9.1: Understanding Earth Through Seismology
Seismology is the study of vibrations within Earth. These vibrations are caused by various events: earthquakes, extraterrestrial impacts, explosions, storm waves hitting the shore, and tidal effects. Of course, seismic techniques have been most widely applied to the detection and study of earthquakes, but there are many other applications, and arguably seismic waves provide the most important information that we have concerning Earth’s interior. Before going any deeper into Earth, however, we need to take a look at the properties of seismic waves. The types of waves that are useful for understanding Earth’s interior are called body waves , meaning that, unlike the surface waves on the ocean, they are transmitted through Earth materials.
Imagine hitting a large block of strong rock (e.g., granite) with a heavy sledgehammer (Figure \(\PageIndex{1}\)). At the point where the hammer strikes it, a small part of the rock will be compressed by a fraction of a millimetre. That compression will transfer to the neighbouring part of the rock, and so on through to the far side of the rock—all in a fraction of a second. This is known as a compression wave, and it can be illustrated by holding a loose spring (like a Slinky) that is attached to something (or someone) at the other end. If you give it a sharp push so the coils are compressed, the compression propagates (travels) along the length of the spring and back (Figure \(\PageIndex{2}\)). You can think of a compression wave as a “push” wave—it’s called a P wave (although the “P” stands for “primary” because P waves arrive first at seismic stations).
When we hit a rock with a hammer, we also create a different type of body wave, one that is characterized by back-and-forth vibrations (as opposed to compressions). This is known as a shear wave ( S wave , where the “S” stands for “secondary”), and an analogy would be what happens when you flick a length of rope with an up-and-down motion. As shown in Figure \(\PageIndex{2}\), a wave will form in the rope, which will travel to the end of the rope and back.
Compression waves and shear waves travel very quickly through geological materials. As shown in Figure \(\PageIndex{3}\), typical P wave velocities are between 0.5 kilometers per second (km/s) and 2.5 km/s in unconsolidated sediments, and between 3.0 km/s and 6.5 km/s in solid crustal rocks. Of the common rocks of the crust, velocities are greatest in basalt and granite. S waves are slower than P waves, with velocities between 0.1 km/s and 0.8 km/s in soft sediments, and between 1.5 km/s and 3.8 km/s in solid rocks.
Imagine that a strong earthquake takes place on Vancouver Island within Strathcona Park (west of Courtenay). Assuming that the crustal average P wave velocity is 5 km per second, how long will it take (in seconds) for the first seismic waves (P waves) to reach you in the following places (distances from the epicentre are shown)?
- Nanaimo (120 km away)
- Surrey (200 km away)
- Kamloops (390 km away)
See Appendix 3 for Exercise 9.1 answers .
Mantle rock is generally denser and stronger than crustal rock and both P- and S-waves travel faster through the mantle than they do through the crust. Moreover, seismic-wave velocities are related to how tightly compressed a rock is, and the level of compression increases dramatically with depth. Finally, seismic waves are affected by the phase state of rock. They are slowed if there is any degree of melting in the rock. If the material is completely liquid, P waves are slowed dramatically and S waves are stopped altogether.
As shown on the right-hand part of Figure \(\PageIndex{4}\), the upper approximately 100 km of the Earth is known as the lithosphere. This includes the rigid upper part of the mantle (or lithospheric mantle) and the crust. The next 150 km is the asthenosphere or low velocity zone (because seismic waves are slowed as they pass through that material). As we’ll see below, that part of the mantle is close to it’s melting point and in some regions may be partially molten.
Accurate seismometers have been used for earthquake studies since the late 1800s, and systematic use of seismic data to understand Earth’s interior started in the early 1900s. The rate of change of seismic waves with depth in Earth (as shown in Figure \(\PageIndex{4}\)) has been determined over the past several decades by analyzing seismic signals from large earthquakes at seismic stations around the world. Small differences in arrival time of signals at different locations have been interpreted to show that:
- Velocities are greater in mantle rock than in the crust.
- Velocities generally increase with pressure, and therefore with depth.
- Velocities slow in the area between a 100 and 250 kilometre depth (called the “low-velocity zone”; equivalent to the asthenosphere).
- Velocities increase dramatically at 660 kilometre depth (because of a mineralogical transition).
- Velocities slow in the region just above the core-mantle boundary (the D” (d-double-prime) layer or “ultra-low-velocity zone”).
- S waves do not pass through the outer part of the core.
- P wave velocities increase dramatically at the boundary between the liquid outer core and the solid inner core.
One of the first discoveries about Earth’s interior made through seismology was in 1909 when Croatian seismologist Andrija Mohorovičić (pronounced Moho-ro-vi-chich ) realized that at certain distances from an earthquake, two separate sets of seismic waves arrived at a seismic station within a few seconds of each other. He reasoned that the waves that went down into the mantle, traveled through the mantle, and then were bent upward back into the crust, reached the seismic station first because although they had farther to go, they traveled faster through mantle rock (as shown in Figure \(\PageIndex{5}\)). The boundary between the crust and the mantle is known as the Mohorovičić discontinuity (or Moho). Its depth is between 30 and 40 kilometers beneath most of the continental crust, and between 5 and 10 kilometers beneath the oceanic crust.
Our current understanding of the patterns of seismic wave transmission through Earth is summarized in Figure \(\PageIndex{6}\). Because of the gradual increase in density (and therefore rock strength) with depth, all waves are refracted (toward the lower density material) as they travel through homogenous parts of Earth and thus tend to curve outward toward the surface. Waves are also refracted at boundaries within Earth, such as at the Moho, at the core-mantle boundary (CMB), and at the outer-core/inner-core boundary.
S waves do not travel through liquids—they are stopped at the CMB—and there is an S wave shadow on the side of Earth opposite a seismic source. The angular distance from the seismic source to the shadow zone is 103° on either side, so the total angular distance of the shadow zone is 154°. We can use this information to infer the depth to the CMB.
P waves do travel through liquids, so they can make it through the liquid part of the core. Because of the refraction that takes place at the CMB, waves that travel through the core are bent away from the surface, and this creates a P wave shadow zone on either side, from 103° to 150°. This information can be used to discover the differences between the inner and outer parts of the core.
We know that other planets must have (or at least did have) liquid cores like ours, and we could use seismic data to find out how big they are. The S wave shadow zones on planets A and B are shown. Using the same method used for Earth (on the left), sketch in the outlines of the cores for these two other planets.
See Appendix 3 for Exercise 9.2 answers .
Using data from many seismometers and hundreds of earthquakes, it is possible to create a two- or three-dimensional image of the seismic properties of part of the mantle. This technique is known as seismic tomography, and an example of the result is shown in Figure \(\PageIndex{8}\).
The Pacific Plate subducts beneath Tonga and appears in Figure \(\PageIndex{8}\) as a 100 kilometre thick slab of cold (blue-colored) oceanic crust that has pushed down into the surrounding hot mantle. The cold rock is more rigid than the surrounding hot mantle rock, so it is characterized by slightly faster seismic velocities. There is volcanism in the Lau spreading centre and also in the Fiji area, and the warm rock in these areas has slower seismic velocities (yellow and red colors).
Image descriptions
| Material | S Wave (kilometers per second) | P Wave (kilometers per second) |
|---|---|---|
| Dry sand | 0.1 to 0.4 | 0.4 to 1.3 |
| Clay | 0.2 to 0.6 | 0.6 to 1.6 |
| Wet sand | 0.7 to 0.8 | 1.5 to 2.2 |
| Till | 0.8 to 1.0 | 1.9 to 2.6 |
| Mudstone | 2.1 to 2.3 | 3.0 to 4.3 |
| Sandstone | 1.4 to 2.5 | 3.0 to 5.0 |
| Limestone | 2.4 to 3.1 | 4.2 to 5.8 |
| Granite | 3.0 to 3.7 | 4.9 to 5.9 |
| Basalt | 3.3 to 4.0 | 5.2 to 6.2 |
[Return to Figure \(\PageIndex{3}\)]
| Layer | Depth from surface (km) | S-Wave velocity (kilometers per second) | P-Wave velocity (kilometers per second) |
|---|---|---|---|
| Crust | 0 to 30 | 3.0 to 4.6 | 5.3 to 7.0 |
| Lithosphere | 30 to 100 | 4.6 to 5.8 | 7.0 to 8.7 |
| Asthenosphere | 100 to 250 | 5.0 to 5.9 | 7.8 to 8.5 |
| Mantle | 250 to 2890 | 5. to 7.0 | 8.2 to 12.6 |
| Outer core | 2890 to 5100 | 0 | 8.0 to 10.1 |
| Inner core | 5100 to 6370 | 0 | 11.8 to 12.0 |
Media Attributions
- Figures 9.1.1, 9.1.2, 9.1.4, 9.1.5, 9.1.6, 9.1.7: © Steven Earle. CC BY.
- Figure \(\PageIndex{3}\): “P Wave Velocity, m/s” and “Shear Wave Velocity, m/s” by the US Environment Protection Agency. Edited by Steven Earle. Public domain.
- Figure \(\PageIndex{8}\): “P-wave Tomography” by D. Zhao, Y. Xu, D.A. Wiens, L. Dorman, J. Hildebrand, and S. Webb. (Science, p. 278, 254-257, 1997). Used with permission.


3.3: Market Equilibrium
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Clearing the Market at Equilibrium Price and Quantity
When a market achieves perfect equilibrium there is no excess supply or demand, which theoretically results in a market clearing.
Learning objectives
Define market equilibrium
The interdependent relationship between supply and demand in the field of economics is inherently designed to identify the ideal price and quantity of a given product or service in a marketplace. This equilibrium point is represented by the intersection of a downward sloping demand line and an upward sloping supply line, with price as the y-axis and quantity as the x-axis. At perfect equilibrium there is no excess demand (represented by ‘A’ in the figure) or excess supply (represented by ‘B’ in the figure), which theoretically results in a market clearing.
Equilibrium Pricing : This chart effectively highlights the various basic implications of a simple supply and demand chart. The equilibrium point is where market clearing will theoretically occur.
Market Clearing Assumptions
A market clearing, by definition, is the economic assumption that the quantity supplied will consistently align with the quantity demanded. This definition requires a variety of assumptions which simplify the complexities of real markets to coincide with a more theoretical framework, most centrally the assumptions of perfect competition and Say’s Law:
- Perfect competition is a market where the price determined for a given good or service is not affected by external forces or competition in a way that allows incumbents (companies) to attain market influence.
- Say’s Law hinges on the concept that capital loses value over time, or that money is essentially perishable. The simplest way to view this law is interest rates. When you invest or owe money, that capital accrues interest due to the fact that there is an opportunity cost in not investing that money elsewhere. This opportunity cost creates the assumption that money will not go unused.
