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11.4: Wave Speed

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Assume that you move one end of a rope up and down just once to generate a wave in the rope. How long will take the wave to travel down the rope to the other end? It depends on the speed of the wave.

The Speed of a Wave

Wave speed is the distance a wave travels in a given amount of time, such as the number of meters it travels per second. Wave speed (and speed in general) can be represented by the equation:

Wave Speed, Wavelength, and Wave Frequency

Wave speed is related to both wavelength and wave frequency. Wavelength is the distance between two corresponding points on adjacent waves. Wave frequency is the number of waves that pass a fixed point in a given amount of time. This equation shows how the three factors are related:

In this equation, wavelength is measured in meters and frequency is measured in hertz (Hz), or number of waves per second. Therefore, wave speed is given in meters per second, which is the SI unit for speed.

Q: If you increase the wavelength of a wave, does the speed of the wave increase as well?

A: Increasing the wavelength of a wave doesn’t change its speed. That’s because when wavelength increases, wave frequency decreases. As a result, the product of wavelength and wave frequency is still the same speed.

Calculating Wave Speed from Wavelength and Wave Frequency

The equation for wave speed can be used to calculate the speed of a wave when both wavelength and wave frequency are known. Consider an ocean wave with a wavelength of 3 meters and a frequency of 1 hertz. The speed of the wave is:

Q: Kim made a wave in a spring by pushing and pulling on one end. The wavelength is 0.1 m, and the wave frequency is 2 hertz. What is the speed of the wave?

A: Substitute these values into the equation for speed:

Calculating Wave Frequency or Wavelength from Wave Speed

The equation for wave speed (above) can be rewritten as:

Therefore, if you know the speed of a wave and either the wavelength or wave frequency, you can calculate the missing value. For example, suppose that a wave is traveling at a speed of 2 meters per second and has a wavelength of 1 meter. Then the frequency of the wave is:

Q: A wave is traveling at a speed of 2 m/s and has a frequency of 2 Hz. What is its wavelength?

A: Substitute these values into the equation for wavelength:

The Medium Matters

The speed of most waves depends on the medium, or the matter through which the waves are traveling. Generally, waves travel fastest through solids and slowest through gases. That’s because particles are closest together in solids and farthest apart in gases. When particles are farther apart, it takes longer for the energy of the disturbance to pass from particle to particle through the medium.

Launch the Doppler Ducks simulation below to visualize how wave speed is related to both wavelength and wave frequency. Adjust the duck velocity slider to change how quickly the duck moves through the water. Remember that positive velocities are rightward and negative velocities are leftward. Try playing around with having the duck go faster or slower than the wave speed, or towards or away from the boat.

Interactive Element

• Wave speed is the distance a wave travels in a given amount of time, such as the number of meters it travels per second.
• Wave speed is related to wavelength and wave frequency by the equation: Speed = Wavelength x Frequency. This equation can be used to calculate wave speed when wavelength and frequency are known.
• The equation for wave speed can be written to solve for wavelength or frequency if the speed and the other value are known.
• The speed of most waves depends on the medium, or the matter through which they are traveling. Generally, waves travel fastest through solids and slowest through gases.
• What is wave speed?
• What is the speed of a wave that has a wavelength of 2 m and a frequency of 1.5 Hz?
• Calculate the frequency of a wave that is traveling at a speed of 3.0 m/s and has a wavelength of 1.2 m.
• Sound energy travels through matter in waves. Do sound waves travel faster through air or water? Explain your answer.

Additional Resources

Study Guide: Waves Study Guide

Video: Speed of Mechanical Waves - Overview

Real World Application: Seismic Shakeup

• Physics Formulas

Wave Speed Formula

A wave is defined as a kind of disturbance in a moving medium, for example, the waves of the ocean move in a medium and can see the movement of wave crest from one point to the other in a given time period. The motion of an object can be described regarding the speed which describes the velocity of the object.

A wave is visualized when a source vibrates and disturbs a particle in the medium. This can usually be seen in the case of tuning fork or ripples in water when a body is dropped, etc.

The number of waves traveled in one second is the frequency and the time period is the reciprocal of the frequency of the waves. Wavelength is the distance between the corresponding points in any two consecutive waves.

Formula of Wave Speed

The Wave speed formula which involves wavelength and frequency is given by:

• v is the velocity of the wave
• f is the frequency of the wave
• λ  is the wavelength

Wave Speed Examples

A light wave travels with a wavelength of 600 nm. Determine its frequency.

Wavelength λ = 600 nm,

The frequency is calculated by,

A sound wave has a wavelength of 1.5 mm. Determine its frequency.

Wavelength λ = 1.5 mm,

Velocity of sound v = 343.2 m/s.

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14.1 Speed of Sound, Frequency, and Wavelength

Section learning objectives.

By the end of this section, you will be able to do the following:

• Relate the characteristics of waves to properties of sound waves
• Describe the speed of sound and how it changes in various media
• Relate the speed of sound to frequency and wavelength of a sound wave

Teacher Support

The learning objectives in this section will help your students master the following standards:

• (A) examine and describe oscillatory motion and wave propagation in various types of media;
• (B) investigate and analyze characteristics of waves, including velocity, frequency, amplitude, and wavelength, and calculate using the relationship between wave speed, frequency, and wavelength;
• (C) compare characteristics and behaviors of transverse waves, including electromagnetic waves and the electromagnetic spectrum, and characteristics and behaviors of longitudinal waves, including sound waves;
• (F) describe the role of wave characteristics and behaviors in medical and industrial applications.

In addition, the High School Physics Laboratory Manual addresses content in this section in the lab titled: Waves, as well as the following standards:

• (B) investigate and analyze characteristics of waves, including velocity, frequency, amplitude, and wavelength, and calculate using the relationship between wave speed, frequency, and wavelength.

Section Key Terms

[BL] [OL] Review waves and types of waves—mechanical and non-mechanical, transverse and longitudinal, pulse and periodic. Review properties of waves—amplitude, period, frequency, velocity and their inter-relations.

Properties of Sound Waves

Sound is a wave. More specifically, sound is defined to be a disturbance of matter that is transmitted from its source outward. A disturbance is anything that is moved from its state of equilibrium. Some sound waves can be characterized as periodic waves, which means that the atoms that make up the matter experience simple harmonic motion .

A vibrating string produces a sound wave as illustrated in Figure 14.2 , Figure 14.3 , and Figure 14.4 . As the string oscillates back and forth, part of the string’s energy goes into compressing and expanding the surrounding air. This creates slightly higher and lower pressures. The higher pressure... regions are compressions, and the low pressure regions are rarefactions . The pressure disturbance moves through the air as longitudinal waves with the same frequency as the string. Some of the energy is lost in the form of thermal energy transferred to the air. You may recall from the chapter on waves that areas of compression and rarefaction in longitudinal waves (such as sound) are analogous to crests and troughs in transverse waves .

The amplitude of a sound wave decreases with distance from its source, because the energy of the wave is spread over a larger and larger area. But some of the energy is also absorbed by objects, such as the eardrum in Figure 14.5 , and some of the energy is converted to thermal energy in the air. Figure 14.4 shows a graph of gauge pressure versus distance from the vibrating string. From this figure, you can see that the compression of a longitudinal wave is analogous to the peak of a transverse wave, and the rarefaction of a longitudinal wave is analogous to the trough of a transverse wave. Just as a transverse wave alternates between peaks and troughs, a longitudinal wave alternates between compression and rarefaction.

The Speed of Sound

[BL] Review the fact that sound is a mechanical wave and requires a medium through which it is transmitted.

[OL] [AL] Ask students if they know the speed of sound and if not, ask them to take a guess. Ask them why the sound of thunder is heard much after the lightning is seen during storms. This phenomenon is also observed during a display of fireworks. Through this discussion, develop the concept that the speed of sound is finite and measurable and is much slower than that of light.

The speed of sound varies greatly depending upon the medium it is traveling through. The speed of sound in a medium is determined by a combination of the medium’s rigidity (or compressibility in gases) and its density. The more rigid (or less compressible) the medium, the faster the speed of sound. The greater the density of a medium, the slower the speed of sound. The speed of sound in air is low, because air is compressible. Because liquids and solids are relatively rigid and very difficult to compress, the speed of sound in such media is generally greater than in gases. Table 14.1 shows the speed of sound in various media. Since temperature affects density, the speed of sound varies with the temperature of the medium through which it’s traveling to some extent, especially for gases.

Misconception Alert

Students might be confused between rigidity and density and how they affect the speed of sound. The speed of sound is slower in denser media. Solids are denser than gases. However, they are also very rigid, and hence sound travels faster in solids. Stress on the fact that the speed of sound always depends on a combination of these two properties of any medium.

[BL] Note that in the table, the speed of sound in very rigid materials such as glass, aluminum, and steel ... is quite high, whereas the speed in rubber, which is considerably less rigid, is quite low.

The Relationship Between the Speed of Sound and the Frequency and Wavelength of a Sound Wave

Sound, like all waves, travels at certain speeds through different media and has the properties of frequency and wavelength . Sound travels much slower than light—you can observe this while watching a fireworks display (see Figure 14.6 ), since the flash of an explosion is seen before its sound is heard.

The relationship between the speed of sound, its frequency, and wavelength is the same as for all waves:

where v is the speed of sound (in units of m/s), f is its frequency (in units of hertz), and λ λ is its wavelength (in units of meters). Recall that wavelength is defined as the distance between adjacent identical parts of a wave. The wavelength of a sound, therefore, is the distance between adjacent identical parts of a sound wave. Just as the distance between adjacent crests in a transverse wave is one wavelength, the distance between adjacent compressions in a sound wave is also one wavelength, as shown in Figure 14.7 . The frequency of a sound wave is the same as that of the source. For example, a tuning fork vibrating at a given frequency would produce sound waves that oscillate at that same frequency. The frequency of a sound is the number of waves that pass a point per unit time.

[BL] [OL] [AL] In musical instruments, shorter strings vibrate faster and hence produce sounds at higher pitches. Fret placements on instruments such as guitars, banjos, and mandolins, are mathematically determined to give the correct interval or change in pitch. When the string is pushed against the fret wire, the string is effectively shortened, changing its pitch. Ask students to experiment with strings of different lengths and observe how the pitch changes in each case.

One of the more important properties of sound is that its speed is nearly independent of frequency. If this were not the case, and high-frequency sounds traveled faster, for example, then the farther you were from a band in a football stadium, the more the sound from the low-pitch instruments would lag behind the high-pitch ones. But the music from all instruments arrives in cadence independent of distance, and so all frequencies must travel at nearly the same speed.

