f (m/sec)
(16.2)
EXAMPLE
A string has a wave traveling along it with a speed of 50 m/sec at a frequency of 100
Hz. Find the wavelength of the wave. From Equation 16.2 we see that
λ = υf = 50 m/sec / 100 Hz
Thus, λ = 0.50 m.
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16.5 Phase
Assume that your hand which is shaking a rope is executing simple harmonic motion,
up and down with a constant amplitude and period. Then an equation for the location
of your hand y is given by
y = A sin ωt =A sin 2πf t
(16.3)
where A is the amplitude (m), ω is the angular frequency (rad/sec), and t is the time
(sec).
If you study the displacement of a segment of rope located a distance x from the
source of the oscillations (your hand), you will notice that the rope is executing the
same type of motion as the source except for the phase of the motion, where phase is
defined as a fraction of a complete cycle from a specified reference, often expressed as
an angle. You see in Figure 16.2 that the segment of the rope located a distance of λ
from your hand is moving in phase with your hand. In general, the phase difference
between two points along the rope is given by 2πx/ λ. If the wave is traveling in the
positive x direction, the phase of the particle at x will be behind the source, and the
motion of the particle is given by
y = A sin (2πf t -2πx/λ) (16.4)
Show that y =A sin (2πf t + 2πx/λ) is a wave traveling in the negative x direction.
16.6 Energy Transfer
Traveling waves transfer energy from one place to another. For matter waves we can
develop an equation for the energy carried by the wave by using our understanding of
simple harmonic motion. Waves in matter can be compared to waves propagated along
a spring. The elastic properties of the medium play the same role as the spring constant,
and the medium has an inertial property that is analogous to the mass on the end of the
spring. The energy per cycle of a traveling wave is given by the maximum potential
energy of a spring system (Equation 15.14)
E = 1/2 kA2 (15.14)
where k is equal to (2πf )2 m and m is the inertial property of the medium,
E = f (2πf )2mA2 = 2π2f2mA2 (16.5)
The power transmitted by the wave is equal to the product of the frequency times
the energy per cycle. Thus,
E = f E = f (2π2f2mA2) = 2π2f3mA2 (16.6)
The intensity of a wave is a measure of the energy transported by the wave through a
unit area in one second, or power per unit area. Hence,
I = P/area = (2π2f2mA2) / area x (υ/λ) = 2π2f 2A2 (m / areaλ) υ
and this can be expressed as follows for a gas, where
m /(area λ) = ρ
the density of the gas. Thus, we can rewrite I in terms of other variables
I = π2f2A2υρ (16.7)
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which is an expression for the intensity of a compression (longitudinal) wave in a gas.
You will notice that the units for this equation are consistent with our previous
definition of power. Also, we can see that the intensity of a wave is proportional to the
square of its amplitude from
I = energy transmitted/(area x time) = 2π2f2A2υρ ∞ (amplitude)2 (16.8)
The transfer of energy from one medium to another via waves is maximized when
the inertial properties of the two media are matched at their interface. This can be
demonstrated by tying two different strings together. The energy transmitted from one
string to the next is determined by the linear densities (mass/meter) of the two strings.
The more closely the linear densities are matched, the more efficient will be the energy
transferred from the first to the second string.
EXAMPLE
A wave of 100 Hz traveling at a speed of 50 m/sec has an amplitude of 2 cm. Find the
power transported by the wave in a string of linear density 20 x 10-3 kg/m.
P = 2π2 x (100 Hz)2 (m / λ) υΑ2 wherem / λ = 20 x 10-3 kg/m
= 2π2 x (100 Hz)2 (2 x 10-2 kg/m)(50 m/sec)(4 x 10-4 m2)
= 2π2 x 2 x 50 x 10-2 x watts
= 2π2 watts
16.7 Reflection and Refraction
If you send a pulse down a string that is tied to a post, you will observe that the pulse
comes back to you(Figure 16.4). We say that the pulse is reflected at the post.
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In this case you will also note that the pulse is inverted upon reflection. This is called a
change in phase. The incoming pulse is called the incident pulse. The return pulse or
wave is known as the reflected pulse. Whenever the inertia parameter increases at an
interface, the reflected wave will be one- half wavelength out of phase with the incident
wave. For example, at a knot connecting a light rope with a heavy rope the wave
traveling from the light to the heavy rope will be reflected exactly out of phase at the
knot. A wave going from larger to smaller density undergoes no phase change.
