9.2: Calorimetry
- Explain the technique of calorimetry
- Calculate and interpret heat and related properties using typical calorimetry data
One technique we can use to measure the amount of heat involved in a chemical or physical process is known as calorimetry . Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter). The change in temperature of the measuring part of the calorimeter is converted into the amount of heat (since the previous calibration was used to establish its heat capacity). The measurement of heat transfer using this approach requires the definition of a system (the substance or substances undergoing the chemical or physical change) and its surroundings (the other components of the measurement apparatus that serve to either provide heat to the system or absorb heat from the system). Knowledge of the heat capacity of the surroundings, and careful measurements of the masses of the system and surroundings and their temperatures before and after the process allows one to calculate the heat transferred as described in this section.
A calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. For example, when an exothermic reaction occurs in solution in a calorimeter, the heat produced by the reaction is absorbed by the solution, which increases its temperature. When an endothermic reaction occurs, the heat required is absorbed from the thermal energy of the solution, which decreases its temperature (Figure \(\PageIndex{1}\)). The temperature change, along with the specific heat and mass of the solution, can then be used to calculate the amount of heat involved in either case.
By convention, q is given a negative (-) sign when the system releases heat to the surroundings (exothermic); q is given a positive (+) sign when the system absorbs heat from the surroundings (endothermic).
Scientists use well-insulated calorimeters that all but prevent the transfer of heat between the calorimeter and its environment. This enables the accurate determination of the heat involved in chemical processes, the energy content of foods, and so on. General chemistry students often use simple calorimeters constructed from polystyrene cups (Figure \(\PageIndex{2}\)). These easy-to-use “coffee cup” calorimeters allow more heat exchange with their surroundings, and therefore produce less accurate energy values.
Commercial solution calorimeters are also available. Relatively inexpensive calorimeters often consist of two thin-walled cups that are nested in a way that minimizes thermal contact during use, along with an insulated cover, handheld stirrer, and simple thermometer. More expensive calorimeters used for industry and research typically have a well-insulated, fully enclosed reaction vessel, motorized stirring mechanism, and a more accurate temperature sensor (Figure \(\PageIndex{3}\)).
Before we practice calorimetry problems involving chemical reactions, consider a simple example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal (M), and a low-temperature substance, such as cool water (W). If we place the metal in the water, heat will flow from M to W. The temperature of M will decrease, and the temperature of W will increase, until the two substances have the same temperature—that is, when they reach thermal equilibrium (Figure \(\PageIndex{4}\)). If this occurs in a calorimeter, ideally all of this heat transfer occurs between the two substances, with no heat gained or lost by either the calorimeter or the calorimeter’s surroundings. Under these ideal circumstances, the net heat change is zero:
\[q_\mathrm{\,substance\: M} + q_\mathrm{\,substance\: W}=0 \label{5.3.1} \]
This relationship can be rearranged to show that the heat gained by substance M is equal to the heat lost by substance W:
\[q_\mathrm{\,substance\: M}=-q_\mathrm{\,substance\: W} \label{5.3.2} \]
The magnitude of the heat (change) is therefore the same for both substances, and the negative sign merely shows that \(q_{substance\, M}\) and \(q_{substance\, W}\) are opposite in direction of heat flow (gain or loss) but does not indicate the arithmetic sign of either q value (that is determined by whether the matter in question gains or loses heat, per definition). In the specific situation described, \(q_{substance\, M}\) is a negative value and \(q_{substance\, W}\) is positive, since heat is transferred from M to W.
A hot 360-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at 24.0 °C. The final temperature of the water is measured as 42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron (Table T4), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings).
Solution
The temperature of the water increases from 24.0 °C to 42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = − heat taken in by water, or:
\[q_\ce{rebar}=−q_\ce{water} \nonumber \]
Since we know how heat is related to other measurable quantities, we have:
\[(c×m×ΔT)_\ce{rebar}=−(c×m×ΔT)_\ce{water} \nonumber \]
Letting f = final and i = initial, in expanded form, this becomes:
\[ c_\ce{rebar}×m_\ce{rebar}×(T_\mathrm{f,rebar}−T_\mathrm{i,rebar})=−c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) \nonumber \]
The density of water is 1.0 g/mL, so 425 mL of water = 425 g. Noting that the final temperature of both the rebar and water is 42.7 °C, substituting known values yields:
\[ \mathrm{(0.449\:J/g\: °C)(360g)(42.7°C−\mathit T_\mathrm{i,rebar})=-(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)} \nonumber \]
\[\mathrm{\mathit T_{i,rebar}=\dfrac{(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)}{(0.449\:J/g\: °C)(360\:g)}+42.7°C} \nonumber \]
Solving this gives Ti,rebar= 248 °C, so the initial temperature of the rebar was 248 °C.
A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water.
- Answer
-
The initial temperature of the copper was 335.6 °C.
A 248-g piece of copper initially at 314 °C is dropped into 390 mL of water initially at 22.6 °C. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature.
- Answer
-
The final temperature (reached by both copper and water) is 38.7 °C.
This method can also be used to determine other quantities, such as the specific heat of an unknown metal.
A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at 22.0 °C. The final temperature is 28.5 °C. Use these data to determine the specific heat of the metal. Use this result to identify the metal.
Solution
Assuming perfect heat transfer, heat given off by metal = −heat taken in by water, or:
\[q_\ce{metal}=−q_\ce{water} \nonumber \]
In expanded form, this is:
\[c_\ce{metal}×m_\ce{metal}×(T_\mathrm{f,metal}−T_\mathrm{i, metal})=−c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) \nonumber \]
Noting that since the metal was submerged in boiling water, its initial temperature was 100.0 °C; and that for water, 60.0 mL = 60.0 g; we have:
\[\mathrm{(\mathit c_{metal})(59.7\:g)(28.5°C−100.0°C)=−(4.18\:J/g\: °C)(60.0\:g)(28.5°C−22.0°C)} \nonumber \]
Solving this:
\[\mathrm{\mathit c_{metal}=\dfrac{−(4.184\:J/g\: °C)(60.0\:g)(6.5°C)}{(59.7\:g)(−71.5°C)}=0.38\:J/g\: °C} \nonumber \]
Comparing this with values in Table T4, our experimental specific heat is closest to the value for copper (0.39 J/g °C), so we identify the metal as copper.
A 92.9-g piece of a silver/gray metal is heated to 178.0 °C, and then quickly transferred into 75.0 mL of water initially at 24.0 °C. After 5 minutes, both the metal and the water have reached the same temperature: 29.7 °C. Determine the specific heat and the identity of the metal. (Note: You should find that the specific heat is close to that of two different metals. Explain how you can confidently determine the identity of the metal).
- Answer
-
\(c_{metal}= 0.13 \;J/g\; °C\)
This specific heat is close to that of either gold or lead. It would be difficult to determine which metal this was based solely on the numerical values. However, the observation that the metal is silver/gray in addition to the value for the specific heat indicates that the metal is lead.
When we use calorimetry to determine the heat involved in a chemical reaction, the same principles we have been discussing apply. The amount of heat absorbed by the calorimeter is often small enough that we can neglect it (though not for highly accurate measurements, as discussed later), and the calorimeter minimizes energy exchange with the surroundings. Because energy is neither created nor destroyed during a chemical reaction, there is no overall energy change during the reaction. The heat produced or consumed in the reaction (the “system”), q reaction , plus the heat absorbed or lost by the solution (the “surroundings”), \(q_{solution}\), must add up to zero:
\[q_\ce{reaction}+q_\ce{solution}=0\ \label{ 5.3.10} \]
This means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution:
\[q_\ce{reaction}=−q_\ce{solution} \label{5.3.11} \]
This concept lies at the heart of all calorimetry problems and calculations.
When 50.0 mL of 0.10 M HCl( aq ) and 50.0 mL of 1.00 M NaOH( aq ), both at 22.0 °C, are added to a coffee cup calorimeter, the temperature of the mixture reaches a maximum of 28.9 °C. What is the approximate amount of heat produced by this reaction?
\[\ce{HCl}(aq)+\ce{NaOH}(aq)⟶\ce{NaCl}(aq)+\ce{H2O}(l) \nonumber \]
S olution
To visualize what is going on, imagine that you could combine the two solutions so quickly that no reaction took place while they mixed; then after mixing, the reaction took place. At the instant of mixing, you have 100.0 mL of a mixture of HCl and NaOH at 22.0 °C. The HCl and NaOH then react until the solution temperature reaches 28.9 °C.
The heat given off by the reaction is equal to that taken in by the solution. Therefore:
\[q_\ce{reaction}=−q_\ce{solution} \nonumber \]
(It is important to remember that this relationship only holds if the calorimeter does not absorb any heat from the reaction, and there is no heat exchange between the calorimeter and its surroundings.)
Next, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperature change:
\[q_\ce{solution}=(c×m×ΔT)_\ce{solution} \nonumber \]
To proceed with this calculation, we need to make a few more reasonable assumptions or approximations. Since the solution is aqueous, we can proceed as if it were water in terms of its specific heat and mass values. The density of water is approximately 1.0 g/mL, so 100.0 mL has a mass of about 1.0 × 10 2 g (two significant figures). The specific heat of water is approximately 4.18 J/g °C, so we use that for the specific heat of the solution. Substituting these values gives:
\[\mathrm{\mathit q_{solution}=(4.184\:J/g\: °C)(1.0×10^2\:g)(28.9°C−22.0°C)=2.89×10^3\:J} \nonumber \]
Finally, since we are trying to find the heat of the reaction, we have:
\[q_\ce{reaction}=−q_\ce{solution}=−2.89×10^3\:J \nonumber \]
The negative sign indicates that the reaction is exothermic. It produces 2.89 kJ of heat.
When 100 mL of 0.200 M NaCl( aq ) and 100 mL of 0.200 M AgNO 3 ( aq ), both at 21.9 °C, are mixed in a coffee cup calorimeter, the temperature increases to 23.5 °C as solid AgCl forms. How much heat is produced by this precipitation reaction? What assumptions did you make to determine your value?
- Answer
-
\(1.34 \times 10^3\; J\); assume no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and that the specific heat and mass of the solution are the same as those for water
When working or playing outdoors on a cold day, you might use a hand warmer to warm your hands (Figure \(\PageIndex{5}\)). A common reusable hand warmer contains a supersaturated solution of NaC 2 H 3 O 2 (sodium acetate) and a metal disc. Bending the disk creates nucleation sites around which the metastable NaC 2 H 3 O 2 quickly crystallizes (a later chapter on solutions will investigate saturation and supersaturation in more detail).
The process \(\ce{NaC2H3O2}(aq)⟶\ce{NaC2H3O2}(s)\) is exothermic, and the heat produced by this process is absorbed by your hands, thereby warming them (at least for a while). If the hand warmer is reheated, the NaC 2 H 3 O 2 redissolves and can be reused.
Another common hand warmer produces heat when it is ripped open, exposing iron and water in the hand warmer to oxygen in the air. One simplified version of this exothermic reaction is
\[\ce{2Fe(s) + 3/2 O2(g) ⟶ Fe2O3(s)}.\ n\nonumber \]
Salt in the hand warmer catalyzes the reaction, so it produces heat more rapidly; cellulose, vermiculite, and activated carbon help distribute the heat evenly. Other types of hand warmers use lighter fluid (a platinum catalyst helps lighter fluid oxidize exothermically), charcoal (charcoal oxidizes in a special case), or electrical units that produce heat by passing an electrical current from a battery through resistive wires.