Combining these two assumptions, in a perfectly competitive market the amount of a product or service that is supplied at a given price will equate to the amount demanded, clearing the market of all goods/services at a given equilibrium point.
Theory and Practice
While this concept of market clearing resonates well in theory, the actual execution of markets is very rarely perfect. Markets demonstrate consistent shifts of supply and shifts of demand based on a wide spectrum of externalities. Even in static markets there is competitive consolidation that allows companies to charge differing price points than that of the equilibrium. The concept of monopolies provides a good example for this experience, as monopolies (see example) can control price and quantity simultaneously.
Another classic criticism of market clearing is the way in which the labor market functions. In the 1930’s, during the worst depression recorded in the United States, the labor market did not clear the way economic theories of market clearing would assume it would. Instead, there seemed to be what John Maynard-Keynes (father of Keynesian Economics) called ‘stickiness,’ which preventing the market from normalizing. The importance of raising these concerns is the understanding that while the concept of market clearing, equilibrium and supply/demand charts are highly useful in understanding the basic functioning of markets, reality does not always conform with these models.
Impacts of Surpluses and Shortages on Market Equilibrium
The existence of surpluses or shortages in supply will result in disequilibrium, or a lack of balance between supply and demand levels.
Learning objectives
Infer the outcomes of departures from equilibrium using the model of supply and demand
In the analysis of market equilibrium, specifically for pricing and volume determinations, a thorough understanding of the supply and demand inputs is critical to economics. Surpluses and shortages on the supply end can have substantial impacts on both the pricing of a specific product or service, alongside the overall quantity sold over time. Shifts such as these in the supply availability results in disequilibrium, or essentially a lack of balance between current supply and demand levels. Surpluses and shortages often result in market inefficiencies due to a shifting market equilibrium.
Surpluses
Surpluses, or excess supply, indicate that the quantity of a good or service exceeds the demand for that particular good at the price in which the producers would wish to sell (equilibrium level). This inefficiency is heavily correlated in circumstances where the price of a good is set too high, resulting in a diminished demand while the quantity available gains excess. There are substantial business risks inherently built into the concept of surpluses, as the general outcome will be either selling off inventory at sub-par prices or leftover unsold inventory. In both scenarios businesses will be forced to minimize margins or incorporate losses on that particular good. Governmental intervention can often create surplus as well, particularly through the utilization of a price floor if it is set at a price above the market equilibrium.
Price Floor : A price floor ensures a minimum price is charged for a specific good, often higher than that what the previous market equilibrium determined. This can result in a surplus.
In a perfectly competitive market, particularly pertaining to goods that are not perishable, excess supply is equivalent to the quantity available in the market beyond the equilibrium point of intersection between supply and demand. In this theoretical scenario the equilibrium point will transition towards a lower price point due to the increased supply, which will in turn motivate consumers to purchase a higher quantity as a result. This allows the economic model of the market to correct itself.
Shortages
Inversely, shortage is a term used to indicate that the supply produced is below that of the quantity being demanded by the consumers. This disparity implies that the current market equilibrium at a given price is unfit for the current supply and demand relationship, noting that the price is set too low. It could also indicate that the desired good has a low level of affordability by the general public, and can be a dangerous societal risk for necessary commodities. Indeed, Garrett Hardin emphasized that a shortage of supply could also be perceived as a ‘longage’ of demand, as the two are inversely related. From this vantage point shortages can be attributed to population growth as much as resource scarcity.
In a perfectly competitive market, a shortage in supply will ultimately result in a shift in the equilibrium point, transitioning towards a higher price point due to the limited supply availability. This will prioritize who receives the good or service based upon their willingness and ability to pay a premium for the specific item in demand, leveraging those along the demand curve who are at higher levels with higher ability and willingness to pay.
Changes in Demand and Supply and Impacts on Equilibrium
Alterations to overall supply or demand dictate the cross-section or equilibrium, ascertaining price and volume for a product or service.
Learning objectives
Illustrate how changes in supply or demand impact the market equilibrium
The interdependent relationship between the supply of a given product or service and the overall demand exercised by interested parties generates a theoretical equilibrium point, dictating the average market price and purchased volume relative to that price. In a static market it would be reasonable to assume that prices and volumes would remain fairly predictable and consistent relative to the population, but realistic markets are not static. Instead, markets are in constant flux as demands and supplies are subjected to varying driving forces and influences. These shifts play a critical role in altering market equilibrium price points and volumes for products and services, requiring constant vigilance and adaptation by providers and consumers. To better understand market variations, it is useful to examine how changes in supply and demand may occur, as well as the impacts and implications of these changes.
Demand Shifts
Demand shifts are defined by more or less of a given product or service being required at a fixed price, resulting in a shift of both price and quantity. As would be assumed, an increase in demand will shift price upwards and volume to the right, increasing the overall value of both metrics relative to the prior equilibrium point. Alternately, a decrease in demand will shift price downwards and volume to the left, decreasing both measurements to realign equilibrium with a reduced demand.
Demand Shifts : In this graph, the demand curve (red) has been affected by an increase in demand. This consequently increases price at a given volume.
Demand shifts can be caused by a wide variety of factors, but largely revolve around drivers of consumer behavior and circumstances. Demand shifts can therefore often be affected by economic factors such as average spending power per person in a given economy or overall average income. Demand can also be affected by cultural changes, demographic shifts, availability of substitutes, environmental factors and concerns (e.g. climate change), politics, and advances in science (e.g. declining demand for unhealthy foods). Demand is particularly malleable in respect to goods that are not necessities, thus are desired or not based upon sociological norms.
Supply Shifts
Supply shifts are defined by more or less of a particular product/service being available to fulfill a given demand, affecting the equilibrium point by shifting the supply curve upwards or downwards. A supply shift to the right, indicating more availability of the specified product or service, will create a lower price point and a higher volume assuming a fixed demand. Alternately, a decrease in supply with a consistent given demand will see an increase in price and a decrease in quantity. This is an intuitive theory underlining the fact that scarcity is relevant to the willingness to pay.
Supply Shifts : In this supply and demand chart we see an increase in the supply provided, shifting quantity to the right and price down. More of a given product, assuming the same demand, will result in lower price points at the equilibrium.
Supply shifts, similar to demand shifts, can ultimately be a result of a wide variety of external factors. As discussed above, scarcity plays a critical role in pricing and thus controlling supply is often even considered a strategic play by companies in specific industries (most notably industries like precious stones, rare earth metals, etc.). Supply shifts can also be a result of technological advances, over-utilization or consumption, globalization, supply-chain efficiency, and economics. For example, the discovery of a new gold deposit, acts as a shock to the supply of gold, shifting the curve right.
Equilibrium
In combining these two potential shifts, equilibrium is constantly subjected to both factors resulting in supply shifts and factors resulting in demand shifts. Due to the demand curve sloping downward and the supply curve sloping upwards, they inadvertently will cross at some given point on any supply/demand chart. This cross-section, or equilibrium, serves as a price and quantity tracking point based upon the consistent inputs of overall demand and supply availability. Any change in either factor will result in immediate impact on equilibrium, balancing the new demand or supply with a corresponding volume and appropriate average price point.
Key Points
- The interdependent relationship between supply and demand in the field of economics is inherently designed to identify the ideal price and quantity of a given product or service in a marketplace.
- A market clearing, by definition, is the economic assumption that the quantity supplied will consistently align with the quantity demanded.
- Market clearing requires a variety of assumptions which simplify the complexities of real markets to coincide with a more theoretical framework, most centrally the assumptions of perfect competition and Say’s Law.
- While this concept of market clearing resonates well in theory, the actual execution of markets is very rarely perfect. The concepts of consolidated markets and ‘sticky’ markets reduces the accuracy of these models.
- Surpluses, or excess supply, essentially indicates that the quantity of a good or service exceeds the demand for that particular good at the price in which the producers would wish to sell ( equilibrium level).
- In a perfectly competitive market, excess supply is equivalent to the quantity available in the market beyond the equilibrium point of intersection between supply and demand. This will result in a shift in market equilibrium towards lower price points.
- Shortage is a term used to indicate that the supply produced is below that of the quantity being demanded by the consumers. This disparity implies that the current market equilibrium at a given price is unfit for the current supply and demand relationship.
- In a perfectly competitive market, a shortage in supply will ultimately result in a shift in the equilibrium point, transitioning towards a higher price point due to the limited supply availability.
- The interdependent relationship between the supply of a given product or service and the overall demand exercised by interested parties generates a theoretical equilibrium point, dictating the average market price and purchase volume relative to that price.
- Markets are in constant flux as demands and supplies are subjected to varying driving forces and influences. These shifts play a critical role, altering market equilibrium price points and volumes for products and services.
- Demand shifts can be caused by a wide variety of factors, but largely revolve around drivers of consumer behavior and circumstances.
- Supply shifts, similar to demand shifts, can ultimately be a result of a wide variety of externalities. Scarcity, or the lack of availability for a particular material, is a core driving force for overall supply.
- Due to a demand curve ‘s sloping downward and a supply curve ‘s sloping upwards, the curves will eventually cross at some point on any supply/demand chart. This point of equilibrium serves as a price and quantity tracking point.
Key Terms
- Say’s Law : The idea that money is perishable.
- Incumbents : A holder of a position as supplier to a market or market segment that allows the holder to earn above-normal profits.
- Opportunity cost : The cost of an opportunity forgone (and the loss of the benefits that could be received from that opportunity); the most valuable forgone alternative.
- Disequilibrium : The loss of equilibrium or stability, especially due to an imbalance of forces.
- surplus : That which remains when use or need is satisfied, or when a limit is reached.
- shortage : Not enough or not sufficient for a given demand.
- scarcity : An insufficiency or lack of availability; a shortage.
- equilibrium : A condition in which competing forces are in balance.
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- Perfect competition. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Perfect_competition . License : CC BY-SA: Attribution-ShareAlike
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17.E: Sound (Exercises)
Conceptual Questions
17.1 Sound Waves
- What is the difference between sound and hearing?
- You will learn that light is an electromagnetic wave that can travel through a vacuum. Can sound waves travel through a vacuum?
- Sound waves can be modeled as a change in pressure. Why is the change in pressure used and not the actual pressure?
17.2 Speed of Sound
- How do sound vibrations of atoms differ from thermal motion?
- When sound passes from one medium to another where its propagation speed is different, does its frequency or wavelength change? Explain your answer briefly.
- A popular party trick is to inhale helium and speak in a high-frequency, funny voice. Explain this phenomenon.
- You may have used a sonic range finder in lab to measure the distance of an object using a clicking sound from a sound transducer. What is the principle used in this device?
- The sonic range finder discussed in the preceding question often needs to be calibrated. During the calibration, the software asks for the room temperature. Why do you suppose the room temperature is required?