Reflection is a phenomenon that is common to all waves at the interface between two
media. Using the conservation of energy, can you write an equation relating the percent
of energy reflected and the percent of energy transmitted at an interface?
A general rule governing reflection is
stated as follows:
The angle of incidence equals the angle of
reflection. (Figure 16.5).
⁄_ i = ⁄_ r
(16.9)
While reflection is a property of wave motion,
it does not distinguish between wave motion
and the motion of particles. A tennis ball
thrown against a wall is also reflected with
the incident angle equal to the reflection angle
in the idealized case where friction and spin
are neglected.
Another common wave phenomenon occurring at an interface is refraction.
Refraction refers to the bending of the wave as it moves from one medium to another.
This bending or change in direction at the interface of two media takes place because
the wave has different velocities in the two media. List some examples of refraction that
you have observed.
The wave in the second medium is bent toward a line perpendicular to the interface
when the wave velocity is less in the second medium than in the first medium. When
the wave velocity is greater in the second medium than it is in the first medium, it is
bent away from this normal line. These two cases are illustrated in Figure 16.6.
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16.8 Superposition of Waves
When two or more waves are propagated in the same region of space, we find the
superposition principle provides us with proper results. The algebraic sum of all
amplitudes present at a given point at any instant gives the correct amplitude of the
resultant wave.
The superposition principle for two waves on a string is shown in Figure 16.7.
Wave C = Wave A + Wave B
The effect of each wave is independent of the other waves present. When more than one
wave is present in the same place, the composite wave displacement is the algebraic
sum of the wave displacements at a given place and time. This algebraic sum of
individual wave displacements depends upon the frequency and phase relations among
the waves. Many important consequences of the superposition of waves occur when the
waves involved have equal frequencies (or multiples of the same frequency) with fixed
phase relationships.
Such conditions result in the unique wave phenomena of interference, diffraction, and
standing waves.
16.9 Interference
Suppose that we have two waves of equal frequency and amplitude in the same phase,
that is, crest to crest and trough to trough, traveling in the same direction. The
displacement of any particle is then the sum of the separate displacements, and the
resultant amplitude is twice the amplitude of either (See Figure 16.8a). This is called
constructive interference. Now, suppose the two waves of the same frequency and
amplitude, and traveling in the same direction, are 180o out of phase-that is, crest
opposite trough. The net displacement of a particle under these conditions is zero. This
condition is called destructive interference (Figure 16.8b). We can summarize by saying
that destructive interference is produced by waves from two coherent sources (phase
relationships remain constant) of equal amplitude and frequency, and out of phase by
180o or π radians. In general, we see that two waves can combine to give an amplitude
greater than either separate wave or they can combine to give zero amplitude. There is
the complete range of possibilities between totally constructive and totally destructive
interference. For another case, consider the constructive and destructive case of waves
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with amplitudes A and (½) A. Here the constructive case yields a maximum amplitude
of (3/2) A and the destructive case an amplitude of (½)A.
Interference phenomena are unique to waves. Particle models cannot account for
experimental observations of interference in nature.
16.10 Diffraction
Diffraction is also uniquely a wave phenomenon. Diffraction is the result of the
superposition of wave fronts produced when a wave encounters an object. Diffraction
patterns are produced by the constructive and destructive interferences of coherent
waves. Sound is diffracted when it passes through an open door, resulting in the
bending of the sound down a hallway. Light shining through the same door casts a
well-defined shadow on the wall. However, when light is incident on a narrow, slit, it is
bent and shows a diffraction pattern. The rule of thumb for significant diffraction
phenomena may be stated as follows: When a wave encounters an object about the same size
as its wavelength, diffraction phenomena can be observed.
The diffraction of water waves passing through a slit is shown in Figure 16.9.
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Christian Huygens provided a construction procedure in 1978 that enables us to picture
diffraction as follows: each point on the front of a wave serves as a source of secondary
waves (traveling with the wave speed in the medium). The superposition of secondary
Huygens wavelets determines succeeding waves. In diffraction, Huygens wavelets are
held back by obstacles, and the results are the bending of waves around these objects.