When solid ammonium nitrate dissolves in water, the solution becomes cold. This is the basis for an “instant ice pack” (Figure \(\PageIndex{5}\)). When 3.21 g of solid NH 4 NO 3 dissolves in 50.0 g of water at 24.9 °C in a calorimeter, the temperature decreases to 20.3 °C.
Calculate the value of q for this reaction and explain the meaning of its arithmetic sign. State any assumptions that you made.
Solution
We assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself), in which case:
\[q_\ce{rxn}=−q_\ce{soln} \nonumber \]
with “rxn” and “soln” used as shorthand for “reaction” and “solution,” respectively.
Assuming also that the specific heat of the solution is the same as that for water, we have:
\[\begin{align*} q_\ce{rxn} &=−q_\ce{soln}=−(c×m×ΔT)_\ce{soln}\\ &=\mathrm{−[(4.184J/g\: °C)×(53.2\:g)×(20.3°C−24.9°C)]}\\ &=\mathrm{−[(4.184J/g\: °C)×(53.2\:g)×(−4.6°C)]}\\ &+\mathrm{1.0×10^3\:J=+1.0\:kJ} \end{align*}\]
The positive sign for q indicates that the dissolution is an endothermic process.
When a 3.00-g sample of KCl was added to 3.00 × 10 2 g of water in a coffee cup calorimeter, the temperature decreased by 1.05 °C. How much heat is involved in the dissolution of the KCl? What assumptions did you make?
- Answer
-
1.33 kJ; assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself) and that the specific heat of the solution is the same as that for water.
If the amount of heat absorbed by a calorimeter is too large to neglect or if we require more accurate results, then we must take into account the heat absorbed both by the solution and by the calorimeter.
The calorimeters described are designed to operate at constant (atmospheric) pressure and are convenient to measure heat flow accompanying processes that occur in solution. A different type of calorimeter that operates at constant volume, colloquially known as a bomb calorimeter , is used to measure the energy produced by reactions that yield large amounts of heat and gaseous products, such as combustion reactions. (The term “bomb” comes from the observation that these reactions can be vigorous enough to resemble explosions that would damage other calorimeters.) This type of calorimeter consists of a robust steel container (the “bomb”) that contains the reactants and is itself submerged in water (Figure \(\PageIndex{6}\)). The sample is placed in the bomb, which is then filled with oxygen at high pressure. A small electrical spark is used to ignite the sample. The energy produced by the reaction is trapped in the steel bomb and the surrounding water. The temperature increase is measured and, along with the known heat capacity of the calorimeter, is used to calculate the energy produced by the reaction. Bomb calorimeters require calibration to determine the heat capacity of the calorimeter and ensure accurate results. The calibration is accomplished using a reaction with a known q , such as a measured quantity of benzoic acid ignited by a spark from a nickel fuse wire that is weighed before and after the reaction. The temperature change produced by the known reaction is used to determine the heat capacity of the calorimeter. The calibration is generally performed each time before the calorimeter is used to gather research data.
Video \(\PageIndex{1}\): Video of view how a bomb calorimeter is prepared for action.
When 3.12 g of glucose, C 6 H 12 O 6 , is burned in a bomb calorimeter, the temperature of the calorimeter increases from 23.8 °C to 35.6 °C. The calorimeter contains 775 g of water, and the bomb itself has a heat capacity of 893 J/°C. How much heat was produced by the combustion of the glucose sample?
Solution
The combustion produces heat that is primarily absorbed by the water and the bomb. (The amounts of heat absorbed by the reaction products and the unreacted excess oxygen are relatively small and dealing with them is beyond the scope of this text. We will neglect them in our calculations.)
The heat produced by the reaction is absorbed by the water and the bomb:
\[\begin{align*}
&q_\ce{rxn}=−(q_\ce{water}+q_\ce{bomb})\\
&=\mathrm{−[(4.184\:J/g\: °C)×(775\:g)×(35.6°C−23.8°C)+893\:J/°C×(35.6°C−23.8°C)]}\\
&=\mathrm{−(38,300\:J+10,500\:J)}\\
&=\mathrm{−48,800\: J=−48.8\: kJ}
\end{align*} \nonumber \]
This reaction released 48.7 kJ of heat when 3.12 g of glucose was burned.
When 0.963 g of benzene, C 6 H 6 , is burned in a bomb calorimeter, the temperature of the calorimeter increases by 8.39 °C. The bomb has a heat capacity of 784 J/°C and is submerged in 925 mL of water. How much heat was produced by the combustion of the glucose sample?
Answer-
39.0 kJ
Since the first one was constructed in 1899, 35 calorimeters have been built to measure the heat produced by a living person. 1 These whole-body calorimeters of various designs are large enough to hold an individual human being. More recently, whole-room calorimeters allow for relatively normal activities to be performed, and these calorimeters generate data that more closely reflect the real world. These calorimeters are used to measure the metabolism of individuals under different environmental conditions, different dietary regimes, and with different health conditions, such as diabetes. In humans, metabolism is typically measured in Calories per day. A nutritional calorie (Calorie) is the energy unit used to quantify the amount of energy derived from the metabolism of foods; one Calorie is equal to 1000 calories (1 kcal), the amount of energy needed to heat 1 kg of water by 1 °C.
In your day-to-day life, you may be more familiar with energy being given in Calories, or nutritional calories, which are used to quantify the amount of energy in foods. One calorie (cal) = exactly 4.184 joules, and one Calorie (note the capitalization) = 1000 cal, or 1 kcal. (This is approximately the amount of energy needed to heat 1 kg of water by 1 °C.)
The macronutrients in food are proteins, carbohydrates, and fats or oils. Proteins provide about 4 Calories per gram, carbohydrates also provide about 4 Calories per gram, and fats and oils provide about 9 Calories/g. Nutritional labels on food packages show the caloric content of one serving of the food, as well as the breakdown into Calories from each of the three macronutrients (Figure \(\PageIndex{7}\)).
For the example shown in (b), the total energy per 228-g portion is calculated by:
\[\mathrm{(5\:g\: protein×4\:Calories/g)+(31\:g\: carb×4\:Calories/g)+(12\:g\: fat×9\:Calories/g)=252\:Calories} \label{5.3.X} \]
So, you can use food labels to count your Calories. But where do the values come from? And how accurate are they? The caloric content of foods can be determined by using bomb calorimetry; that is, by burning the food and measuring the energy it contains. A sample of food is weighed, mixed in a blender, freeze-dried, ground into powder, and formed into a pellet. The pellet is burned inside a bomb calorimeter, and the measured temperature change is converted into energy per gram of food.
Today, the caloric content on food labels is derived using a method called the Atwater system that uses the average caloric content of the different chemical constituents of food, protein, carbohydrate, and fats. The average amounts are those given in the equation and are derived from the various results given by bomb calorimetry of whole foods. The carbohydrate amount is discounted a certain amount for the fiber content, which is indigestible carbohydrate. To determine the energy content of a food, the quantities of carbohydrate, protein, and fat are each multiplied by the average Calories per gram for each and the products summed to obtain the total energy.
Summary
Calorimetry is used to measure the amount of thermal energy transferred in a chemical or physical process. This requires careful measurement of the temperature change that occurs during the process and the masses of the system and surroundings. These measured quantities are then used to compute the amount of heat produced or consumed in the process using known mathematical relations. Calorimeters are designed to minimize energy exchange between the system being studied and its surroundings. They range from simple coffee cup calorimeters used by introductory chemistry students to sophisticated bomb calorimeters used to determine the energy content of food.
Footnotes
- 1 Francis D. Reardon et al. “The Snellen human calorimeter revisited, re-engineered and upgraded: Design and performance characteristics.” Medical and Biological Engineering and Computing 8 (2006)721–28, The Snellen human calorimeter revisited, re-engineered and upgraded: design and performance characteristics [link.springer.com] .
Glossary
- bomb calorimeter
- device designed to measure the energy change for processes occurring under conditions of constant volume; commonly used for reactions involving solid and gaseous reactants or products
- calorimeter
- device used to measure the amount of heat absorbed or released in a chemical or physical process
- calorimetry
- process of measuring the amount of heat involved in a chemical or physical process
- nutritional calorie (Calorie)
- unit used for quantifying energy provided by digestion of foods, defined as 1000 cal or 1 kcal
- surroundings
- all matter other than the system being studied
- system
- portion of matter undergoing a chemical or physical change being studied


10.5: §72. The Perfect Participle Base suffix -URA as Abstract Noun
Alongside these abstract nouns in -io , Latin could form other abstract nouns from the very same perfect participle bases, using the suffix -ura (> E -ure ). Thus, from the perfect participle captus , there developed two nouns, captio and captura , both meaning “the act (or process) of taking.” In English, of course, caption and capture are very different words; but the semantic force of the -io and -ura endings is so similar in Latin that it is hardly worth while trying to see any contrast in connotation between those suffixes. For some perfect participles, we may have as many as three different Latin nouns, all extremely close in meaning. Just consider, from stare , status (“stand”), the forms status , statio , and statura , all of which denote some kind of “standing”; their derivatives status , station , and stature have become usefully differentiated in English. The perfect participle of jungere , junctus (“join”) is the source of English joint ( junctus ), junction ( junctio ), and juncture ( junctura )—three synonymous words in English.
Here are some more familiar examples of this -ura suffix:
| LATIN VERB | LATIN NOUN | ENG. NOUN |
| frangere , fractus (“break”) | fractura (“a breaking”) | fracture |
| gerere , ges t us (“bear”) | gestura (“a bearing”) | gesture |
| legere , lectus (“read”) | lectura (“a reading”) |
lecture
|
| nasci , natus (“be born”) | natura (“a being born”) |
nature
|
| pascere , pastus (“feed,” “tend”) | pastura (“a feeding”) |
pasture
|
| rapere , raptus (“seize”) | raptura (“a seizing”) |
rapture
|
| rumper e, ruptus (“burst”) | ruptura (“a burstng”) | rupture |
| scribere , scriptus (“write”) | scriptura (“a writing”) |
scriptura
|
| struere , structus (“build”) | structura (“a building”) | structure |
From jacere, jactus (“throw”) comes the form conjectura , E conjecture (“a throwing together”). Aperire , apertus (“open”) is the source of apertura , E aperture ; a very different kind of “opening” is the surprising doublet overture. [1]
-
Cf. L
apertus
> F
ouvert
> E
overt
. Similarly, L
co-opertus
(“covered over”) became E
covert
. ↵


9.2: Quotations
What this handout is about
Used effectively, quotations can provide important pieces of evidence and lend fresh voices and perspectives to your narrative. Used ineffectively, however, quotations clutter your text and interrupt the flow of your argument. This handout will help you decide when and how to quote like a pro.
When should I quote?