Recall that v = f λ v = f λ , and in a given medium under fixed temperature and humidity, v is constant. Therefore, the relationship between f and λ λ is inverse: The higher the frequency, the shorter the wavelength of a sound wave.

Teacher Demonstration

Hold a meter stick flat on a desktop, with about 80 cm sticking out over the edge of the desk. Make the meter stick vibrate by pulling the tip down and releasing, while holding the meter stick tight to the desktop. While it is vibrating, move the stick back onto the desktop, shortening the part that is sticking out. Students will see the shortening of the vibrating part of the meter stick, and hear the pitch or number of vibrations go up—an increase in frequency.

The speed of sound can change when sound travels from one medium to another. However, the frequency usually remains the same because it is like a driven oscillation and maintains the frequency of the original source. If v changes and f remains the same, then the wavelength λ λ must change. Since v = f λ v = f λ , the higher the speed of a sound, the greater its wavelength for a given frequency.

[AL] Ask students to predict what would happen if the speeds of sound in air varied by frequency.

Virtual Physics

This simulation lets you see sound waves. Adjust the frequency or amplitude (volume) and you can see and hear how the wave changes. Move the listener around and hear what she hears. Switch to the Two Source Interference tab or the Interference by Reflection tab to experiment with interference and reflection.

Tips For Success

Make sure to have audio enabled and set to Listener rather than Speaker, or else the sound will not vary as you move the listener around.

• Because, intensity of the sound wave changes with the frequency.
• Because, the speed of the sound wave changes when the frequency is changed.
• Because, loudness of the sound wave takes time to adjust after a change in frequency.
• Because it takes time for sound to reach the listener, so the listener perceives the new frequency of sound wave after a delay.
• Yes, the speed of propagation depends only on the frequency of the wave.
• Yes, the speed of propagation depends upon the wavelength of the wave, and wavelength changes as the frequency changes.
• No, the speed of propagation depends only on the wavelength of the wave.
• No, the speed of propagation is constant in a given medium; only the wavelength changes as the frequency changes.

Voice as a Sound Wave

In this lab you will observe the effects of blowing and speaking into a piece of paper in order to compare and contrast different sound waves.

• sheet of paper

Instructions

• Suspend a sheet of paper so that the top edge of the paper is fixed and the bottom edge is free to move. You could tape the top edge of the paper to the edge of a table, for example.
• Gently blow air near the edge of the bottom of the sheet and note how the sheet moves.
• Speak softly and then louder such that the sounds hit the edge of the bottom of the paper, and note how the sheet moves.
• Interpret the results.

Grasp Check

Which sound wave property increases when you are speaking more loudly than softly?

• amplitude of the wave
• frequency of the wave
• speed of the wave
• wavelength of the wave

Worked Example

What are the wavelengths of audible sounds.

Calculate the wavelengths of sounds at the extremes of the audible range, 20 and 20,000 Hz, in conditions where sound travels at 348.7 m/s.

To find wavelength from frequency, we can use v = f λ v = f λ .

(1) Identify the knowns. The values for v and f are given.

(2) Solve the relationship between speed, frequency and wavelength for λ λ .

(3) Enter the speed and the minimum frequency to give the maximum wavelength.

(4) Enter the speed and the maximum frequency to give the minimum wavelength.

Because the product of f multiplied by λ λ equals a constant velocity in unchanging conditions, the smaller f is, the larger λ λ must be, and vice versa. Note that you can also easily rearrange the same formula to find frequency or velocity.

Practice Problems

• 5 × 10 3 m / s
• 3.2 × 10 2 m / s
• 2 × 10 − 4 m/s
• 8 × 10 2 m / s
• 2.0 × 10 7 m
• 1.5 × 10 7 m
• 1.4 × 10 2 m
• 7.4 × 10 − 3 m

Links To Physics

Echolocation.

Echolocation is the use of reflected sound waves to locate and identify objects. It is used by animals such as bats, dolphins and whales, and is also imitated by humans in SONAR—Sound Navigation and Ranging—and echolocation technology.

Bats, dolphins and whales use echolocation to navigate and find food in their environment. They locate an object (or obstacle) by emitting a sound and then sensing the reflected sound waves. Since the speed of sound in air is constant, the time it takes for the sound to travel to the object and back gives the animal a sense of the distance between itself and the object. This is called ranging . Figure 14.8 shows a bat using echolocation to sense distances.

Echolocating animals identify an object by comparing the relative intensity of the sound waves returning to each ear to figure out the angle at which the sound waves were reflected. This gives information about the direction, size and shape of the object. Since there is a slight distance in position between the two ears of an animal, the sound may return to one of the ears with a bit of a delay, which also provides information about the position of the object. For example, if a bear is directly to the right of a bat, the echo will return to the bat’s left ear later than to its right ear. If, however, the bear is directly ahead of the bat, the echo would return to both ears at the same time. For an animal without a sense of sight such as a bat, it is important to know where other animals are as well as what they are; their survival depends on it.

Principles of echolocation have been used to develop a variety of useful sensing technologies. SONAR, is used by submarines to detect objects underwater and measure water depth. Unlike animal echolocation, which relies on only one transmitter (a mouth) and two receivers (ears), manmade SONAR uses many transmitters and beams to get a more accurate reading of the environment. Radar technologies use the echo of radio waves to locate clouds and storm systems in weather forecasting, and to locate aircraft for air traffic control. Some new cars use echolocation technology to sense obstacles around the car, and warn the driver who may be about to hit something (or even to automatically parallel park). Echolocation technologies and training systems are being developed to help visually impaired people navigate their everyday environments.

• The echo would return to the left ear first.
• The echo would return to the right ear first.

Check Your Understanding

Use these questions to assess student achievement of the section’s Learning Objectives. If students are struggling with a specific objective, these questions will help identify which and direct students to the relevant content.

• Rarefaction is the high-pressure region created in a medium when a longitudinal wave passes through it.
• Rarefaction is the low-pressure region created in a medium when a longitudinal wave passes through it.
• Rarefaction is the highest point of amplitude of a sound wave.
• Rarefaction is the lowest point of amplitude of a sound wave.

What sort of motion do the particles of a medium experience when a sound wave passes through it?

• Simple harmonic motion
• Circular motion
• Random motion
• Translational motion

What does the speed of sound depend on?

• The wavelength of the wave
• The size of the medium
• The frequency of the wave
• The properties of the medium

What property of a gas would affect the speed of sound traveling through it?

• The volume of the gas
• The flammability of the gas
• The mass of the gas
• The compressibility of the gas

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13.2 Wave Properties: Speed, Amplitude, Frequency, and Period

Learning objectives.

By the end of this section, you will be able to do the following:

• Define amplitude, frequency, period, wavelength, and velocity of a wave
• Relate wave frequency, period, wavelength, and velocity
• Solve problems involving wave properties

Wave Variables

In the chapter on motion in two dimensions, we defined the following variables to describe harmonic motion:

• Amplitude—maximum displacement from the equilibrium position of an object oscillating around such equilibrium position
• Frequency—number of events per unit of time
• Period—time it takes to complete one oscillation

For waves, these variables have the same basic meaning. However, it is helpful to word the definitions in a more specific way that applies directly to waves:

• Amplitude—distance between the resting position and the maximum displacement of the wave
• Frequency—number of waves passing by a specific point per second
• Period—time it takes for one wave cycle to complete

In addition to amplitude, frequency, and period, their wavelength and wave velocity also characterize waves. The wavelength λ λ is the distance between adjacent identical parts of a wave, parallel to the direction of propagation. The wave velocity v w v w is the speed at which the disturbance moves.

Tips For Success

Wave velocity is sometimes also called the propagation velocity or propagation speed because the disturbance propagates from one location to another.

Consider the periodic water wave in Figure 13.7 . Its wavelength is the distance from crest to crest or from trough to trough. The wavelength can also be thought of as the distance a wave has traveled after one complete cycle—or one period. The time for one complete up-and-down motion is the simple water wave’s period T . In the figure, the wave itself moves to the right with a wave velocity v w . Its amplitude X is the distance between the resting position and the maximum displacement—either the crest or the trough—of the wave. It is important to note that this movement of the wave is actually the disturbance moving to the right, not the water itself; otherwise, the bird would move to the right. Instead, the seagull bobs up and down in place as waves pass underneath, traveling a total distance of 2 X in one cycle. However, as mentioned in the text feature on surfing, actual ocean waves are more complex than this simplified example.

Watch Physics

Amplitude, period, frequency, and wavelength of periodic waves.

This video is a continuation of the video “Introduction to Waves” from the "Types of Waves" section. It discusses the properties of a periodic wave: amplitude, period, frequency, wavelength, and wave velocity.

The crest of a wave is sometimes also called the peak .

The Relationship between Wave Frequency, Period, Wavelength, and Velocity

Since wave frequency is the number of waves per second, and the period is essentially the number of seconds per wave, the relationship between frequency and period is

just as in the case of harmonic motion of an object. We can see from this relationship that a higher frequency means a shorter period. Recall that the unit for frequency is hertz (Hz), and that 1 Hz is one cycle—or one wave—per second.

The speed of propagation v w is the distance the wave travels in a given time, which is one wavelength in a time of one period. In equation form, it is written as

From this relationship, we see that in a medium where v w is constant, the higher the frequency, the smaller the wavelength. See Figure 13.8 .

These fundamental relationships hold true for all types of waves. As an example, for water waves, v w is the speed of a surface wave; for sound, v w is the speed of sound; and for visible light, v w is the speed of light. The amplitude X is completely independent of the speed of propagation v w and depends only on the amount of energy in the wave.

Waves in a Bowl

In this lab, you will take measurements to determine how the amplitude and the period of waves are affected by the transfer of energy from a cork dropped into the water. The cork initially has some potential energy when it is held above the water—the greater the height, the higher the potential energy. When it is dropped, such potential energy is converted to kinetic energy as the cork falls. When the cork hits the water, that energy travels through the water in waves.

• Large bowl or basin
• Cork (or ping pong ball)
• Measuring tape

Instructions

• Fill a large bowl or basin with water and wait for the water to settle so there are no ripples.
• Gently drop a cork into the middle of the bowl.
• Estimate the wavelength and the period of oscillation of the water wave that propagates away from the cork. You can estimate the period by counting the number of ripples from the center to the edge of the bowl while your partner times it. This information, combined with the bowl measurement, will give you the wavelength when the correct formula is used.
• Remove the cork from the bowl and wait for the water to settle again.
• Gently drop the cork at a height that is different from the first drop.
• Repeat Steps 3 to 5 to collect a second and third set of data, dropping the cork from different heights and recording the resulting wavelengths and periods.
• Interpret your results.
• No, only the amplitude is affected.
• Yes, the wavelength is affected.