An opening appears to be a new wave source, and the smaller the opening the more the
opening looks like a new point source of waves.
16.11 Standing Waves
Standing waves result from the superposition
of two traveling waves of equal frequency
moving in opposite directions (Figure 16.10).
You will notice that at time t2, a time later than
t1, the location of the places where the
amplitude is zero has not changed. This fixed
location of zero amplitude is called a node. The
positions of maximum displacement, called
antinodes, are halfway between adjacent
nodes. The appearance of a wave traveling
along the string has disappeared. So these
waves are called standing waves. The energy
remains localized in the form of kinetic and
potential energies of the antinode regions of
the medium. Standing waves result for a
particular physical system at its resonance
conditions. As you recall, resonance occurs
when a system oscillates at the natural
frequency of the system. Standing wave
resonances occur when the wavelengths of the
exciting frequencies match those allowed by
the physical boundary conditions of the
system. For example, a string fixed at both
ends must have nodes at each end and such a
string has resonant frequencies for the length
of the string to be λ/2, λ, 3λ/2, etc., where λ is
the wavelength. Only certain frequencies of
oscillation can produce standing waves in a
given physical system. These frequencies will
satisfy the relation f = υ/λ, where υ is the
velocity of waves in the medium and λ is the
wavelength which satisfies the boundary
conditions. Fixed boundaries will be nodes
and free ends will be antinodes.
A standing wave represents a normal mode of oscillation for a given system. The
lowest normal mode frequency is called the fundamental frequency. All other normal
mode frequencies are integral multiples of the fundamental frequencies of the system
!
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and are called overtones or harmonics. Examples for a string fixed at both ends are
shown in Figure 16.11. The first harmonic is the fundamental frequency, the first
overtone is the second harmonic, and so on.
Any possible wave form of a system can be regarded as a superposition of normal
modes for the system. The normal mode frequencies are natural frequencies for
resonance of a system. This is the physical basis for the geometrical shapes of musical
instruments such as the sounding cavities for string instruments and organ pipes.
16.12 Fourier's Theorem
Fourier's theorem provides the mathematical basis for building up any wave form by the
superposition of sine waves of specific wavelengths and amplitudes. Any wave form,
no matter how complex, can be expressed as the sum of sine waves. These sine waves
each have different frequencies and are called the Fourier components of the complex
wave. Wave form analysis, providing Fourier components of wave forms such as brain
waves, has become an important medical application of this theorem. Sound spectrums
or voice prints based on Fourier analysis are proving to be as dependable as
fingerprints in identifying different people. The first four Fourier components of a
square wave and their resultants are shown in Figure 16.12 .
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If the change in the wave form is abrupt, the high frequency Fourier components
become important in the synthesis of the wave form. The square wave of Figure 16.12
requires a large number of components for its faithful reproduction.
16.13 Dispersion
If the velocity of a wave in a medium is a function of the frequency of the wave, the
wave demonstrates dispersion. The medium producing this phenomenon is called the
dispersive medium. Air is not a dispersive medium for audible sound-all audible
frequencies travel through air with the same speed. What would you expect to happen
if air were a dispersive medium for sound? Water is a dispersive medium for water
waves-low frequency waves are propagated at higher speeds than high frequency water
waves. Glass is a dispersive medium for light, and this is the explanation of Newton's
famous experiment in which he produced the colored spectrum from sunlight using a
glass prism.
16.14 Waves in Elastic Media
We have seen that an elastic medium is any material that tends to preserve its length,
shape, or volume against external forces. Such a material can be said to have a restoring
force that tends to return the material to its original condition after the external force is
removed. The restoring force is characteristic of the material and arises from the
binding forces between the individual atoms or molecules of the material. As we have
seen in Chapter 13, the strength of the restoring force of a material is characterized by a
force constant, or restoring modulus. The larger the restoring modulus of a material, the
greater is the tendency of the material to resist changes in its length, shape, or volume.