17.3 Sound Intensity
- Six members of a synchronized swim team wear earplugs to protect themselves against water pressure at depths, but they can still hear the music and perform the combinations in the water perfectly. One day, they were asked to leave the pool so the dive team could practice a few dives, and they tried to practice on a mat, but seemed to have a lot more difficulty. Why might this be?
- A community is concerned about a plan to bring train service to their downtown from the town’s outskirts. The current sound intensity level, even though the rail yard is blocks away, is 70 dB downtown. The mayor assures the public that there will be a difference of only 30 dB in sound in the downtown area. Should the townspeople be concerned? Why?
17.4 Normal Modes of a Standing Sound Wave
- You are given two wind instruments of identical length. One is open at both ends, whereas the other is closed at one end. Which is able to produce the lowest frequency?
- What is the difference between an overtone and a harmonic? Are all harmonics overtones? Are all overtones harmonics?
- Two identical columns, open at both ends, are in separate rooms. In room A, the temperature is T = 20 °C and in room B, the temperature is T = 25 °C. A speaker is attached to the end of each tube, causing the tubes to resonate at the fundamental frequency. Is the frequency the same for both tubes? Which has the higher frequency?
17.5 Sources of Musical Sound
- 14. How does an unamplified guitar produce sounds so much more intense than those of a plucked string held taut by a simple stick?
- Consider three pipes of the same length (L). Pipe A is open at both ends, pipe B is closed at both ends, and pipe C has one open end and one closed end. If the velocity of sound is the same in each of the three tubes, in which of the tubes could the lowest fundamental frequency be produced? In which of the tubes could the highest fundamental frequency be produced?
- Pipe A has a length L and is open at both ends. Pipe B has a length \(\frac{L}{2}\) and has one open end and one closed end. Assume the speed of sound to be the same in both tubes. Which of the harmonics in each tube would be equal?
- A string is tied between two lab posts a distance L apart. The tension in the string and the linear mass density is such that the speed of a wave on the string is v = 343 m/s. A tube with symmetric boundary conditions has a length L and the speed of sound in the tube is v = 343 m/s. What could be said about the frequencies of the harmonics in the string and the tube? What if the velocity in the string were v = 686 m/s?
17.6 Beats
- Two speakers are attached to variable-frequency signal generator. Speaker A produces a constant-frequency sound wave of 1.00 kHz, and speaker B produces a tone of 1.10 kHz. The beat frequency is 0.10 kHz. If the frequency of each speaker is doubled, what is the beat frequency produced?
- The label has been scratched off a tuning fork and you need to know its frequency. From its size, you suspect that it is somewhere around 250 Hz. You find a 250-Hz tuning fork and a 270-Hz tuning fork. When you strike the 250-Hz fork and the fork of unknown frequency, a beat frequency of 5 Hz is produced. When you strike the unknown with the 270-Hz fork, the beat frequency is 15 Hz. What is the unknown frequency? Could you have deduced the frequency using just the 250-Hz fork?
- Referring to the preceding question, if you had only the 250-Hz fork, could you come up with a solution to the problem of finding the unknown frequency? 21. A “showy” custom-built car has two brass horns that are supposed to produce the same frequency but actually emit 263.8 and 264.5 Hz. What beat frequency is produced?
17.7 The Doppler Effect
- Is the Doppler shift real or just a sensory illusion?
- Three stationary observers observe the Doppler shift from a source moving at a constant velocity. The observers are stationed as shown below. Which observer will observe the highest frequency? Which observer will observe the lowest frequency? What can be said about the frequency observed by observer 3?
- Shown below is a stationary source and moving observers. Describe the frequencies observed by the observers for this configuration
- Prior to 1980, conventional radar was used by weather forecasters. In the 1960s, weather forecasters began to experiment with Doppler radar. What do you think is the advantage of using Doppler radar?
17.8 Shock Waves
- What is the difference between a sonic boom and a shock wave?
- Due to efficiency considerations related to its bow wake, the supersonic transport aircraft must maintain a cruising speed that is a constant ratio to the speed of sound (a constant Mach number). If the aircraft flies from warm air into colder air, should it increase or decrease its speed? Explain your answer.
- When you hear a sonic boom, you often cannot see the plane that made it. Why is that?
Problems
- Consider a sound wave modeled with the equation s(x, t) = 4.00 nm cos(3.66 m −1 x − 1256 s −1 t). What is the maximum displacement, the wavelength, the frequency, and the speed of the sound wave?
- Consider a sound wave moving through the air modeled with the equation s(x, t) = 6.00 nm cos(54.93 m −1 x − 18.84 x 10 3 s −1 t). What is the shortest time required for an air molecule to move between 3.00 nm and –3.00 nm?
- Consider a diagnostic ultrasound of frequency 5.00 MHz that is used to examine an irregularity in soft tissue. (a) What is the wavelength in air of such a sound wave if the speed of sound is 343 m/s? (b) If the speed of sound in tissue is 1800 m/s, what is the wavelength of this wave in tissue?
- A sound wave is modeled as \(\Delta\)P = 1.80 Pa sin(55.41 m −1 x − 18,840 s −1 t). What is the maximum change in pressure, the wavelength, the frequency, and the speed of the sound wave?
- A sound wave is modeled with the wave function \(\Delta\)P = 1.20 Pa sin(kx − 6.28 x 10 4 s −1 t) and the sound wave travels in air at a speed of v = 343.00 m/s. (a) What is the wave number of the sound wave? (b) What is the value for \(\Delta\)P(3.00 m, 20.00 s)?
- The displacement of the air molecules in sound wave is modeled with the wave function s(x, t) = 5.00 nm cos(91.54 m −1 x − 3.14 x 10 4 s −1 t). (a) What is the wave speed of the sound wave? (b) What is the maximum speed of the air molecules as they oscillate in simple harmonic motion? (c) What is the magnitude of the maximum acceleration of the air molecules as they oscillate in simple harmonic motion?
- A speaker is placed at the opening of a long horizontal tube. The speaker oscillates at a frequency f, creating a sound wave that moves down the tube. The wave moves through the tube at a speed of v = 340.00 m/s. The sound wave is modeled with the wave function s(x, t) = s max cos(kx − \(\omega\)t + \(\phi\). At time t = 0.00 s, an air molecule at x = 3.5 m is at the maximum displacement of 7.00 nm. At the same time, another molecule at x = 3.7 m has a displacement of 3.00 nm. What is the frequency at which the speaker is oscillating?
- A 250-Hz tuning fork is struck and begins to vibrate. A sound-level meter is located 34.00 m away. It takes the sound \(\Delta\)t = 0.10 s to reach the meter. The maximum displacement of the tuning fork is 1.00 mm. Write a wave function for the sound.
- A sound wave produced by an ultrasonic transducer, moving in air, is modeled with the wave equation s(x, t) = 4.50 nm cos(9.15 x 10 4 m −1 x − 2\(\pi\)(5.00 MHz)t). The transducer is to be used in nondestructive testing to test for fractures in steel beams. The speed of sound in the steel beam is v = 5950 m/s. Find the wave function for the sound wave in the steel beam.
- Porpoises emit sound waves that they use for navigation. If the wavelength of the sound wave emitted is 4.5 cm, and the speed of sound in the water is v = 1530 m/s, what is the period of the sound?
- Bats use sound waves to catch insects. Bats can detect frequencies up to 100 kHz. If the sound waves travel through air at a speed of v = 343 m/s, what is the wavelength of the sound waves?
- A bat sends of a sound wave 100 kHz and the sound waves travel through air at a speed of v = 343 m/s. (a) If the maximum pressure difference is 1.30 Pa, what is a wave function that would model the sound wave, assuming the wave is sinusoidal? (Assume the phase shift is zero.) (b) What are the period and wavelength of the sound wave?
- Consider the graph shown below of a compression wave. Shown are snapshots of the wave function for t = 0.000 s (blue) and t = 0.005 s (orange). What are the wavelength, maximum displacement, velocity, and period of the compression wave?
- Consider the graph in the preceding problem of a compression wave. Shown are snapshots of the wave function for t = 0.000 s (blue) and t = 0.005 s (orange). Given that the displacement of the molecule at time t = 0.00 s and position x = 0.00 m is s(0.00 m, 0.00 s) = 1.08 mm, derive a wave function to model the compression wave.
- A guitar string oscillates at a frequency of 100 Hz and produces a sound wave. (a) What do you think the frequency of the sound wave is that the vibrating string produces? (b) If the speed of the sound wave is v = 343 m/s, what is the wavelength of the sound wave?
17.2 Speed of Sound
- When poked by a spear, an operatic soprano lets out a 1200-Hz shriek. What is its wavelength if the speed of sound is 345 m/s?
- What frequency sound has a 0.10-m wavelength when the speed of sound is 340 m/s?
- Calculate the speed of sound on a day when a 1500-Hz frequency has a wavelength of 0.221 m.
- (a) What is the speed of sound in a medium where a 100-kHz frequency produces a 5.96-cm wavelength? (b) Which substance in Table 17.1 is this likely to be?
- Show that the speed of sound in 20.0 °C air is 343 m/s, as claimed in the text.
- Air temperature in the Sahara Desert can reach 56.0 °C (about 134 °F). What is the speed of sound in air at that temperature?
- Dolphins make sounds in air and water. What is the ratio of the wavelength of a sound in air to its wavelength in seawater? Assume air temperature is 20.0 °C.
- A sonar echo returns to a submarine 1.20 s after being emitted. What is the distance to the object creating the echo? (Assume that the submarine is in the ocean, not in fresh water.)
- (a) If a submarine’s sonar can measure echo times with a precision of 0.0100 s, what is the smallest difference in distances it can detect? (Assume that the submarine is in the ocean, not in fresh water.) (b) Discuss the limits this time resolution imposes on the ability of the sonar system to detect the size and shape of the object creating the echo.
- Ultrasonic sound waves are often used in methods of nondestructive testing. For example, this method can be used to find structural faults in a steel I-beams used in building. Consider a 10.00 meter long, steel I-beam with a cross-section shown below. The weight of the I-beam is 3846.50 N. What would be the speed of sound through in the I-beam? (Y steel = 200 GPa, \(\beta_{steel}\) = 159 GPa).
- A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.400 s. (a) How far away is the explosion if air temperature is 24.0 °C and if you neglect the time taken for light to reach the physicist? (b) Calculate the distance to the explosion taking the speed of light into account. Note that this distance is negligibly greater.
- During a 4th of July celebration, an M80 firework explodes on the ground, producing a bright flash and a loud bang. The air temperature of the night air is T F = 90.00 °F. Two observers see the flash and hear the bang. The first observer notes the time between the flash and the bang as 1.00 second. The second observer notes the difference as 3.00 seconds. The line of sight between the two observers meet at a right angle as shown below. What is the distance \(\Delta\)x between the two observers?