Links To Physics

Geology: physics of seismic waves.

Geologists rely heavily on physics to study earthquakes since earthquakes involve several types of wave disturbances, including disturbance of Earth’s surface and pressure disturbances under the surface. Surface earthquake waves are similar to surface waves on water. The waves under Earth’s surface have both longitudinal and transverse components. The longitudinal waves in an earthquake are called pressure waves (P-waves) and the transverse waves are called shear waves (S-waves). These two types of waves propagate at different speeds, and the speed at which they travel depends on the rigidity of the medium through which they are traveling. During earthquakes, the speed of P-waves in granite is significantly higher than the speed of S-waves. Both components of earthquakes travel more slowly in less rigid materials, such as sediments. P-waves have speeds of 4 to 7 km/s, and S-waves have speeds of 2 to 5 km/s, but both are faster in more rigid materials. The P-wave gets progressively farther ahead of the S-wave as they travel through Earth’s crust. For that reason, the time difference between the P- and S-waves is used to determine the distance to their source, the epicenter of the earthquake.

We know from seismic waves produced by earthquakes that parts of the interior of Earth are liquid. Shear or transverse waves cannot travel through a liquid and are not transmitted through Earth’s core. In contrast, compression or longitudinal waves can pass through a liquid and they do go through the core.

All waves carry energy, and the energy of earthquake waves is easy to observe based on the amount of damage left behind after the ground has stopped moving. Earthquakes can shake whole cities to the ground, performing the work of thousands of wrecking balls. The amount of energy in a wave is related to its amplitude. Large-amplitude earthquakes produce large ground displacements and greater damage. As earthquake waves spread out, their amplitude decreases, so there is less damage the farther they get from the source.

What is the relationship between the propagation speed, frequency, and wavelength of the S-waves in an earthquake?

• The relationship between the propagation speed, frequency, and wavelength is v w = f λ . v w = f λ .
• The relationship between the propagation speed, frequency, and wavelength is v w = λ f . v w = λ f .

Virtual Physics

Wave on a string.

In this animation, watch how a string vibrates in slow motion by choosing the Slow Motion setting. Select the No End and Manual options, and wiggle the end of the string to make waves yourself. Then switch to the Oscillate setting to generate waves automatically. Adjust the frequency and the amplitude of the oscillations to see what happens. Then experiment with adjusting the damping and the tension.

Which of the settings—amplitude, frequency, damping, or tension—changes the amplitude of the wave as it propagates? What does it do to the amplitude?

• Frequency; it decreases the amplitude of the wave as it propagates.
• Frequency; it increases the amplitude of the wave as it propagates.
• Damping; it decreases the amplitude of the wave as it propagates.
• Damping; it increases the amplitude of the wave as it propagates.

Solving Wave Problems

Worked example, calculate the velocity of wave propagation: gull in the ocean.

Calculate the wave velocity of the ocean wave in the previous figure if the distance between wave crests is 10.0 m and the time for a seagull to bob up and down is 5.00 s.

The values for the wavelength ( λ = 10.0   m ) ( λ = 10.0   m ) and the period ( T = 5.00 s) ( T = 5.00 s) are given and we are asked to find v w v w Therefore, we can use v w = λ T v w = λ T to find the wave velocity.

Enter the known values into v w = λ T v w = λ T

This slow speed seems reasonable for an ocean wave. Note that in the figure, the wave moves to the right at this speed, which is different from the varying speed at which the seagull bobs up and down.

Calculate the Period and the Wave Velocity of a Toy Spring

The woman in Figure 13.3 creates two waves every second by shaking the toy spring up and down. (a)What is the period of each wave? (b) If each wave travels 0.9 meters after one complete wave cycle, what is the velocity of wave propagation?

STRATEGY FOR (A)

To find the period, we solve for T = 1 f T = 1 f , given the value of the frequency ( f = 2 s − 1 ). ( f = 2 s − 1 ).

Enter the known value into T = 1 f T = 1 f

STRATEGY FOR (B)

Since one definition of wavelength is the distance a wave has traveled after one complete cycle—or one period—the values for the wavelength ( λ = 0.9   m ) ( λ = 0.9   m ) as well as the frequency are given. Therefore, we can use v w = f λ v w = f λ to find the wave velocity.

Enter the known values into v w = f λ v w = f λ

v w = f λ = ( 2  s −1 )(0 .9 m) = 1 .8 m/s . v w = f λ = ( 2  s −1 )(0 .9 m) = 1 .8 m/s .

We could have also used the equation v w = λ T v w = λ T to solve for the wave velocity since we already know the value of the period ( T = 0.5 s) ( T = 0.5 s) from our calculation in part (a), and we would come up with the same answer.

Practice Problems

The frequency of a wave is 10 Hz. What is its period?

• The period of the wave is 100 s.
• The period of the wave is 10 s.
• The period of the wave is 0.01 s.
• The period of the wave is 0.1 s.

What is the velocity of a wave whose wavelength is 2 m and whose frequency is 5 Hz?

Check Your Understanding

What is the amplitude of a wave?

• A quarter of the total height of the wave
• Half of the total height of the wave
• Two times the total height of the wave
• Four times the total height of the wave
• The wavelength is the distance between adjacent identical parts of a wave, parallel to the direction of propagation.
• The wavelength is the distance between adjacent identical parts of a wave, perpendicular to the direction of propagation.
• The wavelength is the distance between a crest and the adjacent trough of a wave, parallel to the direction of propagation.
• The wavelength is the distance between a crest and the adjacent trough of a wave, perpendicular to the direction of propagation.
• f = ( 1 T ) 2
• f = ( T ) 2

When is the wavelength directly proportional to the period of a wave?

• When the velocity of the wave is halved
• When the velocity of the wave is constant
• When the velocity of the wave is doubled
• When the velocity of the wave is tripled

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24: Electromagnetic Waves (Exercises)

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Conceptual Questions

24.2: production of electromagnetic waves.

1. The direction of the electric field shown in each part of Figure 1 is that produced by the charge distribution in the wire. Justify the direction shown in each part, using the Coulomb force law and the definition of \(\displaystyle E = F/q\), where \(\displaystyle q\) is a positive test charge.

2. Is the direction of the magnetic field shown in Figure 2a consistent with the right-hand rule for current (RHR-2) in the direction shown in the figure?

3. Why is the direction of the current shown in each part of Figure 2 opposite to the electric field produced by the wire’s charge separation?

4. In which situation shown in the figure will the electromagnetic wave be more successful in inducing a current in the wire? Explain.

Electromagnetic waves approaching long straight wires.

5. In which situation shown in the figure will the electromagnetic wave be more successful in inducing a current in the loop? Explain.

Electromagnetic waves approaching a wire loop.

6. Should the straight wire antenna of a radio be vertical or horizontal to best receive radio waves broadcast by a vertical transmitter antenna? How should a loop antenna be aligned to best receive the signals? (Note that the direction of the loop that produces the best reception can be used to determine the location of the source. It is used for that purpose in tracking tagged animals in nature studies, for example.)

7. Under what conditions might wires in a DC circuit emit electromagnetic waves?

8. Give an example of interference of electromagnetic waves.

9. The figure shows the interference pattern of two radio antennas broadcasting the same signal. Explain how this is analogous to the interference pattern for sound produced by two speakers. Could this be used to make a directional antenna system that broadcasts preferentially in certain directions? Explain.

An overhead view of two radio broadcast antennas sending the same signal, and the interference pattern they produce.

10. Can an antenna be any length? Explain your answer.

24.3: The Electromagnetic Spectrum

11. If you live in a region that has a particular TV station, you can sometimes pick up some of its audio portion on your FM radio receiver. Explain how this is possible. Does it imply that TV audio is broadcast as FM?

12. Explain why people who have the lens of their eye removed because of cataracts are able to see low-frequency ultraviolet.

13. How do fluorescent soap residues make clothing look “brighter and whiter” in outdoor light? Would this be effective in candlelight?

14. Give an example of resonance in the reception of electromagnetic waves.

15. Illustrate that the size of details of an object that can be detected with electromagnetic waves is related to their wavelength, by comparing details observable with two different types (for example, radar and visible light or infrared and X-rays).

16. Why don’t buildings block radio waves as completely as they do visible light?

17. Make a list of some everyday objects and decide whether they are transparent or opaque to each of the types of electromagnetic waves.

18. Your friend says that more patterns and colors can be seen on the wings of birds if viewed in ultraviolet light. Would you agree with your friend? Explain your answer.

19. The rate at which information can be transmitted on an electromagnetic wave is proportional to the frequency of the wave. Is this consistent with the fact that laser telephone transmission at visible frequencies carries far more conversations per optical fiber than conventional electronic transmission in a wire? What is the implication for ELF radio communication with submarines?

20. Give an example of energy carried by an electromagnetic wave.

21. In an MRI scan, a higher magnetic field requires higher frequency radio waves to resonate with the nuclear type whose density and location is being imaged. What effect does going to a larger magnetic field have on the most efficient antenna to broadcast those radio waves? Does it favor a smaller or larger antenna?

22. Laser vision correction often uses an excimer laser that produces 193-nm electromagnetic radiation. This wavelength is extremely strongly absorbed by the cornea and ablates it in a manner that reshapes the cornea to correct vision defects. Explain how the strong absorption helps concentrate the energy in a thin layer and thus give greater accuracy in shaping the cornea. Also explain how this strong absorption limits damage to the lens and retina of the eye.

Problems & Exercises

24.1: maxwell’s equations: electromagnetic waves predicted and observed.

23. Verify that the correct value for the speed of light c is obtained when numerical values for the permeability and permittivity of free space (\(\displaystyle mu_{0}\) and \(\displaystyle \epsilon_{0}\)) are entered into the equation \(\displaystyle c = \frac{1}{\mu_{0}\epsilon_{0}}\).

24. Show that, when SI units for \(\displaystyle \mu_{0}\) and \(\displaystyle \epsilon_{0}\) are entered, the units given by the right-hand side of the equation in the problem above are m/s.

25. What is the maximum electric field strength in an electromagnetic wave that has a maximum magnetic field strength of \(\displaystyle 5.00 \times 10^{-4} T\) (about 10 times the Earth’s)?

Solution 150 kV/m

26. The maximum magnetic field strength of an electromagnetic field is \(\displaystyle 5 \times 10^{-6} T\). Calculate the maximum electric field strength if the wave is traveling in a medium in which the speed of the wave is \(\displaystyle 0.75c\).