You may recall that Young's modulus, Y, is the measure of a material's tendency to
maintain its length against external forces,
Υ = (F/A)/(ΔL/L)
(13.5)
where F is the magnitude of the external force, A is the area over which the force is
acting, ΔL is the change in length, and L is the original length. In a similar way, we
defined the bulk modulus B, the tendency of an object to maintain its volume,
B = (F/A)/ (ΔV/V) = P/(ΔV/V)
(13.7)
where F is the magnitude of the external force, A is the area over which the force is
acting, ΔV is the change in volume and V is the original volume. Some typical values for
Young's modulus and for the bulk modulus are shown in Table 13.1. The general form
of the equation for the velocity of a wave in an elastic medium is as follows:
velocity = SQR RT [restoring factor/inertial variable]
Specifically, we have the following applications: (all in SI units)
WAVE ON A STRING
υ = SQR RT [T/µ]
where T = tension; µ = mass per unit length
(16.10a)
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COMPRESSIONAL WAVES (SOUND) IN A FLUID
υ = SQR RT [B/ρ]
where B = bulk modulus of fluid; ρ = density of fluid
(16.10b)
COMPRESSIONAL WAVES (SOUND) IN A SOLID
υ = f = SQR RT [Υ/ρ] where Υ = Young's modulus of the solid; ρ = density of solid
(16.10c)
EXAMPLES
1. A string 2.00 m long has a mass of 40.0 g. Find the wavelength of the waves on the
string if the tension is 50.0 N and the frequency is 200 Hz.
υ = fλ
λ = υ / f = SQR RT [T/µ]/f = SQR RT [ 50.0 N / 0.020 kg/m ]/200 Hz
= 50.0 /200 m = 25.0 cm
2. The measurement of the velocity of sound is used to determine the Young's modulus
of the solid. Given the velocity of sound to be 5000 m/sec in a solid with a density =
8.90 g/cm3, find Young's modulus for this solid.
Υ = υ2 ρ = (5000)2 (m/sec)2 (8.90 x 10-3 kg/10-6 m3)
= 22.3 x 1010 N/m2
3. A string is vibrating in its fundamental mode with a length of 0.500 m when under a
tension of 180 N. The linear density of the string is 2.00 x 10-3 kg/m. Find the
fundamental frequency of this string.
fo = υ/λo =υ/2L = SQR RT [T/µ]/2L
= SQR RT [180/(2.00 x 10-3)1/2]/2 x 0.5 = 3.00 x 102 Hz
16.15 Intensity of a Wave Emanating from a Point Source
If you are located a great distance from the source of a wave, no matter how large the
source, the source appears to be small and we can represent it by a point in space. For
example, even the largest of stars in the sky appears as a small point source to us.
Consider a source that emits energy at a constant rate, that is, the energy emitted per
unit of time remains constant. It follows from Equation 16.8 that the intensity of the
wave times the area through which the wave is transported remains a constant. For an
isotropic source, the wave spreads out uniformly in all directions. The area through
which the wave passes as it leaves the source will be the surface area of a sphere which
is centered on the point source and which surrounds the source of the wave. As the
distance the wave travels from the source increases, the area through which the emitted
energy passes increases. In fact, the area through which the energy passes increases as
the square of the distance from the source. We can use this fact to derive the inverse
square law for the intensity of a wave emitted from a point source as follows:
energy emitted = intensity x area x time
E = I A t =I(4πr2)t
where the surface area of a sphere of radius r is 4πr2.
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(16.11)
If the time rate of energy emitted from a small isotropic source is constant (E/t = constant), the
intensity is inversely proportional to the square of the distance from the source. Notice that in
Equations 16.11 and 16.8 the SI units of intensity are joules per second per square meter
or watts per square meter
intensity at point A ∝ 1/r2
(16.12)
where r is the distance from the source to point A.
EXAMPLE
Suppose you are trying to read a road map on a dark sidewalk and the map is 48 meters
from the only nearby street light. How far will you have to move to have the intensity
of the light shining on the map increase by four times?
We can solve this problem using proportional reasoning.
intensity at 48 m ∝ 1/482
intensity at new location = 4x (intensity at 48 m) ∝ 1/r2
Now we divide the first proportional relationship by the second one.
intensity at 48 m / intensity at new location = 1/4 = (1/482) / (1/r2) =r2/482
Taking the square root of both sides, then
1/2 = r/48r = 24 m
You would have to move to a position where the map is 24 m from the street light.
SUMMARY
Use these questions to evaluate how well you have achieved the goals of this chapter.
The answers to these questions are given at the end of the summary with the number of
the section where you can find related content material.