- The density of a sample of water is \(\rho\) = 998.00 kg/m 3 and the bulk modulus is \(\beta\) = 2.15 GPa. What is the speed of sound through the sample?
- Suppose a bat uses sound echoes to locate its insect prey, 3.00 m away. (See Figure 17.6.) (a) Calculate the echo times for temperatures of 5.00 °C and 35.0 °C. (b) What percent uncertainty does this cause for the bat in locating the insect? (c) Discuss the significance of this uncertainty and whether it could cause difficulties for the bat. (In practice, the bat continues to use sound as it closes in, eliminating most of any difficulties imposed by this and other effects, such as motion of the prey.)
17.3 Sound Intensity
- What is the intensity in watts per meter squared of a 85.0-dB sound?
- The warning tag on a lawn mower states that it produces noise at a level of 91.0 dB. What is this in watts per meter squared?
- A sound wave traveling in air has a pressure amplitude of 0.5 Pa. What is the intensity of the wave?
- What intensity level does the sound in the preceding problem correspond to?
- What sound intensity level in dB is produced by earphones that create an intensity of 4.00 x 10 −2 W/m 2 ?
- What is the decibel level of a sound that is twice as intense as a 90.0-dB sound? (b) What is the decibel level of a sound that is one-fifth as intense as a 90.0-dB sound?
- What is the intensity of a sound that has a level 7.00 dB lower than a 4.00 x 10 −9 W/m 2 sound? (b) What is the intensity of a sound that is 3.00 dB higher than a 4.00 x 10 −9 -W/m 2 sound?
- People with good hearing can perceive sounds as low as −8.00 dB at a frequency of 3000 Hz. What is the intensity of this sound in watts per meter squared?
- If a large housefly 3.0 m away from you makes a noise of 40.0 dB, what is the noise level of 1000 flies at that distance, assuming interference has a negligible effect?
- Ten cars in a circle at a boom box competition produce a 120-dB sound intensity level at the center of the circle. What is the average sound intensity level produced there by each stereo, assuming interference effects can be neglected?
- The amplitude of a sound wave is measured in terms of its maximum gauge pressure. By what factor does the amplitude of a sound wave increase if the sound intensity level goes up by 40.0 dB?
- If a sound intensity level of 0 dB at 1000 Hz corresponds to a maximum gauge pressure (sound amplitude) of 10 −9 atm, what is the maximum gauge pressure in a 60-dB sound? What is the maximum gauge pressure in a 120-dB sound?
- An 8-hour exposure to a sound intensity level of 90.0 dB may cause hearing damage. What energy in joules falls on a 0.800-cm-diameter eardrum so exposed?
- Sound is more effectively transmitted into a stethoscope by direct contact rather than through the air, and it is further intensified by being concentrated on the smaller area of the eardrum. It is reasonable to assume that sound is transmitted into a stethoscope 100 times as effectively compared with transmission though the air. What, then, is the gain in decibels produced by a stethoscope that has a sound gathering area of 15.0 cm 2 , and concentrates the sound onto two eardrums with a total area of 0.900 cm 2 with an efficiency of 40.0%?
- Loudspeakers can produce intense sounds with surprisingly small energy input in spite of their low efficiencies. Calculate the power input needed to produce a 90.0-dB sound intensity level for a 12.0-cm-diameter speaker that has an efficiency of 1.00%. (This value is the sound intensity level right at the speaker.)
- The factor of 10 -12 in the range of intensities to which the ear can respond, from threshold to that causing damage after brief exposure, is truly remarkable. If you could measure distances over the same range with a single instrument and the smallest distance you could measure was 1 mm, what would the largest be?
- What are the closest frequencies to 500 Hz that an average person can clearly distinguish as being different in frequency from 500 Hz? The sounds are not present simultaneously.
- Can you tell that your roommate turned up the sound on the TV if its average sound intensity level goes from 70 to 73 dB?
- If a woman needs an amplification of 5.0 x 10 5 times the threshold intensity to enable her to hear at all frequencies, what is her overall hearing loss in dB? Note that smaller amplification is appropriate for more intense sounds to avoid further damage to her hearing from levels above 90 dB.
- A person has a hearing threshold 10 dB above normal at 100 Hz and 50 dB above normal at 4000 Hz. How much more intense must a 100-Hz tone be than a 4000-Hz tone if they are both barely audible to this person?
17.4 Normal Modes of a Standing Sound Wave
- (a) What is the fundamental frequency of a 0.672-mlong tube, open at both ends, on a day when the speed of sound is 344 m/s? (b) What is the frequency of its second harmonic?
- What is the length of a tube that has a fundamental frequency of 176 Hz and a first overtone of 352 Hz if the speed of sound is 343 m/s?
- The ear canal resonates like a tube closed at one end. (See Figure 17.19 .) If ear canals range in length from 1.80 to 2.60 cm in an average population, what is the range of fundamental resonant frequencies? Take air temperature to be 37.0 °C, which is the same as body temperature.
- Calculate the first overtone in an ear canal, which resonates like a 2.40-cm-long tube closed at one end, by taking air temperature to be 37.0 °C. Is the ear particularly sensitive to such a frequency? (The resonances of the ear canal are complicated by its nonuniform shape, which we shall ignore.)
- A crude approximation of voice production is to consider the breathing passages and mouth to be a resonating tube closed at one end. (a) What is the fundamental frequency if the tube is 0.240 m long, by taking air temperature to be 37.0 °C ? (b) What would this frequency become if the person replaced the air with helium? Assume the same temperature dependence for helium as for air.
- A 4.0-m-long pipe, open at one end and closed at one end, is in a room where the temperature is T = 22 °C. A speaker capable of producing variable frequencies is placed at the open end and is used to cause the tube to resonate. (a) What is the wavelength and the frequency of the fundamental frequency? (b) What is the frequency and wavelength of the first overtone?
- A 4.0-m-long pipe, open at both ends, is placed in a room where the temperature is T = 25 °C. A speaker capable of producing variable frequencies is placed at the open end and is used to cause the tube to resonate. (a) What are the wavelength and the frequency of the fundamental frequency? (b) What are the frequency and wavelength of the first overtone?
- A nylon guitar string is fixed between two lab posts 2.00 m apart. The string has a linear mass density of \(\mu\) = 7.20 g/m and is placed under a tension of 160.00 N. The string is placed next to a tube, open at both ends, of length L. The string is plucked and the tube resonates at the n = 3 mode. The speed of sound is 343 m/s. What is the length of the tube?
- A 512-Hz tuning fork is struck and placed next to a tube with a movable piston, creating a tube with a variable length. The piston is slid down the pipe and resonance is reached when the piston is 115.50 cm from the open end. The next resonance is reached when the piston is 82.50 cm from the open end. (a) What is the speed of sound in the tube? (b) How far from the open end will the piston cause the next mode of resonance?
- Students in a physics lab are asked to find the length of an air column in a tube closed at one end that has a fundamental frequency of 256 Hz. They hold the tube vertically and fill it with water to the top, then lower the water while a 256-Hz tuning fork is rung and listen for the first resonance. (a) What is the air temperature if the resonance occurs for a length of 0.336 m? (b) At what length will they observe the second resonance (first overtone)?
17.5 Sources of Musical Sound
- If a wind instrument, such as a tuba, has a fundamental frequency of 32.0 Hz, what are its first three overtones? It is closed at one end. (The overtones of a real tuba are more complex than this example, because it is a tapered tube.)
- What are the first three overtones of a bassoon that has a fundamental frequency of 90.0 Hz? It is open at both ends. (The overtones of a real bassoon are more complex than this example, because its double reed makes it act more like a tube closed at one end.)
- How long must a flute be in order to have a fundamental frequency of 262 Hz (this frequency corresponds to middle C on the evenly tempered chromatic scale) on a day when air temperature is 20.0°C ? It is open at both ends.
- What length should an oboe have to produce a fundamental frequency of 110 Hz on a day when the speed of sound is 343 m/s? It is open at both ends.
- (a) Find the length of an organ pipe closed at one end that produces a fundamental frequency of 256 Hz when air temperature is 18.0 °C. (b) What is its fundamental frequency at 25.0 °C?
- An organ pipe (L = 3.00 m) is closed at both ends. Compute the wavelengths and frequencies of the first three modes of resonance. Assume the speed of sound is v = 343.00 m/s.
- An organ pipe (L = 3.00 m) is closed at one end. Compute the wavelengths and frequencies of the first three modes of resonance. Assume the speed of sound is v = 343.00 m/s.
- A sound wave of a frequency of 2.00 kHz is produced by a string oscillating in the n = 6 mode. The linear mass density of the string is \(\mu\) = 0.0065 kg/m and the length of the string is 1.50 m. What is the tension in the string?
- Consider the sound created by resonating the tube shown below. The air temperature is T C = 30.00 °C. What are the wavelength, wave speed, and frequency of the sound produced?
- A student holds an 80.00-cm lab pole one quarter of the length from the end of the pole. The lab pole is made of aluminum. The student strikes the lab pole with a hammer. The pole resonates at the lowest possible frequency. What is that frequency?
- A string on the violin has a length of 24.00 cm and a mass of 0.860 g. The fundamental frequency of the string is 1.00 kHz. (a) What is the speed of the wave on the string? (b) What is the tension in the string?
- By what fraction will the frequencies produced by a wind instrument change when air temperature goes from 10.0 °C to 30.0 °C? That is, find the ratio of the frequencies at those temperatures.
17.6 Beats
- What beat frequencies are present: (a) If the musical notes A and C are played together (frequencies of 220 and 264 Hz)? (b) If D and F are played together (frequencies of 297 and 352 Hz)? (c) If all four are played together?
- What beat frequencies result if a piano hammer hits three strings that emit frequencies of 127.8, 128.1, and 128.3 Hz?
- A piano tuner hears a beat every 2.00 s when listening to a 264.0-Hz tuning fork and a single piano string. What are the two possible frequencies of the string?
- Two identical strings, of identical lengths of 2.00 m and linear mass density of \(\mu\) = 0.0065 kg/m, are fixed on both ends. String A is under a tension of 120.00 N. String B is under a tension of 130.00 N. They are each plucked and produce sound at the n = 10 mode. What is the beat frequency?
- A piano tuner uses a 512-Hz tuning fork to tune a piano. He strikes the fork and hits a key on the piano and hears a beat frequency of 5 Hz. He tightens the string of the piano, and repeats the procedure. Once again he hears a beat frequency of 5 Hz. What happened?