27. Verify the units obtained for magnetic field strength \(\displaystyle B\) in the in-chapter example (using the equation \(\displaystyle B = \frac{E}{c}\)) are in fact teslas (T).

28. (a) Two microwave frequencies are authorized for use in microwave ovens: 900 and 2560 MHz. Calculate the wavelength of each.

(b) Which frequency would produce smaller hot spots in foods due to interference effects?

Solution (a) 33.3 cm (900 MHz) 11.7 cm (2560 MHz) (b) The microwave oven with the smaller wavelength would produce smaller hot spots in foods, corresponding to the one with the frequency 2560 MHz.

29. (a) Calculate the range of wavelengths for AM radio given its frequency range is 540 to 1600 kHz.

(b) Do the same for the FM frequency range of 88.0 to 108 MHz.

30. A radio station utilizes frequencies between commercial AM and FM. What is the frequency of a 11.12-m-wavelength channel?

Solution 26.96 MHz

31. Find the frequency range of visible light, given that it encompasses wavelengths from 380 to 760 nm.

32. Combing your hair leads to excess electrons on the comb. How fast would you have to move the comb up and down to produce red light?

Solution \(\displaystyle 5.0 \times 10^{14}\) Hz

33. Electromagnetic radiation having a \(\displaystyle 15.0 - \mu m\) wavelength is classified as infrared radiation. What is its frequency?

34. Approximately what is the smallest detail observable with a microscope that uses ultraviolet light of frequency \(\displaystyle 1.20 \times 10^{15} Hz\)?

Solution \(\displaystyle \lambda = \frac{c}{f} = \frac{3.00 \times 10^{8} m/s}{1.20 \times 10^{15} Hz} = 2.50 \times 10^{-7} m\)

35. A radar used to detect the presence of aircraft receives a pulse that has reflected off an object \(\displaystyle 6 \times 10^{-5} s\) after it was transmitted. What is the distance from the radar station to the reflecting object?

36. Some radar systems detect the size and shape of objects such as aircraft and geological terrain. Approximately what is the smallest observable detail utilizing 500-MHz radar?

Solution 0.600 m

37. Determine the amount of time it takes for X-rays of frequency \(\displaystyle 3 \times 10^{18} Hz\) to travel (a) 1 mm and (b) 1 cm.

38. If you wish to detect details of the size of atoms (about \(\displaystyle 1 \times 10^{-10} m\)) with electromagnetic radiation, it must have a wavelength of about this size.

(a) What is its frequency?

(b) What type of electromagnetic radiation might this be?

Solution (a) \(\displaystyle f = \frac{c}{\lambda} = \frac{3.00 \times 10^{8} m/s}{1 \times 10^{-10} m} = 3 \times 10^{18} Hz\) (b) X-rays

39. If the Sun suddenly turned off, we would not know it until its light stopped coming. How long would that be, given that the Sun is \(\displaystyle 1.50 \times 10^{11} m\) away?

40. Distances in space are often quoted in units of light years, the distance light travels in one year.

(a) How many meters is a light year?

(b) How many meters is it to Andromeda, the nearest large galaxy, given that it is \(\displaystyle 2.00 \times 10^{6}\) light years away?

(c) The most distant galaxy yet discovered is \(\displaystyle 12.0 \times 10^{9}\) light years away. How far is this in meters?

41. A certain 50.0-Hz AC power line radiates an electromagnetic wave having a maximum electric field strength of 13.0 kV/m.

(a) What is the wavelength of this very low frequency electromagnetic wave?

(b) What is its maximum magnetic field strength?

Solution (a) \(\displaystyle 6.00 \times 10^{6} m\) (b) \(\displaystyle 4.33 \times 10^{-5} T\)

42. During normal beating, the heart creates a maximum 4.00-mV potential across 0.300 m of a person’s chest, creating a 1.00-Hz electromagnetic wave.

(a) What is the maximum electric field strength created?

(b) What is the corresponding maximum magnetic field strength in the electromagnetic wave?

(c) What is the wavelength of the electromagnetic wave?

43. (a) The ideal size (most efficient) for a broadcast antenna with one end on the ground is one-fourth the wavelength (\(\displaystyle \lambda / 4 \) of the electromagnetic radiation being sent out. If a new radio station has such an antenna that is 50.0 m high, what frequency does it broadcast most efficiently? Is this in the AM or FM band?

(b) Discuss the analogy of the fundamental resonant mode of an air column closed at one end to the resonance of currents on an antenna that is one-fourth their wavelength.

Solution (a) \(\displaystyle 1.50 \times 10^{6} Hz\), AM band (b) The resonance of currents on an antenna that is 1/4 their wavelength is analogous to the fundamental resonant mode of an air column closed at one end, since the tube also has a length equal to 1/4 the wavelength of the fundamental oscillation.

44. (a) What is the wavelength of 100-MHz radio waves used in an MRI unit?

(b) If the frequencies are swept over a \(\displaystyle +/- 1.00\) range centered on 100 MHz, what is the range of wavelengths broadcast?

45. (a) What is the frequency of the 193-nm ultraviolet radiation used in laser eye surgery?

(b) Assuming the accuracy with which this EM radiation can ablate the cornea is directly proportional to wavelength, how much more accurate can this UV be than the shortest visible wavelength of light?

Solution (a) \(\displaystyle 1.55 \times 10^{15} Hz\) (b) The shortest wavelength of visible light is 380 nm, so that \[\frac{\lambda_{visible}}{\lambda_{UV}}\] \[= \frac{380 nm}{193 nm}\] \[= 1.97.\] In other words, the UV radiation is 97% more accurate than the shortest wavelength of visible light, or almost twice as accurate!

46. TV-reception antennas for VHF are constructed with cross wires supported at their centers, as shown in Figure 15 The ideal length for the cross wires is one-half the wavelength to be received, with the more expensive antennas having one for each channel. Suppose you measure the lengths of the wires for particular channels and find them to be 1.94 and 0.753 m long, respectively. What are the frequencies for these channels?

A television reception antenna has cross wires of various lengths to most efficiently receive different wavelengths.

47. Conversations with astronauts on lunar walks had an echo that was used to estimate the distance to the Moon. The sound spoken by the person on Earth was transformed into a radio signal sent to the Moon, and transformed back into sound on a speaker inside the astronaut’s space suit. This sound was picked up by the microphone in the space suit (intended for the astronaut’s voice) and sent back to Earth as a radio echo of sorts. If the round-trip time was 2.60 s, what was the approximate distance to the Moon, neglecting any delays in the electronics?

Solution \(\displaystyle 3.90 \times 10^{8} m\)

48. Lunar astronauts placed a reflector on the Moon’s surface, off which a laser beam is periodically reflected. The distance to the Moon is calculated from the round-trip time.

(a) To what accuracy in meters can the distance to the Moon be determined, if this time can be measured to 0.100 ns?

(b) What percent accuracy is this, given the average distance to the Moon is \(\displaystyle 3.84 \times 10^{8} m\)?

49. Radar is used to determine distances to various objects by measuring the round-trip time for an echo from the object.

(a) How far away is the planet Venus if the echo time is 1000 s?

(b) What is the echo time for a car 75.0 m from a Highway Police radar unit?

(c) How accurately (in nanoseconds) must you be able to measure the echo time to an airplane 12.0 km away to determine its distance within 10.0 m?

Solution (a) \(\displaystyle 1.50 \times 10^{11} m\) (b) \(\displaystyle 0.500 \mu s\) (c) 66.7 ns

50. Integrated Concepts:

(a) Calculate the ratio of the highest to lowest frequencies of electromagnetic waves the eye can see, given the wavelength range of visible light is from 380 to 760 nm.

(b) Compare this with the ratio of highest to lowest frequencies the ear can hear.

51. Integrated Concepts:

(a) Calculate the rate in watts at which heat transfer through radiation occurs (almost entirely in the infrared) from \(\displaystyle 1.0 m^{2}\) of the Earth’s surface at night. Assume the emissivity is 0.90, the temperature of the Earth is \(\displaystyle 15^{\circ}C\), and that of outer space is 2.7 K.

(b) Compare the intensity of this radiation with that coming to the Earth from the Sun during the day, which averages about \(\displaystyle 800 W/m^{2}\), only half of which is absorbed.

(c) What is the maximum magnetic field strength in the outgoing radiation, assuming it is a continuous wave?

Solution (a) \(\displaystyle -3.5 \times 10^{2} W/m^{2}\) (b) 88% (c) \(\displaystyle 1.7 \mu T\)

24.4: Energy in Electromagnetic Waves

52. What is the intensity of an electromagnetic wave with a peak electric field strength of 125 V/m?

Solution \(\displaystyle I = \frac{c \epsilon_{0} E_{0}^{2}}{2}\) \(\displaystyle = \frac{\left( 3.00 \times 10^{8} m/s \right) \left( 8.85 \times 10^{-12} C^{2} / N \cdot m^{2} \right) \left(125 V/m \right) ^{2}}{2}\) \(\displaystyle = 20.7 W/m^{2}\)

53. Find the intensity of an electromagnetic wave having a peak magnetic field strength of \(\displaystyle 4.00 \times 10^{-9} T\).

54. Assume the helium-neon lasers commonly used in student physics laboratories have power outputs of 0.250 mW.

(a) If such a laser beam is projected onto a circular spot 1.00 mm in diameter, what is its intensity?

(b) Find the peak magnetic field strength.

(c) Find the peak electric field strength.

Solution (a) \(\displaystyle I = \frac{P}{A} = \frac{P}{\pi r^{2}} = \frac{0.250 \times 10^{-3} W}{ \pi \left( 0.500 \times 10^{-3} m \right) ^{2}} = 318 W/m^{2}\) \(\displaystyle I_{ave} = \frac{c B_{0}^{2}}{2 \mu_{0}} \rightarrow B_{0} = \left( \frac{2 \mu_{0}}{c}\right)^{1/2}\) (b) \(\displaystyle = \left( \frac{2 \left(2 \pi \times 10^{-7} T \cdot m/A \right) \left( 318.3 W/m^{2} \right)}{ 3.00 \times 10^{8} m/s} \right) ^{1/2} \) \(\displaystyle = 1.63 \times 10^{-6} T\) (c) \(\displaystyle E_{0} = c B_{0} = \left( 3.00 \times 10^{8} m/s \right) \left(1.633 \times 10^{-6} T \right)\) \(\displaystyle = 4.90 \times 10^{2} V/m\)

55. An AM radio transmitter broadcasts 50.0 kW of power uniformly in all directions.

(a) Assuming all of the radio waves that strike the ground are completely absorbed, and that there is no absorption by the atmosphere or other objects, what is the intensity 30.0 km away? (Hint: Half the power will be spread over the area of a hemisphere.)