Figure \(\PageIndex{1}\)
Use quotations at strategically selected moments. You have probably been told by teachers to provide as much evidence as possible in support of your thesis. But packing your paper with quotations will not necessarily strengthen your argument. The majority of your paper should still be your original ideas in your own words (after all, it’s your paper). And quotations are only one type of evidence: well-balanced papers may also make use of paraphrases, data, and statistics. The types of evidence you use will depend in part on the conventions of the discipline or audience for which you are writing. For example, papers analyzing literature may rely heavily on direct quotations of the text, while papers in the social sciences may have more paraphrasing, data, and statistics than quotations.
1. Discussing specific arguments or ideas.
Sometimes, in order to have a clear, accurate discussion of the ideas of others, you need to quote those ideas word for word. Suppose you want to challenge the following statement made by John Doe, a well-known historian:
“At the beginning of World War Two, almost all Americans assumed the war would end quickly.”
If it is especially important that you formulate a counterargument to this claim, then you might wish to quote the part of the statement that you find questionable and establish a dialogue between yourself and John Doe:
Historian John Doe has argued that in 1941 “almost all Americans assumed the war would end quickly” (Doe 223). Yet during the first six months of U.S. involvement, the wives and mothers of soldiers often noted in their diaries their fear that the war would drag on for years.
2. Giving added emphasis to a particularly authoritative source on your topic.
There will be times when you want to highlight the words of a particularly important and authoritative source on your topic. For example, suppose you were writing an essay about the differences between the lives of male and female slaves in the U.S. South. One of your most provocative sources is a narrative written by a former slave, Harriet Jacobs. It would then be appropriate to quote some of Jacobs’s words:
Harriet Jacobs, a former slave from North Carolina, published an autobiographical slave narrative in 1861. She exposed the hardships of both male and female slaves but ultimately concluded that “slavery is terrible for men; but it is far more terrible for women.”
In this particular example, Jacobs is providing a crucial first-hand perspective on slavery. Thus, her words deserve more exposure than a paraphrase could provide. Jacobs is quoted in Harriet A. Jacobs, Incidents in the Life of a Slave Girl, ed. Jean Fagan Yellin (Cambridge: Harvard University Press, 1987).
3. Analyzing how others use language.
This scenario is probably most common in literature and linguistics courses, but you might also find yourself writing about the use of language in history and social science classes. If the use of language is your primary topic, then you will obviously need to quote users of that language.
Examples of topics that might require the frequent use of quotations include:
Southern colloquial expressions in William Faulkner’s Light in August
Ms. and the creation of a language of female empowerment
A comparison of three British poets and their use of rhyme
4.
Figure \(\PageIndex{2}\)
In order to lend variety to your prose, you may wish to quote a source with particularly vivid language. All quotations, however, must closely relate to your topic and arguments. Do not insert a quotation solely for its literary merits.
One example of a quotation that adds flair:
Calvin Coolidge’s tendency to fall asleep became legendary. As H. L. Mencken commented in the American Mercury in 1933, “Nero fiddled, but Coolidge only snored.”
How do I set up and follow up a quotation?
Once you’ve carefully selected the quotations that you want to use, your next job is to weave those quotations into your text. The words that precede and follow a quotation are just as important as the quotation itself. You can think of each quote as the filling in a sandwich: it may be tasty on its own, but it’s messy to eat without some bread on either side of it. Your words can serve as the “bread” that helps readers digest each quote easily. Below are four guidelines for setting up and following up quotations.
In illustrating these four steps, we’ll use as our example, Franklin Roosevelt’s famous quotation, “The only thing we have to fear is fear itself.”
1. Provide a context for each quotation.
Do not rely on quotations to tell your story for you. It is your responsibility to provide your reader with a context for the quotation. The context should set the basic scene for when, possibly where, and under what circumstances the quotation was spoken or written. So, in providing a context for our above example, you might write:
When Franklin Roosevelt gave his inaugural speech on March 4, 1933, he addressed a nation weakened and demoralized by economic depression.
2. Attribute each quotation to its source.
Tell your reader who is speaking. Here is a good test: try reading your text aloud. Could your reader determine without looking at your paper where your quotations begin? If not, you need to attribute the quote more noticeably.
Avoid getting into the “he/she said” attribution rut! There are many other ways to attribute quotes besides this construction. Here are a few alternative verbs, usually followed by “that”:
| add | remark | exclaim |
| announce | reply | state |
| comment | respond | estimate |
| write | point out | predict |
| argue | suggest | propose |
| declare | criticize | proclaim |
| note | complain | opine |
| observe | think | note |
Different reporting verbs are preferred by different disciplines, so pay special attention to these in your disciplinary reading. If you’re unfamiliar with the meanings of any of these words or others you find in your reading, consult a dictionary before using them.
3. Explain the significance of the quotation.
Once you’ve inserted your quotation, along with its context and attribution, don’t stop! Your reader still needs your assessment of why the quotation holds significance for your paper. Using our Roosevelt example, if you were writing a paper on the first one-hundred days of FDR’s administration, you might follow the quotation by linking it to that topic:
With that message of hope and confidence, the new president set the stage for his next one-hundred days in office and helped restore the faith of the American people in their government.
4. Provide a citation for the quotation.
All quotations, just like all paraphrases, require a formal citation. For more details about particular citation formats, see the UNC Library’s citation tutorial . In general, you should remember one rule of thumb: Place the parenthetical reference or footnote/endnote number after—not within—the closed quotation mark.
Roosevelt declared, “The only thing we have to fear is fear itself” (Roosevelt, Public Papers 11).
Roosevelt declared, “The only thing we have to fear is fear itself.”1
How much should I quote?
As few words as possible. Remember, your paper should primarily contain your own words, so quote only the most pithy and memorable parts of sources. Here are three guidelines for selecting quoted material judiciously.
1. Excerpt fragments.
Sometimes, you should quote short fragments, rather than whole sentences. Suppose you interviewed Jane Doe about her reaction to John F. Kennedy’s assassination. She commented:
“I couldn’t believe it. It was just unreal and so sad. It was just unbelievable. I had never experienced such denial. I don’t know why I felt so strongly. Perhaps it was because JFK was more to me than a president. He represented the hopes of young people everywhere.”
You could quote all of Jane’s comments, but her first three sentences are fairly redundant. You might instead want to quote Jane when she arrives at the ultimate reason for her strong emotions:
Jane Doe grappled with grief and disbelief. She had viewed JFK, not just as a national figurehead, but as someone who “represented the hopes of young people everywhere.”
2. Excerpt those fragments carefully!
Quoting the words of others carries a big responsibility. Misquoting misrepresents the ideas of others. Here’s a classic example of a misquote:
John Adams has often been quoted as having said: “This would be the best of all possible worlds if there were no religion in it.”
John Adams did, in fact, write the above words. But if you see those words in context, the meaning changes entirely. Here’s the rest of the quotation:
Twenty times, in the course of my late reading, have I been on the point of breaking out, ‘this would be the best of all possible worlds, if there were no religion in it!!!!’ But in this exclamation, I should have been as fanatical as Bryant or Cleverly. Without religion, this world would be something not fit to be mentioned in public company—I mean hell.
As you can see from this example, context matters!
This example is from Paul F. Boller, Jr. and John George, They Never Said It: A Book of Fake Quotes, Misquotes, and Misleading Attributions (Oxford University Press, 1989).
3. Use block quotations sparingly.
There may be times when you need to quote long passages. However, you should use block quotations only when you fear that omitting any words will destroy the integrity of the passage. If that passage exceeds four lines (some sources say five), then set it off as a block quotation.
Here are a few general tips for setting off your block quotation—to be sure you are handling block quotes correctly in papers for different academic disciplines, check the index of the citation style guide you are using:
- Set up a block quotation with your own words followed by a colon.
- Indent. You normally indent 4-5 spaces for the start of a paragraph. When setting up a block quotation, indent the entire paragraph once from the left-hand margin.
- Single space or double space within the block quotation, depending on the style guidelines of your discipline (MLA, CSE, APA, Chicago, etc.).
- Do not use quotation marks at the beginning or end of the block quote—the indentation is what indicates that it’s a quote.
- Place parenthetical citation according to your style guide (usually after the period following the last sentence of the quote).
- Follow up a block quotation with your own words.
So, using the above example from John Adams, here’s how you might include a block quotation:
After reading several doctrinally rigid tracts, John Adams recalled the zealous ranting of his former teacher, Joseph Cleverly, and minister, Lemuel Bryant. He expressed his ambivalence toward religion in an 1817 letter to Thomas Jefferson:
Twenty times, in the course of my late reading, have I been on the point of breaking out, ‘this would be the best of all possible worlds, if there were no religion in it!!!!’ But in this exclamation, I should have been as fanatical as Bryant or Cleverly. Without religion, this world would be something not fit to be mentioned in public company—I mean hell.
Adams clearly appreciated religion, even if he often questioned its promotion.
How do I combine quotation marks with other punctuation marks?
It can be confusing when you start combining quotation marks with other punctuation marks. You should consult a style manual for complicated situations, but the following two rules apply to most cases:
1) Keep periods and commas within quotation marks.
So, for example:
According to Professor Jones, Lincoln “feared the spread of slavery,” but many of his aides advised him to “watch and wait.”
In the above example, both the comma and period were enclosed in the quotation marks. The main exception to this rule involves the use of internal citations, which always precede the last period of the sentence. For example:
According to Professor Jones, Lincoln “feared the spread of slavery,” but many of his aides advised him to “watch and wait” (Jones 143).
Note, however, that the period remains inside the quotation marks when your citation style involved superscript footnotes or endnotes. For example:
According to Professor Jones, Lincoln “feared the spread of slavery,” but many of his aides advised him to “watch and wait.”2
2) Place all other punctuation marks (colons, semicolons, exclamation marks, question marks) outside the quotation marks, except when they were part of the original quotation.
Take a look at the following examples:
The student wrote that the U. S. Civil War “finally ended around 1900”!
The coach yelled, “Run!”
In the first example, the author placed the exclamation point outside the quotation mark because she added it herself to emphasize the absurdity of the student’s comment. The student’s original comment had not included an exclamation mark. In the second example, the exclamation mark remains within the quotation mark because it is indicating the excited tone in which the coach yelled the command. Thus, the exclamation mark is considered to be part of the original quotation.
How do I indicate quotations within quotations?
If you are quoting a passage that contains a quotation, then you use single quotation marks for the internal quotation. Quite rarely, you quote a passage that has a quotation within a quotation. In that rare instance, you would use double quotation marks for the second internal quotation.
Here’s an example of a quotation within a quotation:
In “The Emperor’s New Clothes,” Hans Christian Andersen wrote, “‘But the Emperor has nothing on at all!’ cried a little child.”
Remember to consult your style guide to determine how to properly cite a quote within a quote.
When do I use those three dots ( . . . )?
Whenever you want to leave out material from within a quotation, you need to use an ellipsis, which is a series of three periods, each of which should be preceded and followed by a space. So, an ellipsis in this sentence would look like . . . this. There are a few rules to follow when using ellipses:
1. Be sure that you don’t fundamentally change the meaning of the quotation by omitting material.
Take a look at the following example: “The Writing Center is located on the UNC campus and serves the entire UNC community.”
“The Writing Center . . . serves the entire UNC community.”
The reader’s understanding of the Writing Center’s mission to serve the UNC community is not affected by omitting the information about its location.