Definitions
1. The ___________ of a wave is its period in space, and its appropriate SI units are
____________ .
2. The ___________ of a wave is its period in time, and its units are _________.
3. The ratio of amplitudes for two waves is 1:4. Find the ratio of intensities of these
waves if their frequencies are the same.
Wave Forms
4. Transverse waves are not possible in fluids because _____________ .
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Problems
5. Given two waves of equal
frequency but 90o out of phase,
find the resultant wave amplitude
if A2 = (½)A.
Superposition
6. Given the waves shown in
Figure 16.13, use the superposition
principle to sketch the resultant
wave.
Standing Waves
7. Compare the fundamental frequencies of a string when both ends are fixed and when
one end is free.
Inverse Square Law
8. To reduce by half the intensity of a sound wave reaching you from a point source,
you have to change your distance from the source by how much?
Answers
1. wavelength, m (Section 16.4)
2. frequency = 1/period, Hertz (cycle/sec) (Section 16.4)
3. I1/I2∝ A12/ A22 = 1/16 (Section 16.6)
4. fluids have no transverse restoring force (Section 16.14)
5. SQR RT [5] A/2 (Section 16.8)
6. see Figure 16.7 (Section 16.8)
7. f (fixed)/f (free) = 2 (Section 16.11)
8. increase the distance by 40 percent (Section 16.15)
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ALGORITHMIC PROBLEMS
Listed below are the important equations from this chapter. The problems following the
equations will help you learn to translate words into equations and to solve single
concept problems.
Equations
υ = λf
(16.2)
y = A sin 2πft
(16.3)
y = A sin (2πft -2πx/λ)
2
2 2
E = (½) kA = 2π f mA
2 3
2
2
(16.4)
(16.5)
P = f E = 2π f mA )
(16.6)
I = P / area = 2π2f2A2υρ
(16.7)
I∝A
2
(16.8)
υ = SQR RT [T/µ]
(16.10a)
υ = SQR RT [B/ρ]
(16.10b)
υ = SQR RT [Υ/ρ]
(16.10c)
I ∝ 1/r2
(16.12)
Problems
1. A reference particle is vibrating with an amplitude of 10 cm and a frequency of 100
Hz. Write the equation for the displacement of the particle as a function of time.
2. If a second particle is 100 cm from the reference particle of problem 1 in a
transmitting medium in which the velocity of propagation is 300 m/sec, write the
equation of displacement of the second particle as a function of time.
3. If the particle in problem 1 has a mass of 1 g, what is the energy per cycle?
4. What is the power transmitted by the wave described in the first three problems?
5. Two similar waves differing only amplitude are being propagated through a
medium. One wave has an amplitude of 6 cm and the other has an amplitude of 10
cm. Compare the intensities of the two waves.
6. What is the speed of a transverse wave in a cord that is 200 cm long and has a mass of
10 g. The tension is 5 x 10-2 N.
7. The speed of sound in water is about 1440 m/sec. What is the coefficient of bulk
modulus (volume of elasticity) of water?
8. The elasticity modulus of a certain metal is found to be 16 x 108 N/m2; its density is
7800 kg/m3. What is the speed of a compressional wave in it?
9. If the solar intensity at the earth is 1.35 x 103 watts/m2, what is the solar intensity at
the planet Mercury, whose distance from the sun is 0.387 times the distance from the
sun to the earth?
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Answers
1. y = 10 sin (200πt) cm
2. y = 10 sin 2π[100t - x/3] cm
3. 2 x 107 ergs = 2 J
4. 2 x 109 ergs/sec = 2 x 102 watts
5. ratio = 36/100 #
6. 3.1 x 102 cm/sec
7. 2.07 x 109 N/m2
8. 450 m/sec
9. 9.01 x 103 watts/m2
EXERCISES
These exercises are designed to help you apply the ideas of a section to physics
situations. When appropriate the numerical answer is given in brackets at the end of the
exercise.