(b) What is the maximum electric field strength at this distance?

56. Suppose the maximum safe intensity of microwaves for human exposure is taken to be \(\displaystyle 1.00 W/m^{2}\).

(a) If a radar unit leaks 10.0 W of microwaves (other than those sent by its antenna) uniformly in all directions, how far away must you be to be exposed to an intensity considered to be safe? Assume that the power spreads uniformly over the area of a sphere with no complications from absorption or reflection.

(b) What is the maximum electric field strength at the safe intensity? (Note that early radar units leaked more than modern ones do. This caused identifiable health problems, such as cataracts, for people who worked near them.)

Solution (a) 89.2 cm (b) 27.4 V/m

57. A 2.50-m-diameter university communications satellite dish receives TV signals that have a maximum electric field strength (for one channel) of \(\displaystyle 7.50 \mu V/m\). (See the figure.) (a) What is the intensity of this wave? (b) What is the power received by the antenna? (c) If the orbiting satellite broadcasts uniformly over an area of \(\displaystyle 1.50 \times 10^{13} m^{2}\) (a large fraction of North America), how much power does it radiate?

Satellite dishes receive TV signals sent from orbit. Although the signals are quite weak, the receiver can detect them by being tuned to resonate at their frequency.

58. Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time -- called pulsed lasers. They are used to ignite nuclear fusion, for example. Such a laser may produce an electromagnetic wave with a maximum electric field strength of \(\displaystyle 1.00 \times 10^{11} V/m\) for a time of 1.00 ns.

(a) What is the maximum magnetic field strength in the wave?

(b) What is the intensity of the beam?

(c) What energy does it deliver on a \(\displaystyle 1.00 - mm^{2}\) area?

Solution (a) 333 T (b) \(\displaystyle 1.33 \times 10^{19} W/m^{2}\) (c) 13.3 kJ

59. Show that for a continuous sinusoidal electromagnetic wave, the peak intensity is twice the average intensity (\(\displaystyle I_{0} = 2I_{ave}\)), using either the fact that \(\displaystyle E_{0} = \sqrt{2}E_{rms}\), or \(\displaystyle B_{0} = \sqrt{2}B_{rms}\), where rms means average (actually root mean square, a type of average).

60. Suppose a source of electromagnetic waves radiates uniformly in all directions in empty space where there are no absorption or interference effects.

(a) Show that the intensity is inversely proportional to \(\displaystyle r^{2}\), the distance from the source squared.

(b) Show that the magnitudes of the electric and magnetic fields are inversely proportional to \(\displaystyle r\).

Solution (a) \(\displaystyle I = \frac{P}{A} = \frac{P}{4 \pi r^{2}} \propto \frac{1}{r^{2}}\) (b) \(\displaystyle I \propto E_{0}^{2}, B_{0}^{2} \rightarrow E_{0}^{2}, B_{0}^{2} \propto \frac{1}{r^{2}} \rightarrow E_{0}, B_{0} \propto \frac{1}{r}\)

61. Integrated Concepts

An \(\displaystyle LC\) circuit with a 5.00-pF capacitor oscillates in such a manner as to radiate at a wavelength of 3.30 m.

(a) What is the resonant frequency?

(b) What inductance is in series with the capacitor?

62. Integrated Concepts

What capacitance is needed in series with an \(\displaystyle 800 - \mu H\) inductor to form a circuit that radiates a wavelength of 196 m?

Solution 13.5 pF

63. Integrated Concepts

Police radar determines the speed of motor vehicles using the same Doppler-shift technique employed for ultrasound in medical diagnostics. Beats are produced by mixing the double Doppler-shifted echo with the original frequency. If \(\displaystyle 1.50 \times 10^{9} - Hz\) microwaves are used and a beat frequency of 150 Hz is produced, what is the speed of the vehicle? (Assume the same Doppler-shift formulas are valid with the speed of sound replaced by the speed of light.)

64. Integrated Concepts:

Assume the mostly infrared radiation from a heat lamp acts like a continuous wave with wavelength \(\displaystyle 1.50 \mu m\).

(a) If the lamp’s 200-W output is focused on a person’s shoulder, over a circular area 25.0 cm in diameter, what is the intensity in \(\displaystyle W/m^{12}\)?

(b) What is the peak electric field strength?

(c) Find the peak magnetic field strength.

(d) How long will it take to increase the temperature of the 4.00-kg shoulder by \(\displaystyle 2.00^{\circ}C\), assuming no other heat transfer and given that its specific heat is \(\displaystyle 3.47 \times 10^{3} J/kg \cdot ^{\circ}C\)?

Solution (a) \(\displaystyle 4.07 kW/m^{2}\) (b) 1.75 kV/m (c) \(\displaystyle 5.84 \mu T\) (d) 2 min 19 s

65. Integrated Concepts:

On its highest power setting, a microwave oven increases the temperature of 0.400 kg of spaghetti by \(\displaystyle 45.0^{\circ}C\) in 120 s.

(a) What was the rate of power absorption by the spaghetti, given that its specific heat is \(\displaystyle 3.76 \times 10^{3} J/kg \cdot ^{\circ}C\)?

(b) Find the average intensity of the microwaves, given that they are absorbed over a circular area 20.0 cm in diameter.

(c) What is the peak electric field strength of the microwave? (d) What is its peak magnetic field strength?

66. Integrated Concepts:

Electromagnetic radiation from a 5.00-mW laser is concentrated on a \(\displaystyle 1.00 - mm^{2}\) area.

(a) What is the intensity in \(\displaystyle W/ m^{2}\)?

(b) Suppose a 2.00-nC static charge is in the beam. What is the maximum electric force it experiences?

(c) If the static charge moves at 400 m/s, what maximum magnetic force can it feel?

Solution (a) \(\displaystyle 5.00 \times 10^{3} W/m^{2}\) (b) \(\displaystyle 3.88 \times 10^{-6} N\) (c)\(\displaystyle 5.18 \times 10^{-12} N\)

67. Integrated Concepts:

A 200-turn flat coil of wire 30.0 cm in diameter acts as an antenna for FM radio at a frequency of 100 MHz. The magnetic field of the incoming electromagnetic wave is perpendicular to the coil and has a maximum strength of \(\displaystyle 1.00 \times 10^{-12} T\).

(a) What power is incident on the coil?

(b) What average emf is induced in the coil over one-fourth of a cycle?

(c) If the radio receiver has an inductance of \(\displaystyle 2.50 \mu H\), what capacitance must it have to resonate at 100 MHz?

68. Integrated Concepts:

If electric and magnetic field strengths vary sinusoidally in time, being zero at \(\displaystyle t = 0\), then \(\displaystyle E = E_{0} \sin{2 \pi} ft\) and \(\displaystyle B = B_{0} \sin{2 \pi} ft\). Let \(\displaystyle f = 1.00 GHz\) here.

(a) When are the field strengths first zero?

(b) When do they reach their most negative value?

(c) How much time is needed for them to complete one cycle?

Solution (a) \(\displaystyle t = 0\) (b) \(\displaystyle 7.50 \times 10^{-10} s\) (c) \(\displaystyle 1.00 \times 10^{-9} s\)

69. Unreasonable Results:

A researcher measures the wavelength of a 1.20-GHz electromagnetic wave to be 0.500 m.

(a) Calculate the speed at which this wave propagates.

(b) What is unreasonable about this result?

(c) Which assumptions are unreasonable or inconsistent?

70. Unreasonable Results:

The peak magnetic field strength in a residential microwave oven is \(\displaystyle 9.20 \times 10^{-5} T\).

(a) What is the intensity of the microwave?

(c) What is wrong about the premise?

Solution (a) \(\displaystyle 1.01 \times 10^{6} W/m^{2}\) (b) Much too great for an oven. (c) The assumed magnetic field is unreasonably large.

71. Unreasonable Results:

An \(\displaystyle LC\) circuit containing a 2.00-H inductor oscillates at such a frequency that it radiates at a 1.00-m wavelength.

(a) What is the capacitance of the circuit?

72. Unreasonable Results:

An \(\displaystyle LC\) circuit containing a 1.00-pF capacitor oscillates at such a frequency that it radiates at a 300-nm wavelength.

(a) What is the inductance of the circuit?

Solution (a) \(\displaystyle 2.53 \times 10^{-20} H\) (b) L is much too small. (c) The wavelength is unreasonably small.

73. Create Your Own Problem:

Consider electromagnetic fields produced by high voltage power lines. Construct a problem in which you calculate the intensity of this electromagnetic radiation in \(\displaystyle W/m^{2}\) based on the measured magnetic field strength of the radiation in a home near the power lines. Assume these magnetic field strengths are known to average less than a \(\displaystyle \mu T\). The intensity is small enough that it is difficult to imagine mechanisms for biological damage due to it. Discuss how much energy may be radiating from a section of power line several hundred meters long and compare this to the power likely to be carried by the lines. An idea of how much power this is can be obtained by calculating the approximate current responsible for \(\displaystyle \mu T\) fields at distances of tens of meters.

74. Create Your Own Problem:

Consider the most recent generation of residential satellite dishes that are a little less than half a meter in diameter. Construct a problem in which you calculate the power received by the dish and the maximum electric field strength of the microwave signals for a single channel received by the dish. Among the things to be considered are the power broadcast by the satellite and the area over which the power is spread, as well as the area of the receiving dish.

Contributors and Attributions

Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a  Creative Commons Attribution License (by 4.0) .

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Learning Resource Types

Vibrations and waves problem solving, interference of electromagnetic waves.

There are two points \( A (x = 0, y = 0) \) and \( B(x = L, y = 0) \) in the \( x \) - \( y \) plane in vacuum in which electromagnetic waves propagate with speed \( c \). About point \( A \) a charge \( Q \) moves with constant speed \( v \) in an anticlockwise direction in a circle of radius \( R \).

About point \( B \) a second identical charge moves with constant speed \( v \) in a clockwise direction in a circle of radius \( R \).

At \( t = 0 \) the first charge is at \( \left(x = \dfrac{R}{\sqrt{2}}, y = \dfrac{R}{\sqrt{2}}\right) \) and the second at \( \left(x = L - \dfrac{R}{\sqrt{2}}, y = \dfrac{R}{\sqrt{2}}\right) \) as shown below.