2. Do not use ellipses at the beginning or ending of quotations, unless it’s important for the reader to know that the quotation was truncated.
For example, using the above example, you would NOT need an ellipsis in either of these situations:
“The Writing Center is located on the UNC campus . . .”
The Writing Center ” . . . serves the entire UNC community.”
3. Use punctuation marks in combination with ellipses when removing material from the end of sentences or clauses.
For example, if you take material from the end of a sentence, keep the period in as usual.
“The boys ran to school, forgetting their lunches and books. Even though they were out of breath, they made it on time.”
“The boys ran to school. . . . Even though they were out of breath, they made it on time.”
Likewise, if you excerpt material at the end of clause that ends in a comma, retain the comma.
“The red car came to a screeching halt that was heard by nearby pedestrians, but no one was hurt.”
“The red car came to a screeching halt . . . , but no one was hurt.”
Is it ever okay to insert my own words or change words in a quotation?
Figure \(\PageIndex{3}\)
Sometimes it is necessary for clarity and flow to alter a word or words within a quotation. You should make such changes rarely. In order to alert your reader to the changes you’ve made, you should always bracket the altered words. Here are a few examples of situations when you might need brackets.
1. Changing verb tense or pronouns in order to be consistent with the rest of the sentence.
Suppose you were quoting a woman who, when asked about her experiences immigrating to the United States, commented “nobody understood me.” You might write:
Esther Hansen felt that when she came to the United States “nobody understood [her].”
In the above example, you’ve changed “me” to “her” in order to keep the entire passage in third person. However, you could avoid the need for this change by simply rephrasing:
“Nobody understood me,” recalled Danish immigrant Esther Hansen.
2. Including supplemental information that your reader needs in order to understand the quotation.
For example, if you were quoting someone’s nickname, you might want to let your reader know the full name of that person in brackets.
“The principal of the school told Billy [William Smith] that his contract would be terminated.”
Similarly, if a quotation referenced an event with which the reader might be unfamiliar, you could identify that event in brackets.
“We completely revised our political strategies after the strike [of 1934].”
3. Indicating the use of nonstandard grammar or spelling.
In rare situations, you may quote from a text that has nonstandard grammar, spelling, or word choice. In such cases, you may want to insert [ sic ], which means “thus” or “so” in Latin. Using [ sic ] alerts your reader to the fact that this nonstandard language is not the result of a typo on your part. Always italicize “sic” and enclose it in brackets. There is no need to put a period at the end. Here’s an example of when you might use [ sic ]:
Twelve-year-old Betsy Smith wrote in her diary, “Father is afraid that he will be guilty of beach [ sic ] of contract.”
Here [ sic ] indicates that the original author wrote “beach of contract,” not breach of contract, which is the accepted terminology.
4. Do not overuse brackets!
For example, it is not necessary to bracket capitalization changes that you make at the beginning of sentences. For example, suppose you were going to use part of this quotation:
“We never looked back, but the memory of our army days remained with us the rest of our lives.”
If you wanted to begin a sentence with an excerpt from the middle of this quotation, there would be no need to bracket your capitalization changes.
“The memory of our army days remained with us the rest of our lives,” commented Joe Brown, a World War II veteran.
Not
“[T]he memory of our army days remained with us the rest of our lives,” commented Joe Brown, a World War II veteran.
Works consulted
We consulted these works while writing the original version of this handout. This is not a comprehensive list of resources on the handout’s topic, and we encourage you to do your own research to find the latest publications on this topic. Please do not use this list as a model for the format of your own reference list, as it may not match the citation style you are using. For guidance on formatting citations, please see the UNC Library’s citation tutorial .
Barzun, Jacques and Henry F. Graff. The Modern Researcher. 6th Edition. Harcourt Brace Jovanovich, 2004.
Booth, Wayne C., Gregory G. Colomb, and Joseph M. Williams. The Craft of Research, 2nd Edition. Chicago: University of Chicago Press, 2003.
Gibaldi, Joseph. MLA Handbook for Writers of Research Papers, 6th Edition. New York: The Modern Language Association of America, 2003.
Turabian, Kate L. A Manual for Writers of Term Papers, Theses, and Dissertations. 6th Edition. Chicago: University of Chicago Press, 1996.


<IP_ADDRESS>: Making a Budget
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What It Is
What is a budget?

A budget helps you make sure you will have enough money every month. Without a budget, you might run out of money before your next paycheck.

A budget helps you decide:
- what you must spend your money on
- if you can spend less money on some things and more money on other things
For example, your budget might show that you spend $100 on clothes every month. You might decide you can spend $50 on clothes. You can use the rest of the money to pay bills or to save for something else.
For Example
Why should I try to save money?
You might need money for an emergency. You also might need to buy something more expensive, like a car. Saving money might help you buy a car, put a security deposit on an apartment, or pay for something else expensive.
What To Know
How do I start a budget?
Start a budget by gathering your bills and pay stubs. Think about how you spend money, besides paying your bills. For example, do you buy a cup of coffee every day? After a month, that coffee money could add up to an expense you might write down.
When you have your bills and pay stubs:
- write down your expenses.  This is called income
- subtract your expenses from how much money you make
If the number is less than zero, you are spending more money than you make. Look for things in your budget you can change. Maybe something you do not need, or a way to spend less.
Use this Budget Worksheet to help you.
What if I don’t get paid every month?
Some people do not get paid every month. If you expect things to be like they were last year, do this:
- add all the money you earned last year
- divide that number by 12. This is about how much money you will have for each month
For Example
Last year my paychecks added up to $30,000.
$30,000 ÷ 12 = $2,500
I had about $2,500 each month.
How can I use my budget?
A budget is something you use every month. A written budget will help you:
- see where you spend money
- see where you can save
- make a plan for how to spend and save your money
Your budget can help you save money for the future. You can make savings one of your expenses. You might find ways to spend less money. Then you can put money into savings every month – maybe into a bank or credit union.
Why should I save money?
It can be hard to save money. It is very hard when your expenses go up and your income does not. Here are some reasons to try to save money even when it is not easy.
- Emergencies – Saving small amounts of money now might help you later. Everyone has expenses they do not expect.
- Expensive things – Sometimes, we have to pay for expensive things – like a car, a trip, or a security deposit on an apartment. You will have more choices if you have money to pay for those expensive things.
- Your goals – You might want to pay for college classes. Maybe you need to visit family in another country. You can plan for these goals and save money. Then you might not have to use a credit card or borrow money to pay.
How else can I save money?
You can try these ways to help save money:
- For one month, write down everything you spend. Small expenses, like a cup of coffee, can add up to a lot of money. When you know where you are spending your money, you can decide what you might not want to buy.
- Pay with your credit card only if you can pay the full amount when the bill comes. That way, you do not pay interest on what you owe.
- Pay your bills when they are due. That way, you will not owe late fees or other charges.
- Keep the money you are saving separate from the money you spend.
- Consider opening a savings account in a bank or credit union. Read more about opening a bank account.
- If you keep cash at home, keep the money you are saving separate from your spending money. Keep all your cash someplace safe.
For Example
What I did not buy this month:
| Music downloads | $5.00 |
| Shirt | $30.00 |
| Movie ticket | $10.00 |
| Top off gas tank | $15.00 |
| Cups of coffee | $12.00 |
| What I saved this month: | $72.00 |
What To Do
A budget is a plan that shows you how you can spend your money every month. Making a budget can help you make sure you do not run out of money each month. A budget also will help you save money for your goals or for emergencies.
How do I make a budget?
Write down your expenses. Expenses are what you spend money on. Expenses include:
Bills :
- bills that are the same each month, like rent
- bills that might change each month, like utilities
- bills you pay once or twice a year, like car insurance
Other expenses , like :
- food
- gas
- entertainment
- clothes
- school supplies
- money for family
- unplanned expenses, like car repairs or medical bills
- credit card bills
You might have bills that change every month. Look at what you paid for the same month last year. You might need $200 for your gas bill in January, but $30 in July.
Write down how much money you make. This includes your paychecks and any other money you get, like child support.
Subtract your expenses from how much money you make. This number should be more than zero. If it is less than zero, you are spending more money than you make. Look at your budget to see what you do not need or what you could spend less on.
How do I use my budget?
You can use your budget every month:
- At the beginning of the month, make a plan for how you will spend your money that month. Write what you think you will earn and spend.
- Write down what you spend. Try to do this every day.
- At the end of the month, see if you spent what you planned.
- Use the information to help you plan the next month’s budget.
Contributors and Attributions
- Authored by : Federal Trade Commission . Located at : https://www.consumer.gov/articles/1002-making-budget#!what-to-do . License : Public Domain: No Known Copyright


5.6: Second order systems and applications
Undamped Mass-Spring Systems
While we did say that we will usually only look at first order systems, it is sometimes more convenient to study the system in the way it arises naturally. For example, suppose we have 3 masses connected by springs between two walls. We could pick any higher number, and the math would be essentially the same, but for simplicity we pick 3 right now. Let us also assume no friction, that is, the system is undamped . The masses are \(m_1, m_2\) , and \(m_3\) and the spring constants are \(k_1, k_2, k_3\) , and \(k_4\) . Let \(x_1\) be the displacement from rest position of the first mass, and \(x_2\) and \(x_3\) the displacement of the second and third mass. We will make, as usual, positive values go right (as \(x_1\) grows, the first mass is moving right). See Figure \(\PageIndex{1}\).
This simple system turns up in unexpected places. For example, our world really consists of many small particles of matter interacting together. When we try the above system with many more masses, we obtain a good approximation to how an elastic material behaves. By somehow taking a limit of the number of masses going to infinity, we obtain the continuous one dimensional wave equation (that we study in Section 4.7 ). But we digress.
Let us set up the equations for the three mass system. By Hooke’s law we have that the force acting on the mass equals the spring compression times the spring constant. By Newton’s second law we have that force is mass times acceleration. So if we sum the forces acting on each mass and put the right sign in front of each term, depending on the direction in which it is acting, we end up with the desired system of equations.
\[\begin{array}{lll}{m_1 x''_1=-k_1 x_1+k_2(x_2-x_1)}&{\quad}&{=-(k_1+k_2)x_1+k_2x_2,} \\ {m_2x''_2=-k_2(x_2-x_1)+k_3(x_3-x_2)}&{\quad}&{=k_2x_1-(k_2+k_3)x_2+k_3x_3,} \\ {m_3x''_3=-k_3(x_3-x_2)-k_4x_3}&{\quad}&{=k_3x_2-(k_3+k_4)x_3.}\end{array} \nonumber \]
We define the matrices
\[ M = \left[ \begin{array}{ccc} m_1 & 0 & 0 \\ 0 & m_2 & 0 \\ 0 & 0 & m_3 \end{array} \right] ~~~{\rm{~and~}}~~~ K= \left[ \begin{array}{ccc} -(k_1+k_2) & k_2 & 0 \\ k_2 & -(k_2+k_3) & k_3 \\ 0 & k_3 & -(k_3+k_4) \end{array} \right]. \nonumber \]
We write the equation simply as
\[M \vec{x}'' = K \vec{x}. \nonumber \]
At this point we could introduce 3 new variables and write out a system of 6 first order equations. We claim this simple setup is easier to handle as a second order system. We call \(\vec{x}\) the displacement vector , \(M\) the mass matrix , and \(K\) the stiffness matrix .