- A string with a linear mass density of \(\mu\) = 0.0062 kg/m is stretched between two posts 1.30 m apart. The tension in the string is 150.00 N. The string oscillates and produces a sound wave. A 1024-Hz tuning fork is struck and the beat frequency between the two sources is 52.83 Hz. What are the possible frequency and wavelength of the wave on the string?
- A car has two horns, one emitting a frequency of 199 Hz and the other emitting a frequency of 203 Hz. What beat frequency do they produce?
- The middle C hammer of a piano hits two strings, producing beats of 1.50 Hz. One of the strings is tuned to 260.00 Hz. What frequencies could the other string have?
- Two tuning forks having frequencies of 460 and 464 Hz are struck simultaneously. What average frequency will you hear, and what will the beat frequency be?
- Twin jet engines on an airplane are producing an average sound frequency of 4100 Hz with a beat frequency of 0.500 Hz. What are their individual frequencies?
- Three adjacent keys on a piano (F, F-sharp, and G) are struck simultaneously, producing frequencies of 349, 370, and 392 Hz. What beat frequencies are produced by this discordant combination?
17.7 The Doppler Effect
- (a) What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound on this day is 345 m/s. (b) What frequency does she receive after the ambulance has passed?
- (a) At an air show a jet flies directly toward the stands at a speed of 1200 km/h, emitting a frequency of 3500 Hz, on a day when the speed of sound is 342 m/s. What frequency is received by the observers? (b) What frequency do they receive as the plane flies directly away from them?
- What frequency is received by a mouse just before being dispatched by a hawk flying at it at 25.0 m/s and emitting a screech of frequency 3500 Hz? Take the speed of sound to be 331 m/s.
- A spectator at a parade receives an 888-Hz tone from an oncoming trumpeter who is playing an 880-Hz note. At what speed is the musician approaching if the speed of sound is 338 m/s?
- A commuter train blows its 200-Hz horn as it approaches a crossing. The speed of sound is 335 m/s. (a) An observer waiting at the crossing receives a frequency of 208 Hz. What is the speed of the train? (b) What frequency does the observer receive as the train moves away?
- Can you perceive the shift in frequency produced when you pull a tuning fork toward you at 10.0 m/s on a day when the speed of sound is 344 m/s? To answer this question, calculate the factor by which the frequency shifts and see if it is greater than 0.300%.
- Two eagles fly directly toward one another, the first at 15.0 m/s and the second at 20.0 m/s. Both screech, the first one emitting a frequency of 3200 Hz and the second one emitting a frequency of 3800 Hz. What frequencies do they receive if the speed of sound is 330 m/s?
- Student A runs down the hallway of the school at a speed of v o = 5.00 m/s, carrying a ringing 1024.00-Hz tuning fork toward a concrete wall. The speed of sound is v = 343.00 m/s. Student B stands at rest at the wall. (a) What is the frequency heard by student B? (b) What is the beat frequency heard by student A?
- An ambulance with a siren (f = 1.00kHz) blaring is approaching an accident scene. The ambulance is moving at 70.00 mph. A nurse is approaching the scene from the opposite direction, running at v o = 7.00 m/s. What frequency does the nurse observe? Assume the speed of sound is v = 343.00 m/s.
- The frequency of the siren of an ambulance is 900 Hz and is approaching you.  What is the speed of the ambulance (in mph) if the speed of sound is v = 340.00 m/s?
- What is the minimum speed at which a source must travel toward you for you to be able to hear that its frequency is Doppler shifted? That is, what speed produces a shift of 0.300% on a day when the speed of sound is 331 m/s?
17.8 Shock Waves
- An airplane is flying at Mach 1.50 at an altitude of 7500.00 meters, where the speed of sound is v = 343.00 m/s. How far away from a stationary observer will the plane be when the observer hears the sonic boom?
- A jet flying at an altitude of 8.50 km has a speed of Mach 2.00, where the speed of sound is v = 340.00 m/s. How long after the jet is directly overhead, will a stationary observer hear a sonic boom?
- The shock wave off the front of a fighter jet has an angle of \(\theta\) = 70.00°. The jet is flying at 1200 km/h. What is the speed of sound?
- A plane is flying at Mach 1.2, and an observer on the ground hears the sonic boom 15.00 seconds after the plane is directly overhead. What is the altitude of the plane? Assume the speed of sound is vw = 343.00 m/s.
- A bullet is fired and moves at a speed of 1342 mph. Assume the speed of sound is v = 340.00 m/s. What is the angle of the shock wave produced?
- A speaker is placed at the opening of a long horizontal tube. The speaker oscillates at a frequency of f, creating a sound wave that moves down the tube. The wave moves through the tube at a speed of v = 340.00 m/s. The sound wave is modeled with the wave function s(x, t) = s max cos(kx − \(\omega\)t + \(\phi\)). At time t = 0.00 s, an air molecule at x = 2.3 m is at the maximum displacement of 6.34 nm. At the same time, another molecule at x = 2.7 m has a displacement of 2.30 nm. What is the wave function of the sound wave, that is, find the wave number, angular frequency, and the initial phase shift?
- An airplane moves at Mach 1.2 and produces a shock wave. (a) What is the speed of the plane in meters per second? (b) What is the angle that the shock wave moves?
Additional Problems
- A 0.80-m-long tube is opened at both ends. The air temperature is 26 °C. The air in the tube is oscillated using a speaker attached to a signal generator. What are the wavelengths and frequencies of first two modes of sound waves that resonate in the tube?
- A tube filled with water has a valve at the bottom to allow the water to flow out of the tube. As the water is emptied from the tube, the length L of the air column changes. A 1024-Hz tuning fork is placed at the opening of the tube. Water is removed from the tube until the n = 5 mode of a sound wave resonates. What is the length of the air column if the temperature of the air in the room is 18 °C?
- Consider the following figure. The length of the string between the string vibrator and the pulley is L = 1.00 m. The linear density of the string is \(\mu\) = 0.006 kg/m. The string vibrator can oscillate at any frequency. The hanging mass is 2.00 kg. (a)What are the wavelength and frequency of n = 6 mode? (b) The string oscillates the air around the string. What is the wavelength of the sound if the speed of the sound is v s = 343.00 m/s?
- Early Doppler shift experiments were conducted using a band playing music on a train. A trumpet player on a moving railroad flatcar plays a 320-Hz note. The sound waves heard by a stationary observer on a train platform hears a frequency of 350 Hz. What is the flatcar’s speed in mph? The temperature of the air is T C = 22 °C.
- Two cars move toward one another, both sounding their horns (f s = 800 Hz). Car A is moving at 65 mph and Car B is at 75 mph. What is the beat frequency heard by each driver? The air temperature is T C = 22.00 °C.
- Student A runs after Student B. Student A carries a tuning fork ringing at 1024 Hz, and student B carries a tuning fork ringing at 1000 Hz. Student A is running at a speed of v A = 5.00 m/s and Student B is running at v B = 6.00 m/s. What is the beat frequency heard by each student? The speed of sound is v = 343.00 m/s.
- Suppose that the sound level from a source is 75 dB and then drops to 52 dB, with a frequency of 600 Hz. Determine the (a) initial and (b) final sound intensities and the (c) initial and (d) final sound wave amplitudes. The air temperature is T C = 24.00 °C and the air density is \(\rho\) = 1.184 kg/m 3 .
- The Doppler shift for a Doppler radar is found by f = f R \(\left(\dfrac{1 + v_{c}}{1 − v_{c}}\right)\), where f R is the frequency of the radar, f is the frequency observed by the radar, c is the speed of light, and v is the speed of the target. What is the beat frequency observed at the radar, assuming the speed of the target is much slower than the speed of light?
- A stationary observer hears a frequency of 1000.00 Hz as a source approaches and a frequency of 850.00 Hz as a source departs. The source moves at a constant velocity of 75 mph. What is the temperature of the air?
- A flute plays a note with a frequency of 600 Hz. The flute can be modeled as a pipe open at both ends, where the flute player changes the length with his finger positions. What is the length of the tube if this is the fundamental frequency?
Challenge Problems
- Two sound speakers are separated by a distance d, each sounding a frequency f. An observer stands at one speaker and walks in a straight line a distance x, perpendicular to the the two speakers, until he comes to the first maximum intensity of sound. The speed of sound is v. How far is he from the speaker?
- Consider the beats shown below. This is a graph of the gauge pressure versus time for the position x = 0.00 m. The wave moves with a speed of v = 343.00 m/s. (a) How many beats are there per second? (b) How many times does the wave oscillate per second? (c) Write a wave function for the gauge pressure as a function of time.
- Two speakers producing the same frequency of sound are a distance of d apart. Consider an arc along a circle of radius R, centered at the midpoint of the speakers, as shown below. (a) At what angles will there be maxima? (b) At what angle will there be minima?
- A string has a length of 1.5 m, a linear mass density \(\mu\) = 0.008 kg/m, and a tension of 120 N. If the air temperature is T = 22 °C, what should the length of a pipe open at both ends for it to have the same frequency for the n = 3 mode?
- A string (\(\mu\) = 0.006 kg/m, L = 1.50 m) is fixed at both ends and is under a tension of 155 N. It oscillates in the n = 10 mode and produces sound. A tuning fork is ringing nearby, producing a beat frequency of 23.76 Hz. (a) What is the frequency of the sound from the string? (b) What is the frequency of the tuning fork if the tuning fork frequency is lower? (c) What should be the tension of the string for the beat frequency to be zero?
- A string has a linear mass density \(\mu\), a length L, and a tension of F T , and oscillates in a mode n at a frequency f. Find the ratio of \(\frac{\Delta f}{f}\) for a small change in tension.
- A string has a linear mass density \(\mu\) = 0.007 kg/m, a length L = 0.70 m, a tension of F T = 110 N, and oscillates in a mode n = 3. (a) What is the frequency of the oscillations? (b) Use the result in the preceding problem to find the change in the frequency when the tension is increased by 1.00%.
- A speaker powered by a signal generator is used to study resonance in a tube. The signal generator can be adjusted from a frequency of 1000 Hz to 1800 Hz. First, a 0.75-m-long tube, open at both ends, is studied. The temperature in the room is T F = 85.00 °F. (a) Which normal modes of the pipe can be studied? What are the frequencies and wavelengths? Next a cap is place on one end of the 0.75-meter-long pipe. (b) Which normal modes of the pipe can be studied? What are the frequencies and wavelengths?
- A string on the violin has a length of 23.00 cm and a mass of 0.900 grams. The tension in the string 850.00 N. The temperature in the room is T C = 24.00 °C. The string is plucked and oscillates in the n = 9 mode. (a) What is the speed of the wave on the string? (b) What is the wavelength of the sounding wave produced? (c) What is the frequency of the oscillating string? (d) What is the frequency of the sound produced? (e) What is the wavelength of the sound produced?