• What is the wavelength \( \lambda \) and frequency \( \nu \) of the radiation from each charge?
• Assuming \(R \ll \lambda \) and \( \lambda \ll L \), write an expression for the total radiated electric field \( \mathbf{E}(t) \) and \( \mathbf{B}(t) \) as a function of time \( t \) at the point \( P \left(x = \dfrac{L}{2}, y = \dfrac{L}{2}\right) \).
• Derive an expression for the flow of momentum per unit area and unit time at point \( P \).

› View/Hide Answers

• $$ \lambda = \dfrac{c}{\nu} = \dfrac{2 \pi R c}{v} $$
• \begin{eqnarray} \nonumber \vec{E} &=& \frac{Q v^2 \sin \left(\omega(t - \frac{L}{\sqrt{2}c})\right )}{2 \pi \epsilon_0 R L c^2} \hat{y} \\ \nonumber \text{where} \hspace{10mm} \omega &=& \frac{v}{R} \\ \nonumber \text{and} \hspace{10mm} \vec{B} &=& 0 \end{eqnarray}
• \(\therefore\) Flow of momentum per unit area and unit time at point \( P \) = 0

At positions \( \left (-\dfrac{\pi}{b}, y(t), 0\right ) \) and \( \left (+\dfrac{\pi}{b}, y(t), 0\right ) \) there are two balls of mass \( m \) attached to ideal springs of constant \( k \). The balls, each of which has a charge \( Q \) on it, are constrained to move in the \( \pm y \) direction.

There is an electromagnetic wave propagating in the \( +x \) direction with electric field \( \mathbf{E} = A \mathbf{\hat{y}} \cos (at - bx) \) (with \( a > 0, b > 0\)) which excites the two oscillators.

Write down an expression for the steady state total magnetic field due to only the radiation by the two charges, at the locations

• \( x = -r, ~ y = 0, ~ z = 0 \) at time \( t \) and
• \( x = 0, ~ y = 0, ~ z = r \) at time \( t \)

for \( r \gg \pi/b \) and \( r \gg \) length of the springs.

Assume the system is in vacuum, there is negligible (but not zero) friction and that, from the point of view of motion of the balls, you can ignore energy loss due to radiation and the interaction between the two balls. Furthermore assume there are no electromagnetic interactions with the springs.

› View/Hide Hint

In the steady state the oscillations at the natural frequency \( \omega_0 = \sqrt{\dfrac{k}{m}}\) will have died out.

• $$ \vec{B} = - \dfrac{a^2Q^2A}{m(\omega_0^2 - a^2)(2 \pi \epsilon_0 rc^3)} \cos (at - br) ~\hat{z} $$
• $$ \vec{B} = \dfrac{a^2Q^2A}{m(\omega_0^2 - a^2)(2 \pi \epsilon_0 rc^3)} \cos (at - br) ~\hat{x} $$

$$ \text{with} \hspace{3mm} \omega_0 = \sqrt{\dfrac{k}{m}} $$

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Chapter: 11th Physics : UNIT 11 : Waves

Solved example problems for physics: waves, numerical problems,   1.  the speed of a wave in a certain medium is 900 m/s. if 3000 waves passes over a certain point of the medium in 2 minutes, then compute its wavelength. solution the speed of a wave in medium v = 900 ms -1 freq. of wave = no. of waves passing per  sec (n) = 3000 waves/2min = 3000 / 2×60 = 25s wave length = λ = v = n λ λ  = v / n a = 900/25 =36m a = 36m answer  :  λ  = 36 m   2. consider a mixture of 2 mol of helium and 4 mol of oxygen. compute the speed of sound in this gas mixture at 300 k. solution helium - 2 mole, oxygen - 4 mol he & o 2 are mixed, hence molecular weight of the mixture of gases answer  : 400.9 ms -1   3. a ship in a sea sends sonar waves straight down into the seawater from the bottom of the ship. the signal reflects from the deep bottom bed rock and returns to the ship after 3.5 s. after the ship moves to 100 km it sends another signal which returns back after 2s. calculate the depth of the sea in each case and also compute the difference in height between two cases. solution velocity of sonar waves in water c = 1500 ms -1 time taken by be wave after reflection from the bottom of sea 2t = 3.5s t =1.75s distance covered (d) = c = d/t => d = c.t = 1500 × 1.75 - 2625 m after moving 100km the time taken by the wave = 2t = 2s t  = 2/2 = 1s d = d = 1500 × l = 1500 the different between these two heights . = 2625 - 1500 = 1125m answer  : δd = 1149.75 m   4. a sound wave is transmitted into a tube as shown in figure. the sound wave splits into two waves at the point a which recombine at point b. let r be the radius of the semi-circle which is varied until the first minimum. calculate the radius of the semi-circle if the wavelength of the sound is 50.0 m. answer  :  r  = 21.9  m solution consider the sound waves start from the pt a. the sound waves will meet at the picture b & interfere and can be detected. the path length of sound waves passing through the curve is equal to half the circle having radio centered at c. thus, be path length of be sound waves in curve part is l 1 = πγ the path length of the sound in tube abc is equal to the diameter of the circle having radius γ centered at e. so the path length of the sound waves travelling in it = l 2 =2r the path difference of the sound waves at the picture is δp = l 1 - l 2 = πr— 2r δp = r(π - 2) ...(1) for the sound to be heard minimum at the detected the difference in path length of be sound waves is δp = λ/2    …………(2) from equation (1) r(π- 2) = λ/2 wavelength of sound = λ = 50m radius of be semicircle r = r = 50 / [2 × 3.14 – 4] = 50/2.28 r = 21.9m answer  :  r  = 21.9  m   5.    n tuning forks are arranged in order of increasing frequency and any two successive tuning forks give n beats per second when sounded together. if the last fork gives double the frequency of the first (called as octave), show that the frequency of the first tuning fork is  f  = (n−1) n . solution no.of tuning forks = n the frequency of the first is double to be last the difference of frequency for two successive tuning forks nhz let be frequency of the first tuning fork is f then be frequency of the last tuning fork is f (n-l)n frequency is reduced to are fourth f last = - f /2  an = a + ( n -1) d (i.e.) 2 f  =   f + (n-1)n = (n-1)n f = (n-1)n   6.    let the source propagate a sound wave whose intensity at a point (initially) be  i . suppose we consider a case when the amplitude of the sound wave is doubled and the frequency is reduced to one-fourth. calculate now the new intensity of sound at the same point . solution intensity of sound wave  = i old amplitude is doubled = 2a = a new frequency is produced to are fourth answer:  i new   ∝  ¼ i  old   7. consider two organ pipes of same length in which one organ pipe is closed and another organ pipe is open. if the fundamental frequency of closed pipe is 250 hz. calculate the fundamental frequency of the open pipe. answer:  500hz   8. a police in a siren car moving with a velocity 20 ms -1  chases a thief who is moving in a car with a velocity  v 0 ms -1 . the police car sounds at frequency 300hz, and both of them move towards a stationary siren of frequency 400hz. calculate the speed in which thief is moving. solution the velocity of car = 20ms -1 the frequency of car = 300hz the frequency stationary siren = 400hz the speed in which thief moving = answer:  v thief   = 10 m s -1   9. consider the following function (a)  y  =  x 2  + 2  α t x (b)  y  = ( x  +  vt ) 2 which among the above function can be characterized as a wave . solution (a) y = x 2 + 2 a t x it is not describing wave. (b) y = (x + 2 vx) 2 it satisfies wave equation. answer:  (a) function is not describing wave (b) satisfies wave equation.  , conceptual questions,   1. why is it that transverse waves cannot be produced in a gas. can the transverse waves can be produced in solids and liquids answer: transverse waves travel in the form of crests and troughs and so involve change in shape. as gas has no elasticity of shape, transverse waves cannot be produced in it. 2. why is the roar of our national animal different from the sound of a mosquito answer: roaring of a national animal (tiger) produces a sound of low pitch and high intensity or loudness, whereas the buzzing of mosquito produces a sound of high pitch and low intensity or loudness. 3. a sound source and listener are both stationary and a strong wind is blowing. is there a doppler effect answer: yes. it does not matter whether there is sound i source or transmission media in motion. 4. in an empty room why is it that a tone sounds louder than in the room having things like furniture etc. answer: sound is a form of energy. the furniture which act as obstacles absorbs most of the energy. so the intensity of sound become low but in empty room, due to the absence of obstacles the intensity of sound remain mostly same and we feel it louder. 5. how do animals sense impending danger of hurricane   answer: some animals are believed to be sensitive to low frequency sound waves emitted by hurricanes. they can also detect the slight fall in air pressure, and water pressure that signal a storm's approach. 6. is it possible to realize whether a vessel kept under the tap is about to fill with water answer: the frequency of the note produced by an air column is inversely proportional to its length. as the level of water in the vessel rises, the length of the air column above it decreases. it produce, sound of decreasing frequency, i.e., the sound becomes shorter. from the shrillness of sound, is possible to realize whether the vessel is filled with water. v min = 11.71 ms -1, solved example problems for wave motion.

EXAMPLE 11.1

Which of the following has longer wavelength?

Answer is (c) 

EXAMPLE 11.2

Three waves are shown in the figure below

Write down  

(a) the frequency in ascending order

(b) the wavelength in ascending order

(a)  f c   <  f a   <  f b

(b)  λ b   <  λ a   <  λ c

From the example 11.2, we observe that the frequency is inversely related to the wavelength,  f   ∝  1/λ

Then,  f λ is equal to what?

[(i.e) fλ=?]

A simple dimensional argument will help us to determine this unknown physical quantity.

Dimension of wavelength is, [λ] = L

Frequency  f  = 1/  Time period  , which implies that the dimension of frequency is,

where  v  is known as the  wave velocity  or  phase velocity . This is the velocity with which   the wave propagates.  Wave velocity  is the distance travelled by a wave in one second.

1. The number of cycles (or revolutions) per unit time is called  angular frequency .  Angular frequency,  ω   = 2π/T = 2πf (unit is   radians/second)

2. The number of cycles per unit distance or number of waves per unit distance is called  wave number .

wave number, k = 2 π /λ  (unit is radians/ meter)

The velocity v, angular frequency ω and wave number λ are related as:

Wave numbers and wave vectors play an essential role in optics and scattering theory.

EXAMPLE 11.3

The average range of frequencies at which human beings can hear sound waves varies from 20 Hz to 20 kHz. Calculate the wavelength of the sound wave in these limits. (Assume the speed of sound to be 340 m s -1 .

Therefore, the audible wavelength region is from 0.017 m to 17 m when the velocity of sound in that region is 340 m s -1 .