Repeat this setup for 4 masses (find the matrices \( M\) and \(K\) ). Do it for 5 masses. Can you find a prescription to do it for \(n\) masses?
As with a single equation we want to “divide by \(M\).” This means computing the inverse of \(M\). The masses are all nonzero and \(M\)is a diagonal matrix, so comping the inverse is easy:
\[ M^{-1}= \left[ \begin{array}{ccc} \frac{1}{m_1} & 0 & 0 \\ 0 & \frac{1}{m_2} & 0 \\ 0 & 0 &\frac{1}{m_3}\end{array} \right]. \nonumber \]
This fact follows readily by how we multiply diagonal matrices. As an exercise, you should verify that \(MM^{-1}=M^{-1}M=I.\)
Let \(A = M^{-1}K\). We look at the system \( \vec{x}'' = M^{-1} K \vec{x}\), or
\[ \vec{x}'' = A \vec{x}. \nonumber \]
Many real world systems can be modeled by this equation. For simplicity, we will only talk about the given masses-and-springs problem. We try a solution of the form
\[ \vec{x} =\vec{v} e^{\alpha t}. \nonumber \]
We compute that for this guess, \( \vec{x}'' =\alpha^2 \vec{v} e^{\alpha t}.\) We plug our guess into the equation and get
\[ \alpha^2 \vec{v} e^{\alpha t} =A \vec{v} e^{\alpha t}. \nonumber \]
We divide by \(e^{\alpha t}\) to arrive at \( \alpha^2 \vec{v}=A\vec{v}\). Hence if \(\alpha^2\) is an eigenvalue of \(A\) and \(\vec{v}\) is a corresponding eigenvector, we have found a solution.
In our example, and in other common applications, \(A\) has only real negative eigenvalues (and possibly a zero eigenvalue). So we study only this case. When an eigenvalue \(\lambda\) is negative, it means that \(\alpha^2= \lambda\) is negative. Hence there is some real number \(\omega\) such that \(-\omega^2 = \lambda\). Then \(\alpha= \pm i \omega\). The solution we guessed was
\[ \vec{x}= \vec{v}(\cos(\omega t) +i \sin(\omega t)). \nonumber \]
By taking the real and imaginary parts (note that \(\vec{v}\) is real), we find that \(\vec{v}\cos(\omega t)\) and \(\vec{v}\sin(\omega t)\) are linearly independent solutions.
If an eigenvalue is zero, it turns out that both \(\vec{v}\) and \(\vec{v}t\) are solutions, where \(\vec{v}\) is an eigenvector corresponding to the eigenvalue 0.
Show that if \(A\) has a zero eigenvalue and \(\vec{v}\) is a corresponding eigenvector, then \(\vec{x}=\vec{v}(a+bt)\) is a solution of \(\vec{x}''=A \vec{x}\) for arbitrary constants \(a\) and \(b\) .
Let \(A\) be an \(n \times n\) matrix with \(n\) distinct real negative eigenvalues we denote by \(-\omega_1^2 > -\omega_2^2> \cdots > -\omega_n^2\) , and corresponding eigenvectors by \(\vec{v}_1, \vec{v}_2, \ldots, \vec{v}_n.\) If \(A\) is invertible (that is, if \(\omega_1>0\) ), then
\[ \vec{x}(t)= \sum_{i=1}^{n} \vec{v}_i(a_i \cos (\omega_i t) + b_i \sin (\omega_i t)), \nonumber \]
is the general solution of
\[ \vec{x}''=A \vec{x}, \nonumber \]
for some arbitrary constants \(a_i\) and \(b_i\) . If \(A\) has a zero eigenvalue, that is \( \omega_1=0\) , and all other eigenvalues are distinct and negative, then the general solution can be written as
\[ \vec{x}(t)= \vec{v}_1(a_1+b_1 t)+ \sum_{i=2}^{n} \vec{v}_i(a_i \cos (\omega_i t) + b_i \sin (\omega_i t)). \nonumber \]
We use this solution and the setup from the introduction of this section even when some of the masses and springs are missing. For example, when there are only 2 masses and only 2 springs, simply take only the equations for the two masses and set all the spring constants for the springs that are missing to zero.
Suppose we have the system in Figure \(\PageIndex{2}\), with \(m_1 = 2, m_2=1, k_1 = 4,\) and \(k_2=2.\)
The equations we write down are
\[ \left[ \begin{array}{cc} 2 & 0 \\ 0 & 1 \end{array} \right] \vec{x}''= \left[ \begin{array}{cc} -(4+2) & 2 \\ 2 & -2 \end{array} \right] \vec{x}, \nonumber \]
or
\[ \vec{x}''= \left[ \begin{array}{cc} -3 & 1 \\ 2 & -2 \end{array} \right] \vec{x}. \nonumber \]
We find the eigenvalues of \(A\) to be \(\lambda = -1, -4\) (exercise). We find corresponding eigenvectors to be \(\left[ \begin{array}{c} 1 \\ 2 \end{array} \right]\) and \(\left[ \begin{array}{c} 1 \\ -1 \end{array} \right]\) respectively (exercise).
We check the theorem and note that \(\omega_1=1\) and \(\omega_2=2\). Hence the general solution is
\[ \vec{x}= \left[ \begin{array}{c} 1 \\ 2 \end{array} \right] (a_1 \cos (t) + b_1 \sin (t)) + \left[ \begin{array}{c} 1 \\ -1 \end{array} \right] (a_2 \cos (2t) + b_2 \sin (2t)). \nonumber \]
The two terms in the solution represent the two so-called natural or normal modes of oscillation . And the two (angular) frequencies are the natural frequencies . The first natural frequency is \(1\), and second natural frequency is \(2\). The two modes are plotted in Figure \(\PageIndex{3}\).
Let us write the solution as
\[ \vec{x}= \left[ \begin{array}{c} 1 \\ 2 \end{array} \right] c_1 \cos (t-\alpha_1) + \left[ \begin{array}{c} 1 \\ -1 \end{array} \right] c_2 \cos (2t- \alpha_2). \nonumber \]
The first term,
\[ \left[ \begin{array}{c} 1 \\ 2 \end{array} \right] c_1 \cos (t-\alpha_1) = \left[ \begin{array}{c} c_1 \cos (t-\alpha_1) \\ 2c_1 \cos (t-\alpha_1) \end{array} \right], \nonumber \]
corresponds to the mode where the masses move synchronously in the same direction.
The second term,
\[ \left[ \begin{array}{c} 1 \\ -1 \end{array} \right] c_2 \cos (2t-\alpha_2) = \left[ \begin{array}{c} c_2 \cos (2t-\alpha_2) \\ -c_2 \cos (2t-\alpha_2) \end{array} \right], \nonumber \]
corresponds to the mode where the masses move synchronously but in opposite directions.
The general solution is a combination of the two modes. That is, the initial conditions determine the amplitude and phase shift of each mode. As an example, suppose we have initial conditions
\[\vec{x}(0) = \begin{bmatrix} 1 \\ -1 \end{bmatrix} , \qquad \vec{x}'(0) = \begin{bmatrix} 0 \\ 6 \end{bmatrix} . \nonumber \]
We use the \(a_j, b_j\) constants to solve for initial conditions. First
\[\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \vec{x}(0) = \begin{bmatrix} 1 \\ 2 \end{bmatrix} a_1 + \begin{bmatrix} 1 \\ -1 \end{bmatrix} a_2 = \begin{bmatrix} a_1+a_2 \\2a_1 - a_2 \end{bmatrix} . \nonumber \]
We solve (exercise) to find \(a_1 = 0\), \(a_2 = 1\). To find the \(b_1\) and \(b_2\), we differentiate first: \[{\vec{x}}' = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \bigl( - a_1 \sin (t) + b_1 \cos (t) \bigr) + \begin{bmatrix} 1 \\ -1 \end{bmatrix} \bigl( - 2a_2 \sin (2t) + 2 b_2 \cos (2t) \bigr) . \nonumber \]
Now we solve: \[\begin{bmatrix} 0 \\ 6 \end{bmatrix} = {\vec{x}}'(0) = \begin{bmatrix} 1 \\ 2 \end{bmatrix} b_1 + \begin{bmatrix} 1 \\ -1 \end{bmatrix} 2 b_2 = \begin{bmatrix} b_1+2b_2 \\ 2b_1-2b_2 \end{bmatrix} . \nonumber \]
Again solve (exercise) to find \(b_1 = 2\), \(b_2 = -1\). So our solution is \[\vec{x} = \begin{bmatrix} 1 \\ 2 \end{bmatrix} 2 \sin (t) + \begin{bmatrix} 1 \\ -1 \end{bmatrix} \bigl( \cos (2t) - \sin (2t) \bigr) = \begin{bmatrix} 2 \sin (t) + \cos(2t)- \sin(2t) \\ 4 \sin (t) - \cos(2t) + \sin(2t) \end{bmatrix} . \nonumber \]
The graphs of the two displacements, \(x_1\) and \(x_2\) of the two carts is in Figure \(\PageIndex{4}\).
Below is a video on coupled oscillators.
We have two toy rail cars. Car 1 of mass \(2\text{ kg}\) is traveling at \(3\:\frac{\text{m}}{\text{s}}\) towards the second rail car of mass \(1\text{ kg}\). There is a bumper on the second rail car that engages at the moment the cars hit (it connects to two cars) and does not let go. The bumper acts like a spring of spring constant \(k=2\:\frac{\text{N}}{\text{m}}\). The second car is \(10\) meters from a wall. See Figure \(\PageIndex{5}\).
We want to ask several questions. At what time after the cars link does impact with the wall happen? What is the speed of car 2 when it hits the wall?