Section 16.5
1. Given y1 = 2 sin (0.628x - 314 t)m,
a. find the amplitude, frequency, and wavelength of the wave
b. determine whether the wave is transverse or longitudinal
c. find the speed of the wave
[a. 2 m, 50 Hz, 10 cm, b. transverse; c. 500 m/sec]
2. A piano string emits a sound with a frequency of 341 Hz. The velocity of sound
through air at this particular time is 340 m/sec. What is the wavelength of the sound
wave in air? If the observer is 4.00 m from the source, what is the phase difference
between the source and the wave at the observer? [1 m, 8π]
3. A string is attached to a vibrating source and the displacement of the source is given
by y = 2 sin 60π t cm. The velocity of wave in the string is 100 cm/sec.
a. Sketch the form of the string for the first 10 cm at time t = 0.5 sec.
b. Sketch the displacement of a particle 5 cm from the source for 0.1 sec. after the
wave reaches the particle.
4. A traveling transverse wave is described by the equation y = 2 sin (512 π t + π x/20)
cm.
a. What are the amplitude, frequency, and wavelength of the wave?
b. What is the velocity of the wave?
c. What are the displacement and velocity of a particle at x = 30 cm and t = 0.4 sec?
[a. 2 cm, 256 Hz, 40 cm; b.-102 m/sec; c. y = 2 sin 54o = 1.6 cm; velocity = 18.4 m/sec]
5. A sinusoidal wave travels along a string. If the time required for a given particle to
travel from maximum displacement to 0 displacement is 0.10 sec, what are the
period and the frequency? If the wavelength is 1.2 m, what is the velocity of
propagation? [0.4 sec, 2.5 Hz, 3 m/sec]
Physics Including Human Applications
351
6. Given the wave in Figure 16.14 with a
speed of 10m/sec, sketch the wave 0.15
sec after the figure shown.
Section 16.6
7. Find the ratio of the intensities of two compressional waves of equal amplitude
traveling in a gas if one has a frequency of 500 Hz and the other 1000 Hz. [I (500
Hz)/I (1000 Hz) = 1/4]
8. If two waves of the same frequency are found to have an intensity ratio of I1/I2 =0.16
in air, find the ratio of their amplitudes. [A1/A2= 0.4]
Section 16.8
9. Suppose a wave given by y2 = 1 sin (0.628 x -314t + 90o) cm is traveling in the same
medium as wave y1 from exercise 1. Find the resultant wave using the superposition
principle. [y = SQR RT [5] sin (0.628x - 314t + 26.50¯)]
10. Two triangular waves are traveling on a
string as shown at t = 0 sec in Figure
16.15; the speed of each is 0.5 m/sec.
Sketch the shape of the string at t = 1.0
sec, 2.0, and 3.0 sec.
Section 16.11
11. The frequency of the fundamental tone of
a violin string is 256 Hz. What are
frequencies of second and third
harmonics? [512 Hz, 768 Hz]
Section 16.14
12. Find the tension necessary to make the fundamental frequency of a 0.5 m string (ρ =
2 x 10-3 kg/m) equal to 150 Hz. [45 N]
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352
Section 16.15
13. An ultraviolet lamp is designed to provide a typical fair- skinned person with a mild
sunburn in 10 minutes if the person is 3.0 m from the lamp. How long should the
person stay under the lamp when it is at a distance of 5.0 m? [28 minutes]
PROBLEMS
Each of the following problems may involve more than one physical concept.
Numerical answers are given in brackets at the end of the problem.
14. The relative index of refraction of two media is defined as ratio of the wave
velocities in the media and is related to the angles of incidence and refraction by
Snell's law:
Snell's law:
n = v1/v2 = sin i / sin r
where i = angle of incidence and r = angle of defraction (see Figure 16.16). If the
velocity in medium 2 is three-fourths that in medium 1 and the angle of incidence is
30o, what is the angle of refraction? [sinr = 3/8, r = 22 o]
15. Trace a wave through the system shown
in See Figure 16.16 if the index of
refraction is 3/2. (See problem 14.) If
medium 2 is 10 cm thick, what is the
lateral displacement of the ray in passing
through medium 2? [1.94 cm]
16. A string fastened at both ends is resonant
with wavelengths of 0.16 m and 0.20 m.
What is the minimum possible string
length? [0.4 m]
17. A wire is vibrating in its first overtone
mode with a frequency of 250 Hz. If the
tension on the wire is 250 N and its linear
density is 10-3 kg/m, find the length of the
wire. [2.0 m]
18. A stretched string 75 cm long is vibrating in its third overtone at a frequency of 300
Hz. Find the velocity of the traveling waves that produce this standing wave by
superposition. [113 m/sec]