EXAMPLE 11.4

A man saw a toy duck on a wave in an ocean. He noticed that the duck moved up and down 15 times per minute. He roughly measured the wavelength of the ocean wave as 1.2 m. Calculate the time taken by the toy duck for going one time up and down and also the velocity of the ocean wave.

Given that the number of times the toy duck moves up and down is 15 times per minute. This information gives us frequency (the number of times the toy duck moves up and down)

But one minute is 60 second, therefore, expressing time in terms of second

The time taken by the toy duck for going one time up and down is time period which is inverse of frequency

The velocity of ocean wave is  v

v = λf  = 1.2 ×0.25 = 0.3 m s -1 .

EXAMPLE 11.5

Consider a string whose one end is attached to a wall. Then compute the following in both situations given in figure (assume waves crosses the distance in one second)

(a)  Wavelength,

(b)  Frequency and 

(c)  Velocity

This means that the speed of the wave along a string is a constant. Higher the frequency, shorter the wavelength and vice versa, and their product is velocity which remains the same.

Solved Example Problems for Velocity of Waves in Different Media

Velocity of transverse waves in a stretched string, example 11.6.

Calculate the velocity of the travelling pulse as shown in the figure below. The linear mass density of pulse is 0.25 kg m -1 . Further, compute the time taken by the travelling pulse to cover a distance of 30 cm on the string.

The tension in the string is  T  =  m g  = 1.2 × 9.8 = 11.76 N

The mass per unit length is  μ  = 0.25 kg m -1 . Therefore, velocity of the wave pulse is

The time taken by the pulse to cover the distance of 30 cm is

ms = milli second.

Velocity of longitudinal waves in an elastic medium

Example 11.7.

Calculate the speed of sound in a steel rod whose Young’s modulus  Y  = 2 × 10 11  N m -2  and  ρ  = 7800  kg m -3 .

Therefore, longitudinal waves travel faster in a solid than in a liquid or a gas. Now you may understand why a shepherd checks before crossing railway track by keeping his ears on the rails to safegaurd his cattle.

EXAMPLE 11.8

An increase in pressure of 100  k Pa causes a certain volume of water to decrease by 0.005% of its original volume.

(a) Calculate the bulk modulus of water?.

(b) Compute the speed of sound (compressional waves) in water?.

(a) Bulk modulus

(b) Speed of sound in water is

Solved Example Problems for Propagation of Sound Waves

Example 11.9.

The ratio of the densities of oxygen and nitrogen is 16:14. Calculate the temperature when the speed of sound in nitrogen gas at 17°C is equal to the speed of sound in oxygen gas.

From equation (11.25), we have

Where, R is the universal gas constant and M is the molecular mass of the gas. The speed of sound in nitrogen gas at 17°C is

Similarly, the speed of sound in oxygen gas at  t  in  K  is

Given that the value of  γ  is same for both the gases, the two speeds must be equal. Hence, equating equation (1) and (2), we get

Squaring on both sides and cancelling  γ R  term and rearranging, we get

Since the densities of oxygen and nitrogen is 16:14,

Substituting equation (5) in equation (3), we get

Solved Example Problems for Reflection of Sound Waves

EXAMPLE 11.10

Suppose a man stands at a distance from a cliff and claps his hands. He receives an echo from the cliff after 4 second. Calculate the distance between the man and the cliff. Assume the speed of sound to be 343 m s-1.

The time taken by the sound to come back as echo is 2 t  = 4  ⇒   t  = 2 s

∴ The distance is   d   =   vt   =(343 m s -1 )(2 s)   = 686 m.

Solved Example Problems for Progressive Waves (or) Travelling Waves

Example 11.11.

Sketch  y  =  x  − a  for different values of  a .

This implies, when increasing the value of  a , the line shifts towards right side. For  a  =  vt ,  y  =  x  −  vt  satisfies the differential   equation. Though this function satisfies the differential equation, it is not finite for all values of  x  and  t . Hence, it does not represent a wave.

EXAMPLE 11.12

How does the wave  y  = sin( x  −  a ) for  a  = 0, a = π/4, a = π/2, a = 3π/2, a = π look like?. Sketch this wave.

From the above picture we observe that  y = sin (x−a) for a = 0, a = π/4, a = π/2, a = 3π/2 and a = π, the function y = sin (x−a)  shifts towards right. Further, we can take   a = vt and v = π/4 ,  and sketching for different times  t  = 0 s ,  t  = 1  s ,  t  = 2 s  etc., we once again observe that  y  = sin( x − vt ) moves towards the right. Hence,  y  = sin( x − vt ) is a travelling (or progressive) wave moving towards the right. If  y  = sin( x + vt ) then the travelling (or progressive) wave moves towards the left. Thus, any arbitrary function of type  y  =  f ( x − vt ) characterising the wave must   move towards right and similarly, any arbitrary function of type  y  =  f ( x + vt ) characterizing the wave must move towards left.

EXAMPLE 11.13

Check the dimensional of the wave  y  = sin( x − vt ). If it is dimensionally wrong, write the above equation in the correct form.

Dimensionally it is not correct. we know that  y  = sin( x − vt ) must be a dimensionless quantity but  x − vt  has dimension. The correct equation is  y  = sin ( k x − ωt ), where  k  and  ω  have the dimensions of inverse   of length and inverse of time respectively. The sine functions and cosine functions are periodic functions with period 2π. Therefore, the correct expression is y = sin ([2π/λ]x, [2π/T]t) where λ and T are wavelength and time period, respectively.

In general,  y ( x , t )=A sin( k x − ωt ).

EXAMPLE 11.14

The wavelength of two sine waves are  λ 1   = 1m and  λ 2   = 6m. Calculate the   corresponding wave numbers.

EXAMPLE 11.15

A mobile phone tower transmits a wave signal of frequency 900MHz. Calculate the length of the waves transmitted from the mobile phone tower.

Frequency,  f   =  900  MHz   =  900  × 10 6   Hz

The speed of wave is  c  = 3 × 10 8 m s −1

Solved Example Problems for Waves Superposition Principle

Interference of waves, example 11.16.

Consider two sources A and B as shown in the figure below. Let the two sources emit simple harmonic waves of same frequency but of different amplitudes, and both are in phase (same phase). Let O be any point equidistant from A and B as shown in the figure. Calculate the intensity at points O, Y and X. (X and Y’ are not equidistant from A & B)

The distance between OA and OB are the same and hence, the waves starting from A and B reach O after covering equal distances (equal path lengths). Thus, the path difference between two waves at O is zero.

OA  −  OB  = 0

Since the waves are in the same phase, at the point O, the phase difference between two waves is also zero. Thus, the resultant intensity at the point O is maximum.

Consider a point Y, such that the path difference between two waves is  λ . Then the phase difference at Y is

Therefore, at the point Y, the two waves from A and B are in phase, hence, the intensity will be maximum.

Consider a point X, and let the path difference the between two waves be λ/2.

Then the phase difference at X is

Therefore, at the point X, the waves meet and are in out of phase, Hence, due to destructive interference, the intensity will be minimum.

EXAMPLE 11.17

Two speakers C and E are placed 5 m apart and are driven by the same source. Let a man stand at A which is 10 m away from the mid point O of C and E. The man walks towards the point O which is at 1 m (parallel to OC) as shown in the figure. He receives the first minimum in sound intensity at B. Then calculate the frequency of the source.

(Assume speed of sound = 343 m s -1 )

The first minimum occurs when the two waves reaching the point B are 180° (out of phase). The path difference ∆x = λ/2.

In order to calculate the path difference, we have to find the path lengths  x 1  and  x 2 .

In a right triangle BDC,

The path difference ∆ x  =  x 2  −  x 1  = 10.6 m−10.1 m = 0.5 m. Required that this path difference

Formation of beats

Example 11.18.

Consider two sound waves with wavelengths 5  m  and 6  m . If these two waves propagate in a gas with velocity 330  ms -1. Calculate the number of beats per second.

Given  λ 1  = 5 m  and  λ 2  = 6 m

Velocity of sound waves in a gas is  v  = 330  ms -1

The relation between wavelength and velocity is  v  =  λf  =>  f   = v/ λ

The number of beats per second is

|  f 1  −  f 2 | = |66 − 55| = 11 beats per sec

EXAMPLE 11.19

Two vibrating tuning forks produce waves whose equation is given by  y 1  = 5 sin(240 π t ) and  y 2  = 4 sin(244 πt ). Compute the number of beats per second.

Given  y 1  = 5 sin(240 π t ) and  y 2  = 4 sin(244 πt )

Comparing with  y  =  A  sin(2 π f 1 t ), we get

2 πf 1  = 240 π   ⇒   f 1  = 120 Hz

2 πf 2  = 244 π   ⇒   f 2  = 122 Hz

The number of beats produced is |  f 1  −  f 2 | = |120 − 122| = |− 2|=2 beats per sec

Solved Example Problems for Stationary waves

Example 11.20.

Compute the distance between anti-node and neighbouring node.

For  nth  mode, the distance between anti-node and neighbouring node is

EXAMPLE 11.21

Let  f  be the fundamental frequency of the string. If the string is divided into three segments  l 1 ,  l 2  and  l 3  such that the fundamental frequencies of each segments be  f 1 ,  f 2  and  f 3 , respectively. Show that

For a fixed tension  T  and mass density µ, frequency is inversely proportional to the string length i.e.

EXAMPLE 11.22

Consider a string in a guitar whose length is 80 cm and a mass of 0.32 g with tension 80 N is plucked. Compute the first four lowest frequencies produced when it is plucked.

The velocity of the wave

The length of the string,  L  = 80 cm=0.8 m The mass of the string, m = 0.32  g  = 0.32 × 10 -3 kg

Therefore, the linear mass density,

The tension in the string,  T  = 80  N

The wavelength corresponding to the fundamental frequency  f 1  is  λ 1  =  2L  = 2 × 0.8 = 1.6 m

The fundamental frequency  f 1  corresponding to the wavelength  λ 1

Similarly, the frequency corresponding to the second harmonics, third harmonics and fourth harmonics are

f 2   = 2 f 1   = 559 Hz

f 3   = 3 f 1   = 838.5 Hz

f 4   = 4 f 1   = 1118 Hz

Solved Example Problems for Intensity and Loudness of sound

Example 11.23.

A baby cries on seeing a dog and the cry is detected at a distance of 3.0 m such that the intensity of sound at this distance is 10 -2   W m -2 . Calculate the intensity of the baby’s cry at a distance 6.0  m .