OK, let us first set the system up. Let \(t=0\) be the time when the two cars link up. Let \(x_1\) be the displacement of the first car from the position at \(t=0\), and let \(x_2\) be the displacement of the second car from its original location. Then the time when \(x_2(t)=10\) is exactly the time when impact with wall occurs. For this \(t,x'_2(t)\) is the speed at impact. This system acts just like the system of the previous example but without \(k_1\). Hence the equation is
\[ \left[ \begin{array}{cc} 2 & 0 \\ 0 & 1 \end{array} \right] \vec{x}''= \left[ \begin{array}{cc} -2 & 2 \\ 2 & -2 \end{array} \right] \vec{x}. \nonumber \]
or
\[ \vec{x}''= \left[ \begin{array}{cc} -1 & 1 \\ 2 & -2 \end{array} \right] \vec{x}. \nonumber \]
We compute the eigenvalues of \(A\). It is not hard to see that the eigenvalues are \(0\) and \(-3\) (exercise). Furthermore, eigenvectors are \(\left[ \begin{array}{c} 1 \\ 1 \end{array} \right]\) and \(\left[ \begin{array}{c} 1 \\ -2 \end{array} \right]\) respectively (exercise). Then \(\omega_2= \sqrt{3}\) and by the second part of the theorem we find our general solution to be
\[ \begin{align}\begin{aligned} \vec{x}&= \left[ \begin{array}{c} 1 \\ 1 \end{array} \right](a_1+b_1t) + \left[ \begin{array}{c} 1 \\ -2 \end{array} \right](a_2 \cos(\sqrt3 t)+ b_2 \sin(\sqrt3 t)) \\ &=\left[ \begin{array}{c} a_1 + b_1t+a_2 \cos(\sqrt3 t)+ b_2 \sin(\sqrt3 t) \\ a_1+ b_1t- 2a_2 \cos(\sqrt3 t)- 2b_2 \sin(\sqrt3 t) \end{array} \right]\end{aligned}\end{align} \nonumber \]
We now apply the initial conditions. First the cars start at position \(0\) so \(x_1(0)=0\) and \(x_2(0)=0\). The first car is traveling at \(3\:\frac{\text{m}}{\text{s}}\), so \(x'_1(0)=3\) and the second car starts at rest, so \(x'_2(0)=0\). The first conditions says
\[ \vec{0} = \vec{x}(0)= \left[ \begin{array}{c} a_1+ a_2 \\ a_1-2a_2 \end{array} \right]. \nonumber \]
It is not hard to see that \(a_1=a_2=0\). We set \(a_1=0\) and \(a_2=0\) in \(\vec{x}(t)\) and differentiate to get
\[ \vec{x}'(t)= \left[ \begin{array}{c} b_1+ \sqrt3 b_2 \cos(\sqrt3 t)\\ b_1- 2 \sqrt3 b_2 \cos(\sqrt3 t) \end{array} \right]. \nonumber \]
So
\[ \left[ \begin{array}{c} 3\\ 0 \end{array} \right] = \vec{x}'(0)= \left[ \begin{array}{c} b_1+ \sqrt3 b_2 \\ b_1- 2 \sqrt3 b_2 \end{array} \right]. \nonumber \]
Solving these two equations we find \(b_1 = 2\) and \(b_2 = \frac{1}{\sqrt3}\). Hence the position of our cars is (until the impact with the wall)
\[ \vec{x}= \left[ \begin{array}{c} 2t +\frac{1}{\sqrt3} \sin(\sqrt3 t) \\ 2t -\frac{2}{\sqrt3} \sin(\sqrt3 t)\end{array} \right] . \nonumber \]
Note how the presence of the zero eigenvalue resulted in a term containing \(t\). This means that the carts will be traveling in the positive direction as time grows, which is what we expect.
What we are really interested in is the second expression, the one for \(x_2\). We have \(x_2(t)= 2t -\frac{2}{\sqrt3} \sin(\sqrt3 t)\). See Figure \(\PageIndex{6}\) for the plot of \(x_2\) versus time.
Just from the graph we can see that time of impact will be a little more than 5 seconds from time zero. For this we have to solve the equation \(10=x_{2}(t)=2t-\frac{2}{\sqrt{3}}\sin (\sqrt{3}t)\) . Using a computer (or even a graphing calculator) we find that \(t_{\text{impact}}\approx 5.22\) seconds.
As for the speed we note that \(x'_2 = 2-2 \cos(\sqrt3 t)\). At time of impact (\(5.22\) seconds from \(t=0\)) we get that \( x'_2(t_{\rm{impact}}) \approx 3.85\).
The maximum speed is the maximum of \(2-2 \cos(\sqrt3 t)\), which is \(4\). We are traveling at almost the maximum speed when we hit the wall.
 Tiana has a Martini in her hand and would like not to spill it. Let us suppose Tiana would not spill her Martini when the first car links up with car 2, but if car 2 hits the wall at any speed greater than zero, Tiana will spill her drink. Suppose Tiana can move car 2 a few meters towards or away from the wall (he cannot go all the way to the wall, nor can she get out of the way of the first car). Is there a “safe” distance for her to be at? A distance such that the impact with the wall is at zero speed?
The answer is yes. Looking at Figure \(\PageIndex{6}\), we note the “plateau” between \(t=3\) and \(t=4\). There is a point where the speed is zero. To find it we need to solve \(x'_2(t)=0\). This is when \(\cos(\sqrt3 t) = 1\) or in other words when \(t=\frac{2 \pi}{\sqrt3}, \frac{4 \pi}{\sqrt3}, \ldots\) and so on. We plug in the first value to obtain \( x_2 \left( \frac{2 \pi}{\sqrt3} \right)=\frac{4 \pi}{\sqrt3} \approx 7.26 \). So a “safe” distance is about 7 and a quarter meters from the wall.
Alternatively Tiana could move away from the wall towards the incoming car 2 where another safe distance is \( \frac{8 \pi}{\sqrt3 } \approx 14.51\) and so on, using all the different \(t\) such that \(x'_2(t)=0\). Of course \(t=0\) is always a solution here, corresponding to \(x_2=0\), but that means standing right at the wall.
Below is a video on normal modes.
Below is another video on normal modes.
Forced Oscillations
Finally we move to forced oscillations. Suppose that now our system is
\[\label{eq:30} \vec{x}'' = A \vec{x} + \vec{F} \cos(\omega t). \]
That is, we are adding periodic forcing to the system in the direction of the vector \(\vec{F}\).
As before, this system just requires us to find one particular solution \(\vec{x}_p\), add it to the general solution of the associated homogeneous system \(\vec{x}_c\), and we will have the general solution to \(\eqref{eq:30}\). Let us suppose that \(\omega\) is not one of the natural frequencies of \( \vec{x}'' = A \vec{x}\), then we can guess
\[ \vec{x}_p= \vec{c} \cos(\omega t), \nonumber \]
where \(\vec{c}\) is an unknown constant vector. Note that we do not need to use sine since there are only second derivatives. We solve for \(\vec{c}\) to find \(\vec{x}_p\). This is really just the method of undetermined coefficients for systems. Let us differentiate \(\vec{x}_p\) twice to get
\[ \vec{x}''_p= -\omega^2 \vec{c} \cos(\omega t). \nonumber \]
Plug \(\vec{x}_p\) and \(\vec{x}''_p\) into the equation \(\eqref{eq:30}\):
\[\overbrace{ -\omega^2 \vec{c} \cos (\omega t) }^{{\vec{x}_p}''} = \overbrace{ A \vec{c} \cos (\omega t) }^{A \vec{x}_p} + \vec{F} \cos (\omega t) . \nonumber \]
We cancel out the cosine and rearrange the equation to obtain
\[ (A + \omega^2 I) \vec{c} = - \vec{F}. \nonumber \]
So
\[ \vec{c} = (A + \omega^2 I)^{-1} (- \vec{F}). \nonumber \]
Of course this is possible only if \( (A + \omega^2 I)= (A -(- \omega^2) I)\) is invertible. That matrix is invertible if and only if \(- \omega^2\) is not an eigenvalue of \(A\). That is true if and only if \(\omega\) is not a natural frequency of the system.
We simplified things a little bit. If we wish to have the forcing term to be in the units of force, say Newtons, then we must write \[M \vec{x}'' = K \vec{x} + \vec{G} \cos(\omega t) . \nonumber \]
If we then write things in terms of \(A = M^{-1} K\), we have \[\vec{x}'' = M^{-1}K \vec{x} + M^{-1} \vec{G} \cos(\omega t) \qquad \text{or} \qquad \vec{x}'' = A \vec{x} + \vec{F} \cos(\omega t) , \nonumber \]
where \(\vec{F} = M^{-1} \vec{G}\).
Let us take the example in Figure \(\PageIndex{2}\) with the same parameters as before: \(m_1=2, m_2=1, k_1= 4,\) and \(k_2=2\). Now suppose that there is a force \( 2 \cos(3t)\) acting on the second cart.
The equation is
\[\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} {\vec{x}}'' = \begin{bmatrix} -4 & 2 \\ 2 & -2 \end{bmatrix} \vec{x} + \begin{bmatrix} 0 \\ 2 \end{bmatrix} \cos (3 t) \qquad \text{or} \qquad {\vec{x}}'' = \begin{bmatrix} -3 & 1 \\ 2 & -2 \end{bmatrix} \vec{x} + \begin{bmatrix} 0 \\ 2 \end{bmatrix} \cos (3 t) . \nonumber \]
We solved the associated homogeneous equation before and found the complementary solution to be
\[ \vec{x}_c= \left[ \begin{array}{c} 1 \\ 2 \end{array} \right] (a_1 \cos(t) + b_1 \sin(t)) + \left[ \begin{array}{c} 1 \\ -1 \end{array} \right] (a_2 \cos(2t) + b_2 \sin(2t)). \nonumber \]
The natural frequencies are \(1\) and \(2\). Hence as \(3\) is not a natural frequency, we can try \( \vec{c} \cos(3t)\). We invert \((A+3^2 I)\):
\[ \left( \left[ \begin{array}{cc} -3 & 1 \\ 2 & -2 \end{array} \right] + 3^2 I \right)^{-1}= \left[ \begin{array}{cc} 6 & 1 \\ 2 & 7 \end{array} \right]^{-1}= \left[ \begin{array}{cc} \frac{7}{40} & \frac{-1}{40} \\ \frac{-1}{20} & \frac{3}{20} \end{array} \right]. \nonumber \]
Hence,
\[ \vec{c}= (A+\omega^2 I)^{-1} (- \vec{F})= \left[ \begin{array}{cc} \frac{7}{40} & \frac{-1}{40} \\ \frac{-1}{20} & \frac{3}{20} \end{array} \right] \left[ \begin{array}{c} 0 \\ -2 \end{array} \right] = \left[ \begin{array}{c} \frac{1}{20} \\ \frac{-3}{10} \end{array} \right]. \nonumber \]
Combining with what we know the general solution of the associated homogeneous problem to be, we get that the general solution to \(\vec{x}''= A \vec{x} +\vec{F} \cos(\omega t)\) is
\[ \vec{x} = \vec{x}_c + \vec{x}_p = \left[ \begin{array}{c} 1 \\ 2 \end{array} \right] (a_1 \cos(t) + b_1 \sin(t)) + \left[ \begin{array}{c} 1 \\ -1 \end{array} \right] (a_2 \cos(2t) + b_2 \sin(2t)) + \left[ \begin{array}{c} \frac{1}{20} \\ \frac{-3}{10} \end{array} \right] \cos(3t). \nonumber \]
The constants \(a_1, a_2, b_1,\) and \(b_2\) must then be solved for given any initial conditions.
Note that given force \(\vec{f}\), we write the equation as \(M\vec{x}''=K\vec{x}+\vec{f}\) to get the units right. Then we write \(\vec{x}''=M^{-1}K\vec{x}+M^{-1}\vec{f}\). The term \(\vec{g}=M^{-1}\vec{f}\) in \(\vec{x}''=A\vec{x}+\vec{g}\) is in units of force per unit mass.
If \(\omega\) is a natural frequency of the system resonance occurs because we will have to try a particular solution of the form
\[ \vec{x}_p = \vec{c}t \sin(\omega t)+ \vec{d}t \cos(\omega t) . \nonumber \]
That is assuming that the eigenvalues of the coefficient matrix are distinct. Next, note that the amplitude of this solution grows without bound as \(t\) grows.