Contributors and Attributions
Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a
Creative Commons Attribution License (by 4.0)
.


28.5B: Classes of Echinoderms
- Differentiate among the classes of echinoderms
The phylum echinoderms is divided into five extant classes: Asteroidea (sea stars), Ophiuroidea (brittle stars), Echinoidea (sea urchins and sand dollars), Crinoidea (sea lilies or feather stars), and Holothuroidea (sea cucumbers).
The most well-known echinoderms are members of class Asteroidea, or sea stars. They come in a large variety of shapes, colors, and sizes, with more than 1,800 species known so far. The key characteristic of sea stars that distinguishes them from other echinoderm classes includes thick arms (ambulacra; singular: ambulacrum) that extend from a central disk where organs penetrate into the arms. Sea stars use their tube feet not only for gripping surfaces, but also for grasping prey. Sea stars have two stomachs, one of which can protrude through their mouths and secrete digestive juices into or onto prey, even before ingestion. This process can essentially liquefy the prey, making digestion easier.
Brittle stars belong to the class Ophiuroidea. Unlike sea stars, which have plump arms, brittle stars have long, thin arms that are sharply demarcated from the central disk. Brittle stars move by lashing out their arms or wrapping them around objects and pulling themselves forward. Of all echinoderms, the Ophiuroidea may have the strongest tendency toward 5-segment radial (pentaradial) symmetry. Ophiuroids are generally scavengers or detritivores. Small organic particles are moved into the mouth by the tube feet. Ophiuroids may also prey on small crustaceans or worms. Some brittle stars, such as the six-armed members of the family Ophiactidae, are fissiparous (divide though fission), with the disk splitting in half. Regrowth of both the lost part of the disk and the arms occur, yielding an animal with three large arms and three small arms during the period of growth.
Sea urchins and sand dollars are examples of Echinoidea. These echinoderms do not have arms, but are hemispherical or flattened with five rows of tube feet that help them in slow movement; tube feet are extruded through pores of a continuous internal shell called a test. Like other echinoderms, sea urchins are bilaterans. Their early larvae have bilateral symmetry, but they develop fivefold symmetry as they mature. This is most apparent in the “regular” sea urchins, which have roughly spherical bodies, with five equally-sized parts radiating out from their central axes. Several sea urchins, however, including the sand dollars, are oval in shape, with distinct front and rear ends, giving them a degree of bilateral symmetry. In these urchins, the upper surface of the body is slightly domed, but the underside is flat, while the sides are devoid of tube feet. This “irregular” body form has evolved to allow the animals to burrow through sand or other soft materials.
Sea lilies and feather stars are examples of Crinoidea. Both of these species are suspension feeders. They live both in shallow water and in depths as great as 6,000 meters. Sea lilies refer to the crinoids which, in their adult form, are attached to the sea bottom by a stalk. Feather stars or comatulids refer to the unstalked forms. Crinoids are characterized by a mouth on the top surface that is surrounded by feeding arms. They have a U-shaped gut; their anus is located next to the mouth. Although the basic echinoderm pattern of fivefold symmetry can be recognized, most crinoids have many more than five arms. Crinoids usually have a stem used to attach themselves to a substrate, but many live attached only as juveniles and become free-swimming as adults.
Sea cucumbers of class Holothuroidea are extended in the oral-aboral axis and have five rows of tube feet. These are the only echinoderms that demonstrate “functional” bilateral symmetry as adults because the uniquely-extended oral-aboral axis compels the animal to lie horizontally rather than stand vertically. Like all echinoderms, sea cucumbers have an endoskeleton just below the skin: calcified structures that are usually reduced to isolated microscopic ossicles joined by connective tissue. In some species these can sometimes be enlarged to flattened plates, forming armor. In pelagic species, such as Pelagothuria natatrix, the skeleton and a calcareous ring are absent.
Key Points
- Sea stars have thick arms called ambulacra that are used for gripping surfaces and grabbing hold of prey.
- Brittle stars have thin arms that wrap around prey or objects to pull themselves forward.
- Sea urchins and sand dollars embody flattened discs that do not have arms, but do have rows of tube feet they use for movement.
- Sea cucumbers demonstrate “functional” bilateral symmetry as adults because they actually lie horizontally rather than stand vertically.
- Sea lilies and feather stars are suspension feeders.
Key Terms
- ossicle : a small bone (or bony structure), especially one of the three of the middle ear
- fissiparous : of cells that reproduce through fission, splitting into two
- ambulacrum : a row of pores for the protrusion of appendages such as tube feet.


2.1.1: Birth of the Solar System
Earth is the third planet from the Sun, one of eight "classical" planets recognized by the International Astronomical Union in our solar system . A planet must orbit the Sun, be big enough that gravity pulls it into a round ball, and must have other things out of its orbital path. Pluto, defined as a planet until 2006 but was declassified as it orbits among the icy bits of the asteroid belt. In 2008 it was classified as a "plutoid", a kind of dwarf planet .
It is thought that the planet like Earth originated from the accretion of bits of solid matter left from the massive explosion of a star or supernova. Bits of material were scattered into space forming a slowly rotating cosmic gas cloud. Gravity slowly gathered the thinly spread atoms of the cosmic cloud. As the atoms moved closer together the gas became hotter and more dense. A new sun was born as hydrogen eventually became so tightly compressed and temperatures so high that nuclear burning began. A flattened rotating disc of gas and dust surrounded the young sun. Outer cooler parts of the disc or star nebulae began to condense to form the building blocks of future planets.
Video : "Birth of the Solar System" Courtesy National Geographic


2.2: Histograms, Ogives, and Frequency Polygons
For most of the work you do in this book, you will use a histogram to display the data. One advantage of a histogram is that it can readily display large data sets. A rule of thumb is to use a histogram when the data set consists of 100 values or more.
A histogram consists of contiguous (adjoining) boxes. It has both a horizontal axis and a vertical axis. The horizontal axis is labeled with what the data represents (for instance, distance from your home to school). The vertical axis is labeled either frequency or relative frequency (or percent frequency or probability). The graph will have the same shape with either label. The histogram (like the stemplot) can give you the shape of the data, the center, and the spread of the data.
The graph for quantitative data looks similar to a bar graph, except there are some major differences. First, in a bar graph the categories can be put in any order on the horizontal axis. There is no set order for these data values. You can’t say how the data is distributed based on the shape, since the shape can change just by putting the categories in different orders. With quantitative data, the data are in specific orders, since you are dealing with numbers. With quantitative data, you can talk about a distribution, since the shape only changes a little bit depending on how many categories you set up. This is called a frequency distribution .
This leads to the second difference from bar graphs. In a bar graph, the categories that you made in the frequency table were determined by you. In quantitative data, the categories are numerical categories, and the numbers are determined by how many categories (or what are called classes) you choose. If two people have the same number of categories, then they will have the same frequency distribution. Whereas in qualitative data, there can be many different categories depending on the point of view of the author.
The third difference is that the categories touch with quantitative data, and there will be no gaps in the graph. The reason that bar graphs have gaps is to show that the categories do not continue on, like they do in quantitative data. Since the graph for quantitative data is different from qualitative data, it is given a new name. The name of the graph is a histogram . To create a histogram, you must first create the frequency distribution. The idea of a frequency distribution is to take the interval that the data spans and divide it up into equal subintervals called classes.
Summary of the steps involved in making a frequency distribution:
- Find the range = largest value – smallest value
- Pick the number of classes to use. Usually the number of classes is between five and twenty. Five classes are used if there are a small number of data points and twenty classes if there are a large number of data points (over 1000 data points). (Note: categories will now be called classes from now on.)
- Class width = \(\frac{\text { range }}{\# \text { classes }}\) Always round up to the next integer (even if the answer is already a whole number go to the next integer). If you don’t do this, your last class will not contain your largest data value, and you would have to add another class just for it. If you round up, then your largest data value will fall in the last class, and there are no issues.
- Create the classes. Each class has limits that determine which values fall in each class. To find the class limits, set the smallest value as the lower class limit for the first class. Then add the class width to the lower class limit to get the next lower class limit. Repeat until you get all the classes. The upper class limit for a class is one less than the lower limit for the next class.
- In order for the classes to actually touch, then one class needs to start where the previous one ends. This is known as the class boundary. To find the class boundaries, subtract 0.5 from the lower class limit and add 0.5 to the upper class limit.
-
Sometimes it is useful to find the class midpoint. The process is
Midpoint \(=\frac{\text { lower limit +upper limit }}{2}\) - To figure out the number of data points that fall in each class, go through each data value and see which class boundaries it is between. Utilizing tally marks may be helpful in counting the data values. The frequency for a class is the number of data values that fall in the class.
Note
The above description is for data values that are whole numbers. If you data value has decimal places, then your class width should be rounded up to the nearest value with the same number of decimal places as the original data. In addition, your class boundaries should have one more decimal place than the original data. As an example, if your data have one decimal place, then the class width would have one decimal place, and the class boundaries are formed by adding and subtracting 0.05 from each class limit.
Example \(\PageIndex{1}\) creating a frequency table
Table 2.2.1 contains the amount of rent paid every month for 24 students from a statistics course. Make a relative frequency distribution using 7 classes.
| 1500 | 1350 | 350 | 1200 | 850 | 900 |
| 1500 | 1150 | 1500 | 900 | 1400 | 1100 |
| 1250 | 600 | 610 | 960 | 890 | 1325 |
| 900 | 800 | 2550 | 495 | 1200 | 690 |
Table 2.2.1 : Data of Monthly Rent
Solution :
-
Find the range:
largest value - smallest value \(= 2550-350=2200\) -
Pick the number of classes:
The directions to say to use 7 classes. -
Find the class width:
width \(=\frac{\text { range }}{7}=\frac{2200}{7} \approx 314.286\)
Round up to 315
Always round up to the next integer even if the width is already an integer. -
Find the class limits:
Start at the smallest value. This is the lower class limit for the first class. Add the width to get the lower limit of the next class. Keep adding the width to get all the lower limits.
\(350+315=665,665+315=980,980+315=1295 \rightleftharpoons\),
The upper limit is one less than the next lower limit: so for the first class the upper class limit would be \(665-1=664\).
When you have all 7 classes, make sure the last number, in this case the 2550, is at least as large as the largest value in the data. If not, you made a mistake somewhere. -
Find the class boundaries:
Subtract 0.5 from the lower class limit to get the class boundaries. Add 0.5 to the upper class limit for the last class's boundary.