I 1   is the intensity of sound detected at a   distance 3.0 m and it is given as 10 -2   W m -2 . Let  I 2  be the intensity of sound detected at a distance 6.0 m. Then,

  r 1  = 3.0 m,

r 2   = 6.0 m

the power output does not depend on the observer and depends on the baby. Therefore,

I 2   = 0.25 × 10 -2  W m -2

EXAMPLE 11.24

The sound level from a musical instrument playing is 50 dB. If three identical musical instruments are played together then compute the total intensity. The intensity of the sound from each instrument is 10 -12  Wm -2

I 1   = 10 -7  Wm -2

Since three musical instruments are played, therefore,  I total  = 3 I 1  = 3 × 10 -7   Wm -2 .

Solved Example Problems for Vibrations of Air Column

Example 11.25.

If a flute sounds a note with 450 Hz  , what are the frequencies of the second, third, and fourth harmonics of this pitch?. If the clarinet sounds with a same note as 450 Hz , then what are the frequencies of the lowest three harmonics produced ?.

For a flute which is an open pipe, we have

Second harmonics           f 2  = 2  f 1  = 900 Hz

Third harmonics    f 3  = 3  f 1         = 1350 Hz

Fourth harmonics  f 4  = 4  f 1         = 1800 Hz

For a clarinet which is a closed pipe, we have

Second harmonics           f   2  = 3  f 1       = 1350 Hz

Third harmonics    f   3  = 5  f 1        = 2250 Hz

Fourth harmonics  f   4  = 7  f 1        = 3150 Hz

EXAMPLE 11.26

If the third harmonics of a closed organ pipe is equal to the fundamental frequency of an open organ pipe, compute the length of the open organ pipe if the length of the closed organ pipe is 30 cm.

Let  l 2  be the length of the open organ pipe, with  l 1  =30  cm  the length of the closed organ pipe.

It is given that the third harmonic of closed organ pipe is equal to the fundamental frequency of open organ pipe.

The third harmonic of a closed organ pipe is

The fundamental frequency of open organ pipe is

EXAMPLE 11.27

A frequency generator with fixed frequency of 343 Hz is allowed to vibrate above a 1.0 m high tube. A pump is switched on to fill the water slowly in the tube. In order to get resonance, what must be the minimum height of the water?. (speed of sound in air is 343 m s −1 )

Let the length of the resonant columns be  L 1 ,  L 2   and  L 3 . The first resonance occurs at   length  L 1

The second resonance occurs at length  L 2

Since total length of the tube is 1.0 m the third and other higher resonances do not occur. Therefore, the minimum height of water H min  for resonance is,

H min  = 1.0 m − 0.75 m = 0.25 m

EXAMPLE 11.28

A student performed an experiment to determine the speed of sound in air using the resonance column method. The length of the air column that resonates in the fundamental mode with a tuning fork is 0.2 m. If the length is varied such that the same tuning fork resonates with the first overtone at 0.7 m. Calculate the end correction.

End correction

EXAMPLE 11.29

Consider a tuning fork which is used to produce resonance in an air column. A resonance air column is a glass tube whose length can be adjusted by a variable piston. At room temperature, the two successive resonances observed are at 20 cm and 85 cm of the column length. If the frequency of the length is 256 Hz, compute the velocity of the sound in air at room temperature.

Given two successive length (resonance) to be L 1   =  20 cm and L 2  = 85 cm

The frequency is  f  = 256 Hz

v  =  f λ  = 2 f ∆ L  = 2 f  ( L 2   −  L 1 )

= 2 × 256 × (85 − 20) × 10 −2  m s −1

v = 332.8 cm −1

Solved Example Problems for Doppler Effect

Example 11.30.

A sound of frequency 1500 Hz is emitted by a source which moves away from an observer and moves towards a cliff at a speed of 6 ms -1 .

(a) Calculate the frequency of the sound which is coming directly from the source.

(b) Compute the frequency of sound heard by the observer reflected off the cliff. Assume the speed of sound in air is 330 m s -1

(a) Source is moving away and observer is stationary, therefore, the frequency of sound heard directly from source is

(b) Sound is reflected from the cliff and reaches observer, therefore,

EXAMPLE 11.31

An observer observes two moving trains, one reaching the station and other leaving the station with equal speeds of 8 m s −1 . If each train sounds its whistles with frequency 240 Hz, then calculate the number of beats heard by the observer.

Observer is stationary

(i) Source (train) is moving towards an observer:

Apparent frequency due to train arriving station is

(ii) Source (train) is moving away form an observer:

Apparent frequency due to train leaving station is

So the number of beats = |  f in  - f out | = (246-234) = 12

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Simulation of Overturning a Wave When a Tsunami Comes Ashore

• Conference paper
• First Online: 01 April 2024
• Cite this conference paper

• Sergey L. Derybin   ORCID: orcid.org/0000-0003-3730-0966 3 &
• Alexey V. Mezentsev   ORCID: orcid.org/0000-0002-5678-8701 3  

Part of the book series: Springer Proceedings in Physics ((SPPHY,volume 1067))

Included in the following conference series:

• All Russian Scientific Conference on Current Issues of Continuum Mechanics and Celestial Mechanics

The paper constructs an exact solution for a system of equations describing one-dimensional and two-dimensional water flows that occur after a wave overturns when a tsunami comes ashore. It is shown that a technique for solving the problem of the breakup of a special discontinuity can be used to describe the wave coming ashore when the depth of the liquid turns to zero at the edge line. Using this technique, solutions of the initial boundary value problem in the vicinity of the edge line in the form of a power series are obtained. The solution of the initial boundary value problem is constructed in the form of power series. The coefficients of the series are found when integrating ordinary differential equations. The theorem on the existence and uniqueness of the solution of the initial boundary value problem is proved.

• Overturning the Wave
• Water Edge Line
• Shallow Water Equation System
• Exact Solution
• Power Series

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Deryabin, S.L., Mezentsev, A.V.: Numerical and analytical modeling of two-dimensional water flows arising after the dam failure. In: Orlov, M.Y., Visakh, P.M. (eds.) Behavior of materials under impact, explosion, high pressures and dynamic strain rates, pp. 113–128. Springer International Publishing, Cham (2023). https://doi.org/10.1007/978-3-031-17073-7_8

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Derybin, S.L., Mezentsev, A.V. (2024). Simulation of Overturning a Wave When a Tsunami Comes Ashore. In: Orlov, M.Y., Visakh, P.M. (eds) Proceedings of the XII All Russian Scientific Conference on Current Issues of Continuum Mechanics and Celestial Mechanics. CICMCM 2023. Springer Proceedings in Physics, vol 1067. Springer, Singapore. https://doi.org/10.1007/978-981-97-1872-6_12

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Letters to the Editor: Speed humps won’t solve the problem of reckless driving

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To the editor: Installing speed bumps on residential streets may slow traffic, but the bumps designed to slow scofflaws also slow/impede emergency vehicles and first responders. Be careful what you wish for. ( “Why does it have to be so hard to get a speed hump in Los Angeles?” , Opinion, March 20)

Perhaps more or better traffic law enforcement might help? Ticket the speeders; cite the folks who ignore traffic signs. It never ceases to amaze me how many of the drivers in my area think those eight-sided red signs with the four white letters don’t apply to them.

Beryl Arbit, Encino

To the editor: Speed humps are the ultimate in NIMBY-ism. One person on a block can get them. There is no forum for people opposed to a particular installation. It does not consider all the stakeholders. They just appear.

Harlan Levinson, Los Angeles

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Semiconductors at scale: New processor achieves remarkable speedup in problem solving

A nnealing processors are designed specifically for addressing combinatorial optimization problems, where the task is to find the best solution from a finite set of possibilities. This holds implications for practical applications in logistics, resource allocation, and the discovery of drugs and materials.

In the context of CMOS (a type of semiconductor technology), it is necessary for the components of annealing processors to be fully "coupled." However, the complexity of this coupling directly affects the scalability of the processors.

In a new IEEE Access study led by Professor Takayuki Kawahara from Tokyo University of Science, researchers have developed and successfully tested a scalable processor that divides the calculation into multiple LSI chips. The innovation was also presented in IEEE 22nd World Symposium on Applied Machine Intelligence and Informatics (SAMI 2024) on 25 January 2024.

According to Prof. Kawahara, "We want to achieve advanced information processing directly at the edge, rather than in the cloud or performing preprocessing at the edge for the cloud. Using the unique processing architecture announced by the Tokyo University of Science in 2020, we have realized a fully coupled LSI (Large Scale Integration) on one chip using 28nm CMOS technology. Furthermore, we devised a scalable method with parallel-operating chips and demonstrated its feasibility using FPGAs (Field-Programmable Gate Arrays) in 2022."

The team created a scalable annealing processor. It used 36 22nm CMOS calculation LSI (Large Scale Integration) chips and one control FPGA. This technology enables the construction of large-scale, fully coupled semiconductor systems following the Ising model (a mathematical model of magnetic systems) with 4096 spins.

The processor incorporates two distinct technologies developed at the Tokyo University of Science. This includes a "spin thread method" that enables 8 parallel solution searches, coupled with a technique that reduces chip requirements by about half compared to conventional methods. Its power needs are also modest, operating at 10MHz with a power consumption of 2.9W (1.3W for the core part). This was practically confirmed using a vertex cover problem with 4096 vertices.

In terms of power performance ratio, the processor outperformed simulating a fully coupled Ising system on a PC (i7, 3.6GHz) using annealing emulation by 2,306 times. Additionally, it surpassed the core CPU and arithmetic chip by 2,186 times.

The successful machine verification of this processor suggests the possibility of enhanced capacity. According to Prof. Kawahara, who holds a vision for the social implementation of this technology (such as initiating a business, joint research, and technology transfer), "In the future, we will develop this technology for a joint research effort targeting an LSI system with the computing power of a 2050-level quantum computer for solving combinatorial optimization problems."

"The goal is to achieve this without the need for air conditioning, large equipment, or cloud infrastructure using current semiconductor processes. Specifically, we would like to achieve 2M (million) spins by 2030 and explore the creation of new digital industries using this."

In summary, researchers have developed a scalable, fully coupled annealing processor incorporating 4096 spins on a single board with 36 CMOS chips. Key innovations, including chip reduction and parallel operations for simultaneous solution searches, played a crucial role in this development.

More information: Taichi Megumi et al, Scalable Fully-Coupled Annealing Processing System Implementing 4096 Spins Using 22nm CMOS LSI, IEEE Access (2024). DOI: 10.1109/ACCESS.2024.3360034

Provided by Tokyo University of Science