13.4: Refraction
Economists normally consider free markets to be the natural way of judging the monetary value of something, but social scientists also use questionnaires to gauge the relative value of privileges, disadvantages, or possessions that cannot be bought or sold. They ask people to imagine that they could trade one thing for another and ask which they would choose. One interesting result is that the average light-skinned person in the U.S. would rather lose an arm than suffer the racist treatment routinely endured by African-Americans. Even more impressive is the value of sight. Many prospective parents can imagine without too much fear having a deaf child, but would have a far more difficult time coping with raising a blind one.
So great is the value attached to sight that some have imbued it with mystical aspects. Joan of Arc saw visions, and my college has a “vision statement.” Christian fundamentalists who perceive a conflict between evolution and their religion have claimed that the eye is such a perfect device that it could never have arisen through a process as helter-skelter as evolution, or that it could not have evolved because half of an eye would be useless. In fact, the structure of an eye is fundamentally dictated by physics, and it has arisen separately by evolution somewhere between eight and 40 times, depending on which biologist you ask. We humans have a version of the eye that can be traced back to the evolution of a light-sensitive “eye spot” on the head of an ancient invertebrate. A sunken pit then developed so that the eye would only receive light from one direction, allowing the organism to tell where the light was coming from. (Modern flatworms have this type of eye.) The top of the pit then became partially covered, leaving a hole, for even greater directionality (as in the nautilus). At some point the cavity became filled with jelly, and this jelly finally became a lens, resulting in the general type of eye that we share with the bony fishes and other vertebrates. Far from being a perfect device, the vertebrate eye is marred by a serious design flaw due to the lack of planning or intelligent design in evolution: the nerve cells of the retina and the blood vessels that serve them are all in front of the light-sensitive cells, blocking part of the light. Squids and other molluscs, whose eyes evolved on a separate branch of the evolutionary tree, have a more sensible arrangement, with the light-sensitive cells out in front.
12.4.1 Refraction
The fundamental physical phenomenon at work in the eye is that when light crosses a boundary between two media (such as air and the eye's jelly), part of its energy is reflected, but part passes into the new medium. In the ray model of light, we describe the original ray as splitting into a reflected ray and a transmitted one (the one that gets through the boundary). Of course the reflected ray goes in a direction that is different from that of the original one, according to the rules of reflection we have already studied. More surprisingly --- and this is the crucial point for making your eye focus light --- the transmitted ray is bent somewhat as well. This bending phenomenon is called refraction . The origin of the word is the same as that of the word “fracture,” i.e., the ray is bent or “broken.” (Keep in mind, however, that light rays are not physical objects that can really be “broken.”) Refraction occurs with all waves, not just light waves.
Figure \(\PageIndex{1}\) : A human eye. Figure \(\PageIndex{2}\): The anatomy of the eye.
The actual anatomy of the eye, Figure \(\PageIndex{2}\), is quite complex, but in essence it is very much like every other optical device based on refraction. The rays are bent when they pass through the front surface of the eye, Figure \(\PageIndex{3}\). Rays that enter farther from the central axis are bent more, with the result that an image is formed on the retina. There is only one slightly novel aspect of the situation. In most human-built optical devices, such as a movie projector, the light is bent as it passes into a lens, bent again as it reemerges, and then reaches a focus beyond the lens. In the eye, however, the “screen” is inside the eye, so the rays are only refracted once, on entering the jelly, and never emerge again.
A common misconception is that the “lens” of the eye is what does the focusing. All the transparent parts of the eye are made of fairly similar stuff, so the dramatic change in medium is when a ray crosses from the air into the eye (at the outside surface of the cornea). This is where nearly all the refraction takes place. The lens medium differs only slightly in its optical properties from the rest of the eye, so very little refraction occurs as light enters and exits the lens. The lens, whose shape is adjusted by muscles attached to it, is only meant for fine-tuning the focus to form images of near or far objects.
Refractive Properties of Media
What are the rules governing refraction? The first thing to observe is that just as with reflection, the new, bent part of the ray lies in the same plane as the normal (perpendicular) and the incident ray, Figure \(\PageIndex{4}\).
If you try shooting a beam of light at the boundary between two substances, say water and air, you'll find that regardless of the angle at which you send in the beam, the part of the beam in the water is always closer to the normal line, Figure \(\PageIndex{5}\).
It doesn't matter if the ray is entering the water or leaving, so refraction is symmetric with respect to time-reversal, Figure \(\PageIndex{6}\).
If, instead of water and air, you try another combination of substances, say plastic and gasoline, again you'll find that the ray's angle with respect to the normal is consistently smaller in one and larger in the other. Also, we find that if substance A has rays closer to normal than in B, and B has rays closer to normal than in C, then A has rays closer to normal than C. This means that we can rank-order all materials according to their refractive properties. Isaac Newton did so, including in his list many amusing substances, such as “Danzig vitriol” and “a pseudo-topazius, being a natural, pellucid, brittle, hairy stone, of a yellow color.” Several general rules can be inferred from such a list:
- Vacuum lies at one end of the list. In refraction across the interface between vacuum and any other medium, the other medium has rays closer to the normal.
- Among gases, the ray gets closer to the normal if you increase the density of the gas by pressurizing it more.
- The refractive properties of liquid mixtures and solutions vary in a smooth and systematic manner as the proportions of the mixture are changed.
- Denser substances usually, but not always, have rays closer to the normal.
The second and third rules provide us with a method for measuring the density of an unknown sample of gas, or the concentration of a solution. The latter technique is very commonly used, and the CRC Handbook of Physics and Chemistry, for instance, contains extensive tables of the refractive properties of sugar solutions, cat urine, and so on.
Snell's Law
The numerical rule governing refraction was discovered by Snell, who must have collected experimental data something like what is shown on this graph and then attempted by trial and error to find the right equation. The equation he came up with was
The value of the constant would depend on the combination of media used. For instance, any one of the data points in the graph would have sufficed to show that the constant was 1.3 for an air-water interface (taking air to be substance 1 and water to be substance 2).
Snell further found that if media A and B gave a constant \(K_{AB}\) and media B and C gave a constant \(K_{BC}\), then refraction at an interface between A and C would be described by a constant equal to the product, \(K_{AC}=K_{AB}K_{BC}\). This is exactly what one would expect if the constant depended on the ratio of some number characterizing one medium to the number characteristic of the second medium. This number is called the index of refraction of the medium, written as \(n\) in equations. Since measuring the angles would only allow him to determine the ratio of the indices of refraction of two media, Snell had to pick some medium and define it as having \(n=1\). He chose to define vacuum as having \(n=1\). (The index of refraction of air at normal atmospheric pressure is 1.0003, so for most purposes it is a good approximation to assume that air has \(n=1\).) He also had to decide which way to define the ratio, and he chose to define it so that media with their rays closer to the normal would have larger indices of refraction. This had the advantage that denser media would typically have higher indices of refraction, and for this reason the index of refraction is also referred to as the optical density. Written in terms of indices of refraction, Snell's equation becomes
but rewriting it in the form
[relationship between angles of rays at the interface between media with indices of refraction \(n_1\) and \(n_2\); angles are defined with respect to the normal] makes us less likely to get the 1's and 2's mixed up, so this the way most people remember Snell's law. A few indices of refraction are given in the back of the book.
Exercise \(\PageIndex{1}\)
- What would the graph look like for two substances with the same index of refraction?
- Based on the graph, when does refraction at an air-water interface change the direction of a ray most strongly?
(answer in the back of the PDF version of the book)
Example \(\PageIndex{1}\): Finding an angle using Snell's law
A submarine shines its searchlight up toward the surface of the water. What is the angle \(\alpha \) shown in Figure \(\PageIndex{8}\)?
Figure \(\PageIndex{8}\): Example 10 .
Solution
The tricky part is that Snell's law refers to the angles with respect to the normal . Forgetting this is a very common mistake. The beam is at an angle of \(30°\) with respect to the normal in the water. Let's refer to the air as medium 1 and the water as 2. Solving Snell's law for \(\theta_1\), we find
As mentioned above, air has an index of refraction very close to 1, and water's is about 1.3, so we find \(\theta_1=40°\). The angle \(\alpha \) is therefore \(50°\).
What neither Snell nor Newton knew was that there is a very simple interpretation of the index of refraction. This may come as a relief to the reader who is taken aback by the complex reasoning involving proportionalities that led to its definition. Later experiments showed that the index of refraction of a medium was inversely proportional to the speed of light in that medium. Since \(c\) is defined as the speed of light in vacuum, and \(n=1\) is defined as the index of refraction of vacuum, we have
[\(n=\) medium's index of refraction, \(v=\) speed of light in that medium, \(c=\) speed of light in a vacuum]
Many textbooks start with this as the definition of the index of refraction, although that approach makes the quantity's name somewhat of a mystery, and leaves students wondering why \(c/v\) was used rather than \(v/c\). It should also be noted that measuring angles of refraction is a far more practical method for determining \(n\) than direct measurement of the speed of light in the substance of interest.
A mechanical model of Snell's law
Why should refraction be related to the speed of light? The mechanical model shown in the figure may help to make this more plausible. Suppose medium 2 is thick, sticky mud, which slows down the car. The car's right wheel hits the mud first, causing the right side of the car to slow down. This will cause the car to turn to the right until is moves far enough forward for the left wheel to cross into the mud. After that, the two sides of the car will once again be moving at the same speed, and the car will go straight.
Of course, light isn't a car. Why should a beam of light have anything resembling a “left wheel” and “right wheel?” After all, the mechanical model would predict that a motorcycle would go straight, and a motorcycle seems like a better approximation to a ray of light than a car. The whole thing is just a model, not a description of physical reality.
A Derivation of Snell's law
However intuitively appealing the mechanical model may be, light is a wave, and we should be using wave models to describe refraction. In fact Snell's law can be derived quite simply from wave concepts. Figure \(\PageIndex{9}\) shows the refraction of a water wave. The water in the upper left part of the tank is shallower, so the speed of the waves is slower there, and their wavelengths is shorter. The reflected part of the wave is also very faintly visible.
In the close-up view on the right, the dashed lines are normals to the interface. The two marked angles on the right side are both equal to \(\theta_1\), and the two on the left to \(\theta_2\).
Trigonometry gives
Eliminating \(h\) by dividing the equations, we find
The frequencies of the two waves must be equal or else they would get out of step, so by \(v=f\lambda \) we know that their wavelengths are proportional to their velocities. Combining \(\lambda\propto v\) with \(v\propto 1/n\) gives \(\lambda\propto 1/n\), so we find
which is one form of Snell's law.
Example \(\PageIndex{1}\): Ocean waves near and far from shore
Ocean waves are formed by winds, typically on the open sea, and the wavefronts are perpendicular to the direction of the wind that formed them. At the beach, however, you have undoubtedly observed that waves tend come in with their wavefronts very nearly (but not exactly) parallel to the shoreline. This is because the speed of water waves in shallow water depends on depth: the shallower the water, the slower the wave. Although the change from the fast-wave region to the slow-wave region is gradual rather than abrupt, there is still refraction, and the wave motion is nearly perpendicular to the normal in the slow region.
Color and Refraction
In general, the speed of light in a medium depends both on the medium and on the wavelength of the light. Another way of saying it is that a medium's index of refraction varies with wavelength. This is why a prism can be used to split up a beam of white light into a rainbow. Each wavelength of light is refracted through a different angle.