\(350-0.5=349.5, \quad 665-0.5=664.5,\quad 980-0.5=979.5, \quad 1295-0.5=1294.5 \rightleftharpoons\)
Every value in the data should fall into exactly one of the classes. No data values should fall right on the boundary of two classes. -
Find the class midpoints:
midpoint \(=\frac{\text { lower limit }+\text { upper limit }}{2}\)
\(\frac{350+664}{2}=507, \frac{665+979}{2}=822, \rightleftharpoons\) -
Tally and find the frequency of the data:
Go through the data and put a tally mark in the appropriate class for each piece of data by looking to see which class boundaries the data value is between. Fill in the frequency by changing each of the tallies into a number.
| Class Limits | Class Boundaries | Class Midpoint | Tally | Frequency |
|---|---|---|---|---|
| 350-664 | 349.5-664.5 | 507 | |||| | 4 |
| 665-979 | 664.5-979.5 | 822 | \(\cancel{||||}\) ||| | 8 |
| 980-1294 | 979.5-1294.5 | 1137 | \(\cancel{||||}\) | 5 |
| 1295-1609 | 1294.5-1609.5 | 1452 | \(\cancel{||||}\) | | 6 |
| 1610-1924 | 1609.5-1924.5 | 1767 | 0 | |
| 1925-2239 | 1924.5-2239.5 | 2082 | 0 | |
| 2240-2554 | 2239.5-2554.5 | 2397 | | | 1 |
Table 2.2.2 : Frequency Distribution for Monthly Rent
Make sure the total of the frequencies is the same as the number of data points.
It is difficult to determine the basic shape of the distribution by looking at the frequency distribution. It would be easier to look at a graph. The graph of a frequency distribution for quantitative data is called a frequency histogram or just histogram for short.
Definition \(\PageIndex{1}\)
Histogram : a graph of the frequencies on the vertical axis and the class boundaries on the horizontal axis. Rectangles where the height is the frequency and the width is the class width are drawn for each class.
Example \(\PageIndex{2}\) drawing a histogram
Draw a histogram for the distribution from Example 2.2.1 .
Solution :
The class boundaries are plotted on the horizontal axis and the frequencies are plotted on the vertical axis. You can plot the midpoints of the classes instead of the class boundaries. Graph 2.2.1 was created using the midpoints because it was easier to do with the software that created the graph.
Graph 2.2.1 : Histogram for Monthly Rent
Notice the graph has the axes labeled, the tick marks are labeled on each axis, and there is a title. It is important that your graphs (all graphs) are clearly labeled.
Reviewing the graph you can see that most of the students pay around $750 per month for rent, with about $1500 being the other common value. You can see from the graph, that most students pay between $600 and $1600 per month for rent. Of course, these values are just estimates from the graph. There is a large gap between the $1500 class and the highest data value. This seems to say that one student is paying a great deal more than everyone else. This value could be considered an outlier. An outlier is a data value that is far from the rest of the values. It may be an unusual value or a mistake. It is a data value that should be investigated. In this case, the student lives in a very expensive part of town, thus the value is not a mistake, and is just very unusual. There are other aspects that can be discussed, but first some other concepts need to be introduced.
Frequencies are helpful, but understanding the relative size each class is to the total is also useful. To find this you can divide the frequency by the total to create a relative frequency. If you have the relative frequencies for all of the classes, then you have a relative frequency distribution.
Definition \(\PageIndex{2}\)
Relative Frequency Distribution
A variation on a frequency distribution is a relative frequency distribution. Instead of giving the frequencies for each class, the relative frequencies are calculated.
Relative frequency \(=\frac{\text { frequency }}{\# \text { of data points }}\)
This gives you percentages of data that fall in each class.
Example \(\PageIndex{3}\) creating a relative frequency table
Find the relative frequency for the grade data.
Solution :
From Example 2.2.1 , the frequency distribution is reproduced in Table 2.2.2 .
| Class Limits | Class Boundaries | Class Midpoint | Frequency |
|---|---|---|---|
| 350-664 | 349.5-664.5 | 507 | 4 |
| 665-979 | 664.5-979.5 | 822 | 8 |
| 980-1294 | 979.5-1294.5 | 1127 | 5 |
| 1295-1609 | 1294.5-1609.5 | 1452 | 6 |
| 1610-1924 | 1609.5-1924.5 | 1767 | 0 |
| 1925-2239 | 1924.5-2239.5 | 2082 | 0 |
| 2240-2554 | 2239.5-2554.5 | 2397 | 1 |
Table 2.2.2 : Frequency Distribution for Monthly Rent
Divide each frequency by the number of data points.
\(\frac{4}{24}=0.17, \frac{8}{24}=0.33, \frac{5}{24}=0.21, \rightleftharpoons\)
| Class Limits | Class Boundaries | Class Midpoint | Frequency | Relative Frequency |
|---|---|---|---|---|
| 350-664 | 349.5-664.5 | 507 | 4 | 0.17 |
| 665-979 | 664.5-979.5 | 822 | 8 | 0.33 |
| 980-1294 | 979.5-1294.5 | 1127 | 5 | 0.21 |
| 1295-1609 | 1294.5-1609.5 | 1452 | 6 | 0.25 |
| 1610-1924 | 1609.5-1924.5 | 1767 | 0 | 0 |
| 1925-2239 | 1924.5-2239.5 | 2082 | 0 | 0 |
| 2240-2554 | 2239.5-2554.5 | 2397 | 1 | 0.04 |
| Total | 24 | 1 |
Table 2.2.3 : Relative Frequency Distribution for Monthly Rent
The relative frequencies should add up to 1 or 100%. (This might be off a little due to rounding errors.)
The graph of the relative frequency is known as a relative frequency histogram. It looks identical to the frequency histogram, but the vertical axis is relative frequency instead of just frequencies.
Example \(\PageIndex{4}\) drawing a relative frequency histogram
Draw a relative frequency histogram for the grade distribution from Example 2.2.1 .
Solution :
The class boundaries are plotted on the horizontal axis and the relative frequencies are plotted on the vertical axis. (This is not easy to do in R, so use another technology to graph a relative frequency histogram.)
Graph 2.2.2 : Relative Frequency Histogram for Monthly Rent
Notice the shape is the same as the frequency distribution.
Another useful piece of information is how many data points fall below a particular class boundary. As an example, a teacher may want to know how many students received below an 80%, a doctor may want to know how many adults have cholesterol below 160, or a manager may want to know how many stores gross less than $2000 per day. This is known as a cumulative frequency . If you want to know what percent of the data falls below a certain class boundary, then this would be a cumulative relative frequency . For cumulative frequencies you are finding how many data values fall below the upper class limit.
To create a cumulative frequency distribution , count the number of data points that are below the upper class boundary, starting with the first class and working up to the top class. The last upper class boundary should have all of the data points below it. Also include the number of data points below the lowest class boundary, which is zero.
Example \(\PageIndex{5}\) creating a cumulative frequency distribution
Create a cumulative frequency distribution for the data in Example 2.2.1 .
Solution :
The frequency distribution for the data is in Table 2.2.2 .
| Class Limits | Class Boundaries | Class Midpoint | Frequency |
|---|---|---|---|
| 350-664 | 349.5-664.5 | 507 | 4 |
| 665-979 | 664.5-979.5 | 822 | 8 |
| 980-1294 | 979.5-1294.5 | 1127 | 5 |
| 1295-1609 | 1294.5-1609.5 | 1452 | 6 |
| 1610-1924 | 1609.5-1924.5 | 1767 | 0 |
| 1925-2239 | 1924.5-2239.5 | 2082 | 0 |
| 2240-2554 | 2239.5-2554.5 | 2397 | 1 |
Table 2.2.2 : Frequency Distribution for Monthly Rent
Now ask yourself how many data points fall below each class boundary. Below 349.5, there are 0 data points. Below 664.5 there are 4 data points, below 979.5, there are 4 + 8 = 12 data points, below 1294.5 there are 4 + 8 + 5 = 17 data points, and continue this process until you reach the upper class boundary. This is summarized in Table 2.2.4 .
| Class Limits | Class Boundaries | Class Midpoint | Frequency | Cumulative Frequency |
|---|---|---|---|---|
| 350-664 | 349.5-664.5 | 507 | 4 | 4 |
| 665-979 | 664.5-979.5 | 822 | 8 | 12 |
| 980-1294 | 979.5-1294.5 | 1127 | 5 | 17 |
| 1295-1609 | 1294.5-1609.5 | 1452 | 6 | 23 |
| 1610-1924 | 1609.5-1924.5 | 1767 | 0 | 23 |
| 1925-2239 | 1924.5-2239.5 | 2082 | 0 | 23 |
| 2240-2554 | 2239.5-2554.5 | 2397 | 1 | 24 |
Table 2.2.4 : Cumulative Distribution for Monthly Rent
Again, it is hard to look at the data the way it is. A graph would be useful. The graph for cumulative frequency is called an ogive (o-jive). To create an ogive, first create a scale on both the horizontal and vertical axes that will fit the data. Then plot the points of the class upper class boundary versus the cumulative frequency. Make sure you include the point with the lowest class boundary and the 0 cumulative frequency. Then just connect the dots.
Example \(\PageIndex{6}\) drawing an ogive
Draw an ogive for the data in Example 2.2.1 .
Solution :
Using the upper class boundary and its corresponding cumulative frequency, plot the points as ordered pairs on the axes. Then connect the dots. You should have a line graph that rises as you move from left to right.
Graph 2.2.3 : Ogive for Monthly Rent
The usefulness of a ogive is to allow the reader to find out how many students pay less than a certain value, and also what amount of monthly rent is paid by a certain number of students. As an example, suppose you want to know how many students pay less than $1500 a month in rent, then you can go up from the $1500 until you hit the graph and then you go over to the cumulative frequency axes to see what value corresponds to this value. It appears that around 20 students pay less than $1500. (See Graph 2.2.4 .)
Graph 2.2.4 : Ogive for Monthly Rent with Example
Also, if you want to know the amount that 15 students pay less than, then you start at 15 on the vertical axis and then go over to the graph and down to the horizontal axis where the line intersects the graph. You can see that 15 students pay less than about $1200 a month. (See Graph 2.2.5 .)
Graph 2.2.5 : Ogive for Monthly Rent with Example
If you graph the cumulative relative frequency then you can find out what percentage is below a certain number instead of just the number of people below a certain value.
Shapes of the distribution:
When you look at a distribution, look at the basic shape. There are some basic shapes that are seen in histograms. Realize though that some distributions have no shape. The common shapes are symmetric, skewed, and uniform. Another interest is how many peaks a graph may have. This is known as modal.
Symmetric means that you can fold the graph in half down the middle and the two sides will line up. You can think of the two sides as being mirror images of each other. Skewed means one “tail” of the graph is longer than the other. The graph is skewed in the direction of the longer tail (backwards from what you would expect). A uniform graph has all the bars the same height.
Modal refers to the number of peaks. Unimodal has one peak and bimodal has two peaks. Usually if a graph has more than two peaks, the modal information is not longer of interest.
Other important features to consider are gaps between bars, a repetitive pattern, how spread out is the data, and where the center of the graph is.
Examples of Graphs :
This graph is roughly symmetric and unimodal:
Graph 2.2.6 : Symmetric, Unimodal Graph
This graph is symmetric and bimodal:
Graph 2.2.7 : Symmetric, Bimodal Graph
This graph is skewed to the right:
Graph 2.2.8 : Skewed Right Graph
This graph is skewed to the left and has a gap:
Graph 2.2.9 : Skewed Left Graph
This graph is uniform since all the bars are the same height:
Graph 2.2.10 : Uniform Graph
Example \(\PageIndex{7}\) creating a frequency distribution, histogram, and ogive
The following data represents the percent change in tuition levels at public, fouryear colleges (inflation adjusted) from 2008 to 2013 (Weissmann, 2013). Create a frequency distribution, histogram, and ogive for the data.
| 19.5% | 40.8% | 57.0% | 15.1% | 17.4% | 5.2% | 13.0% |
| 15.6% | 51.5% | 15.6% | 14.5% | 22.4% | 19.5% | 31.3% |
| 21.7% | 27.0% | 13.1% | 26.8% | 24.3% | 38.0% | 21.1% |
| 9.3% | 46.7% | 14.5% | 78.4% | 67.3% | 21.1% | 22.4% |
| 5.3% | 17.3% | 17.5% | 36.6% | 72.0% | 63.2% | 15.1% |
| 2.2% | 17.5% | 36.7% | 2.8% | 16.2% | 20.5% | 17.8% |
| 30.1% | 63.6% | 17.8% | 23.2% | 25.3% | 21.4% | 28.5% |
| 9.4% |
Table 2.2.5 : Data of Tuition Levels at Public, Four-Year Colleges
Solution :
-
Find the range:
largest value - smallest value = \(78.4\)% \(-2.2\)% \(=76.2\)% -
Pick the number of classes:
Since there are 50 data points, then around 6 to 8 classes should be used. Let's use 8. -
Find the class width:
width \(=\frac{\text { range }}{8}=\frac{76.2 \%}{8} \approx 9.525 \%\)
Since the data has one decimal place, then the class width should round to one decimal place. Make sure you round up.
width \(=9.6\)% -
Find the class limits:
\(2.2 \%+9.6 \%=11.8 \%, 11.8 \%+9.6 \%=21.4 \%, 21.4 \%+9.6 \%=31.0 \%, \leftrightharpoons\) -
Find the class boundaries:
Since the data has one decimal place, the class boundaries should have two decimal places, so subtract 0.05 from the lower class limit to get the class boundaries. Add 0.05 to the upper class limit for the last class’s boundary.
\(2.2-0.05=2.15 \%, 11.8-0.05=11.75 \%, 21.4-0.05=21.35 \% \leftrightharpoons\)
Every value in the data should fall into exactly one of the classes. No data values should fall right on the boundary of two classes. -
Find the class midpoints:
midpoint \(=\frac{\text { lower limt }+\text { upper limit }}{2}\)
\(\frac{2.2+11.7}{2}=6.95 \%, \frac{11.8+21.3}{2}=16.55 \%, \leftrightharpoons\) - Tally and find the frequency of the data:
| Class Limits | Class Boundaries | Class Midpoint | Tally | Frequency | Relative Frequency | Cumulative Frequency |
|---|---|---|---|---|---|---|
| 2.2-11.7 | 2.15-11.75 | 6.95 | \(\cancel{||||} |\) | 6 | 0.12 | 6 |
| 11.8-21.3 | 11.75-21.35 | 16.55 | \(\cancel{||||} \cancel{||||} \cancel{||||} \cancel{||||}\) | 20 | 0.40 | 26 |
| 21.4-30.9 | 21.35-30.95 | 26.15 | \(\cancel{||||} \cancel{||||} |\) | 11 | 0.22 | 37 |
| 31.0-45.0 | 30.95-40.55 | 35.75 | \( |||| \) | 4 | 0.08 | 41 |
| 40.6-50.1 | 40.55-50.15 | 45.35 | \( || \) | 2 | 0.04 | 43 |
| 50.2-59.7 | 50.15-59.75 | 54.95 | \( || \) | 2 | 0.04 | 45 |
| 59.8-69.3 | 59.75-69.35 | 64.55 | \( ||| \) | 3 | 0.06 | 48 |
| 69.4-78.9 | 69.35-78.95 | 74.15 | \( || \) | 2 | 0.04 | 50 |
Table 2.2.6 : Frequency Distribution for Tuition Levels at Public, Four-Year Colleges
Make sure the total of the frequencies is the same as the number of data points.
Graph 2.2.11 : Histogram for Tuition Levels at Public, Four-Year Colleges
This graph is skewed right, with no gaps. This says that most percent increases in tuition were around 16.55%, with very few states having a percent increase greater than 45.35%.
Graph 2.2.12 : Ogive for Tuition Levels at Public, Four-Year Colleges
Looking at the ogive, you can see that 30 states had a percent change in tuition levels of about 25% or less.
There are occasions where the class limits in the frequency distribution are predetermined. Example 2.2.8 demonstrates this situation.
Example \(\PageIndex{8}\) creating a frequency distribution and histogram
The following are the percentage grades of 25 students from a statistics course. Make a frequency distribution and histogram.
| 62 | 87 | 81 | 69 | 87 | 62 | 45 | 95 | 76 | 76 |
| 62 | 71 | 65 | 67 | 72 | 80 | 40 | 77 | 87 | 58 |
| 84 | 73 | 93 | 64 | 89 |
Table 2.2.7 : Data of Test Grades
Solution :
Since this data is percent grades, it makes more sense to make the classes in multiples of 10, since grades are usually 90 to 100%, 80 to 90%, and so forth. It is easier to not use the class boundaries, but instead use the class limits and think of the upper class limit being up to but not including the next classes lower limit. As an example the class 80 – 90 means a grade of 80% up to but not including a 90%. A student with an 89.9% would be in the 80-90 class.
| Class Limit | Class Midpoint | Tally | Freqeuncy |
|---|---|---|---|
| 40-50 | 45 | \( || \) | 2 |
| 50-60 | 55 | \( | \) | 1 |
| 60-70 | 65 | \( \cancel{||||} || \) | 7 |
| 70-80 | 75 | \( \cancel{||||} | \) | 6 |
| 80-90 | 85 | \( \cancel{||||} || \) | 7 |
| 90-100 | 95 | \( || \) | 2 |
Table 2.2.8 : Frequency Distribution for Test Grades
Graph 2.2.13 : Histogram for Test Grades
It appears that most of the students had between 60 to 90%. This graph looks somewhat symmetric and also bimodal. The same number of students earned between 60 to 70% and 80 to 90%.
There are other types of graphs for quantitative data. They will be explored in the next section.


4.4: Webster’s Method
Daniel Webster (1782-1852) proposed a method similar to Jefferson’s in 1832. It was adopted by Congress in 1842, but replaced by Hamilton’s method in 1852. It was then adopted again in 1901. The difference is that Webster rounds the quotas to the nearest whole number rather than dropping the decimal parts. If that doesn’t produce the desired results at the beginning, he says, like Jefferson, to adjust the divisor until it does. (In Jefferson’s case, at least the first adjustment will always be to make the divisor smaller. That is not always the case with Webster’s method.)
- Determine how many people each representative should represent. Do this by dividing the total population of all the states by the total number of representatives. This answer is called the standard divisor .
- Divide each state’s population by the divisor to determine how many representatives it should have. Record this answer to several decimal places. This answer is called the quota .
- Round all the quotas to the nearest whole number (but don’t forget what the decimals were). Add up the remaining whole numbers.
- If the total from Step 3 was less than the total number of representatives, reduce the divisor and recalculate the quota and allocation. If the total from step 3 was larger than the total number of representatives, increase the divisor and recalculate the quota and allocation. Continue doing this until the total in Step 3 is equal to the total number of representatives. The divisor we end up using is called the modified divisor or adjusted divisor .
Again, Delaware, with an initial divisor of \(21,900.82927\):
\(\begin{array}{lrrc}
\text { County } & \text { Population } & \text{ Quota } & \text{ Initial } \\
\hline
\text { Kent } & 166,310 & 7.4111 & 7 \\
\text { New Castle } & 538,479 & 24.5872 & 25 \\
\text { Sussex } & 197,145 & 9.0017 & 9 \\
\textbf{ Total } & \bf{ 897,934 } & & \bf{ 41 }\end{array}\)
Solution
This gives the required total, so we’re done.
Again, Rhode Island, with an initial divisor of \(14,034.22667\):
\(\begin{array}{lrrc}
\text { County } & \text { Population } & \text{ Quota } & \text{ Initial } \\
\hline \text { Bristol } & 49,875 & 3.5538 & 4 \\
\text { Kent } & 166,158 & 11.8395 & 12 \\
\text { Newport } & 82,888 & 5.9061 & 6 \\
\text { Providence } & 626,667 & 44.6528 & 45 \\
\text { Washington } & 126,979 & 9.0478 & 9\\
\textbf{ Total } & \bf{ 1,052,567 } & & \bf{ 76 }\end{array}\)
Solution
This is too many, so we need to increase the divisor. Let’s try \(14,100\):
\(\begin{array}{lrrc}
\text { County } & \text { Population } & \text{ Quota } & \text{ Initial } \\
\hline \text { Bristol } & 49,875 & 3.5372 & 4 \\
\text { Kent } & 166,158 & 11.7843 & 12 \\
\text { Newport } & 82,888 & 5.8786 & 6 \\
\text { Providence } & 626,667 & 5.8786 & 44 \\
\text { Washington } & 126,979 & 9.0056 & 9\\
\textbf{ Total } & \bf{ 1,052,567 } & & \bf{ 75 }\end{array}\)
This works, so we’re done.
Like Jefferson’s method, Webster’s method carries a bias in favor of states with large populations, but rounding the quotas to the nearest whole number greatly reduces this bias. (Notice that Providence County, the largest, is the one that gets a representative trimmed because of the increased quota.) Also like Jefferson’s method, Webster’s method does not always follow the quota rule, but it follows the quota rule much more often than Jefferson’s method does. (In fact, if Webster’s method had been applied to every apportionment of Congress in all of American history, it would have followed the quota rule every single time.)
In 1980, two mathematicians, Peyton Young and Mike Balinski, proved what we now call the Balinski-Young Impossibility Theorem.
The Balinski-Young Impossibility Theorem shows that any apportionment method which always follows the quota rule will be subject to the possibility of paradoxes like the Alabama, New States, or Population paradoxes. In other words, we can choose a method that avoids those paradoxes, but only if we are willing to give up the guarantee of following the quota rule.