How much light is reflected, and how much is transmitted?
In section 6.2 we developed an equation for the percentage of the wave energy that is transmitted and the percentage reflected at a boundary between media. This was only done in the case of waves in one dimension, however, and rather than discuss the full three dimensional generalization it will be more useful to go into some qualitative observations about what happens. First, reflection happens only at the interface between two media, and two media with the same index of refraction act as if they were a single medium. Thus, at the interface between media with the same index of refraction, there is no reflection, and the ray keeps going straight. Continuing this line of thought, it is not surprising that we observe very little reflection at an interface between media with similar indices of refraction.
The next thing to note is that it is possible to have situations where no possible angle for the refracted ray can satisfy Snell's law. Solving Snell's law for \(\theta_2\), we find
and if \(n_1\) is greater than \(n_2\), then there will be large values of \(\theta_1\) for which the quantity \((n_1/n_2)\sin\theta \) is greater than one, meaning that your calculator will flash an error message at you when you try to take the inverse sine. What can happen physically in such a situation? The answer is that all the light is reflected, so there is no refracted ray. This phenomenon is known as total internal reflection , and is used in the fiber-optic cables that nowadays carry almost all long-distance telephone calls.
The electrical signals from your phone travel to a switching center, where they are converted from electricity into light. From there, the light is sent across the country in a thin transparent fiber. The light is aimed straight into the end of the fiber, and as long as the fiber never goes through any turns that are too sharp, the light will always encounter the edge of the fiber at an angle sufficiently oblique to give total internal reflection. If the fiber-optic cable is thick enough, one can see an image at one end of whatever the other end is pointed at.
Alternatively, a bundle of cables can be used, since a single thick cable is too hard to bend. This technique for seeing around corners is useful for making surgery less traumatic. Instead of cutting a person wide open, a surgeon can make a small “keyhole” incision and insert a bundle of fiber-optic cable (known as an endoscope) into the body.
Since rays at sufficiently large angles with respect to the normal may be completely reflected, it is not surprising that the relative amount of reflection changes depending on the angle of incidence, and is greatest for large angles of incidence.
Discussion Questions
◊ What index of refraction should a fish have in order to be invisible to other fish?
◊ Does a surgeon using an endoscope need a source of light inside the body cavity? If so, how could this be done without inserting a light bulb through the incision?
◊ A denser sample of a gas has a higher index of refraction than a less dense sample (i.e., a sample under lower pressure), but why would it not make sense for the index of refraction of a gas to be proportional to density?
◊ The earth's atmosphere gets thinner and thinner as you go higher in altitude. If a ray of light comes from a star that is below the zenith, what will happen to it as it comes into the earth's atmosphere?
◊ Does total internal reflection occur when light in a denser medium encounters a less dense medium, or the other way around? Or can it occur in either case?
12.4.2 Lenses
Figures n/1 and n/2 show examples of lenses forming images. There is essentially nothing for you to learn about imaging with lenses that is truly new. You already know how to construct and use ray diagrams, and you know about real and virtual images. The concept of the focal length of a lens is the same as for a curved mirror. The equations for locating images and determining magnifications are of the same form. It's really just a question of flexing your mental muscles on a few examples. The following self-checks and discussion questions will get you started.
Exercise \(\PageIndex{1}\)
- In figures Figure \(\PageIndex{13; part 1}\) and Figure \(\PageIndex{13; part 2}\), classify the images as real or virtual.
- Glass has an index of refraction that is greater than that of air. Consider the topmost ray in Figure \(\PageIndex{13; part 1}\). Explain why the ray makes a slight left turn upon entering the lens, and another left turn when it exits.
- If the flame in Figure \(\PageIndex{13; part 2}\) were moved closer to the lens, what would happen to the location of the image?
- Answer
-
(answer in the back of the PDF version of the book)
Discussion Questions
◊ In figures n/1 and n/2, the front and back surfaces are parallel to each other at the center of the lens. What will happen to a ray that enters near the center, but not necessarily along the axis of the lens? Draw a BIG ray diagram, and show a ray that comes from off axis.
In discussion questions B-F, don't draw ultra-detailed ray diagrams as in A.
◊ Suppose you wanted to change the setup in figure n/1 so that the location of the actual flame in the figure would instead be occupied by an image of a flame. Where would you have to move the candle to achieve this? What about in n/2?
◊ There are three qualitatively different types of image formation that can occur with lenses, of which figures n/1 and n/2 exhaust only two. Figure out what the third possibility is. Which of the three possibilities can result in a magnification greater than one? Cf. problem 10, p. 797.
◊ Classify the examples shown in figure o according to the types of images delineated in discussion question C.
◊ In figures n/1 and n/2, the only rays drawn were those that happened to enter the lenses. Discuss this in relation to figure o.
◊ In the right-hand side of figure o, the image viewed through the lens is in focus, but the side of the rose that sticks out from behind the lens is not. Why?
The Lensmaker's Equation
The focal length of a spherical mirror is simply \(r/2\), but we cannot expect the focal length of a lens to be given by pure geometry, since it also depends on the index of refraction of the lens. Suppose we have a lens whose front and back surfaces are both spherical. (This is no great loss of generality, since any surface with a sufficiently shallow curvature can be approximated with a sphere.) Then if the lens is immersed in a medium with an index of refraction of 1, its focal length is given approximately by
where \(n\) is the index of refraction and \(r_1\) and \(r_2\) are the radii of curvature of the two surfaces of the lens. Equation \ref{LM} is known as the lensmaker's equation . In my opinion it is not particularly worthy of memorization. The positive sign is used when both surfaces are curved outward or both are curved inward; otherwise a negative sign applies. The proof of this equation is left as an exercise to those readers who are sufficiently brave and motivated.
12.4.3 Dispersion
For most materials, we observe that the index of refraction depends slightly on wavelength, being highest at the blue end of the visible spectrum and lowest at the red. For example, white light disperses into a rainbow when it passes through a prism, q.
Even when the waves involved aren't light waves, and even when refraction isn't of interest, the dependence of wave speed on wavelength is referred to as dispersion. Dispersion inside spherical raindrops is responsible for the creation of rainbows in the sky, and in an optical instrument such as the eye or a camera it is responsible for a type of aberration called chromatic aberration (subsection 12.3.3 and problem 28). As we'll see in subsection 13.3.2, dispersion causes a wave that is not a pure sine wave to have its shape distorted as it travels, and also causes the speed at which energy and information are transported by the wave to be different from what one might expect from a naive calculation. The microscopic reasons for dispersion of light in matter are discussed in optional subsection 12.4.6.
The principle of least time for refraction
We have seen previously how the rules governing straight-line motion of light and reflection of light can be derived from the principle of least time. What about refraction? In the figure, it is indeed plausible that the bending of the ray serves to minimize the time required to get from a point A to point B. If the ray followed the unbent path shown with a dashed line, it would have to travel a longer distance in the medium in which its speed is slower. By bending the correct amount, it can reduce the distance it has to cover in the slower medium without going too far out of its way. It is true that Snell's law gives exactly the set of angles that minimizes the time required for light to get from one point to another. The proof of this fact is left as an exercise (problem 38, p. 802).
Microscopic description of refraction
Given that the speed of light is different in different media, we've seen two different explanations (on p. 774 and in subsection 12.4.5 above) of why refraction must occur. What we haven't yet explained is why the speed of light does depend on the medium.
A good clue as to what's going on comes from the figure s. The relatively minor variation of the index of refraction within the visible spectrum was misleading. At certain specific frequencies, \(n\) exhibits wild swings in the positive and negative directions. After each such swing, we reach a new, lower plateau on the graph. These frequencies are resonances. For example, the visible part of the spectrum lies on the left-hand tail of a resonance at about \(2\times10^{15}\ \text{Hz}\), corresponding to the ultraviolet part of the spectrum. This resonance arises from the vibration of the electrons, which are bound to the nuclei as if by little springs. Because this resonance is narrow, the effect on visible-light frequencies is relatively small, but it is stronger at the blue end of the spectrum than at the red end. Near each resonance, not only does the index of refraction fluctuate wildly, but the glass becomes nearly opaque; this is because the vibration becomes very strong, causing energy to be dissipated as heat. The “staircase” effect is the same one visible in any resonance, e.g., figure k on p. 180: oscillators have a finite response for \(f \ll f_0\), but the response approaches zero for \(f \gg f_0\).
So far, we have a qualitative explanation of the frequency-variation of the loosely defined “strength” of the glass's effect on a light wave, but we haven't explained why the effect is observed as a change in speed, or why each resonance is an up-down swing rather than a single positive peak. To understand these effects in more detail, we need to consider the phase response of the oscillator. As shown in the bottom panel of figure j on p. 181, the phase response reverses itself as we pass through a resonance.
Suppose that a plane wave is normally incident on the left side of a thin sheet of glass, t/1, at \(f \ll f_0\). The light wave observed on the right side consists of a superposition of the incident wave consisting of \(\mathbf{E}_0\) and \(\mathbf{B}_0\) with a secondary wave \(\mathbf{E}^*\) and \(\mathbf{B}^*\) generated by the oscillating charges in the glass. Since the frequency is far below resonance, the response \(q\mathbf{x}\) of a vibrating charge \(q\) is in phase with the driving force \(\mathbf{E}_0\). The current is the derivative of this quantity, and therefore 90 degrees ahead of it in phase. The magnetic field generated by a sheet of current has been analyzed in subsection 11.2.1, and the result, shown in figure e on p. 664, is just what we would expect from the right-hand rule. We find, t/1, that the secondary wave is 90 degrees ahead of the incident one in phase. The incident wave still exists on the right side of the sheet, but it is superposed with the secondary one. Their addition is shown in t/2 using the complex number representation introduced in subsection 10.5.7. The superposition of the two fields lags behind the incident wave, which is the effect we would expect if the wave had traveled more slowly through the glass.
In the case \(f \gg 0\), the same analysis applies except that the phase of the secondary wave is reversed. The transmitted wave is advanced rather than retarded in phase. This explains the dip observed in figure s after each spike.
All of this is in accord with our understanding of relativity, ch. 7, in which we saw that the universal speed \(c\) was to be understood fundamentally as a conversion factor between the units used to measure time and space --- not as the speed of light. Since \(c\) isn't defined as the speed of light, it's of no fundamental importance whether light has a different speed in matter than it does in vacuum. In fact, the picture we've built up here is one in which all of our electromagnetic waves travel at \(c\); propagation at some other speed is only what appears to happen because of the superposition of the \((\mathbf{E}_0,\mathbf{B}_0)\) and \((\mathbf{E}^*,\mathbf{B}^*)\) waves, both of which move at \(c\).
But it is worrisome that at the frequencies where \(n\lt1\), the speed of the wave is greater than \(c\). According to special relativity, information is never supposed to be transmitted at speeds greater than \(c\), since this would produce situations in which a signal could be received before it was transmitted! This difficulty is resolved in subsection 13.3.2, where we show that there are two different velocities that can be defined for a wave in a dispersive medium, the phase velocity and the group velocity. The group velocity is the velocity at which information is transmitted, and it is always less than \(c\).
Contributors and Attributions
Benjamin Crowell (Fullerton College). Conceptual Physics is copyrighted with a CC-BY-SA license.