Wavelength: Equation (5.2) can also be written as
Dv=f (5.5)
Interpretation: All cars travel on the road at exactly the same speed, v. The speed is measured in miles per hour. The cars are sent out from a starting point at a rate of f cars per hour. After a few hours, the cars are separated by equal distances; that distance is, as determined by Eq. (5.5).
Equation (5.5) can be applied directly to Fig.5.7. There is a sine wave with a frequency of 800 Hz. The speed of sound is 344 m/s, and so the wavelength is 344=800D0:43m.
It may look as though Eqs. (5.2)–(5.5) require you to memorize four equations.
In fact, there are only two, v D f andf D 1=T. The other equations can be derived from these two. Please try to think about it that way. It will help your math skills.
Changing Media Equations (5.3), (5.4), and (5.5) describe the relationship be- tween frequency, wavelength, and the speed of sound. What happens when a sound wave goes from one medium to another so that the speed of sound changes?
Somehow those equations need to be satisfied. Does the wavelength change, or does the frequency change, or do they both change? The answer is that the frequency stays the same and the wavelength changes to accommodate the new speed of sound.
When a sound wave travels from air into water, where the speed of sound is five times greater, the wavelength increases by a factor of 5. (To be more complete, when a sound wave in air hits the water, most of the wave is reflected. But that part of the wave that does enter the water has its wavelength increased by a factor of 5.)
5.4 Sound Waves in More Than One Dimension of Space
The section called “Sound waves in space and time” considered only one dimension of space. That description applies well to a tin-can telephone, where sound waves travel on a stretched string, but otherwise you know that there are three spatial dimensions. From any point you can move forward–backward, left–right, or up–
down. That makes three. We need to deal with this problem of spatial dimensions.
One-Dimension We begin by trying to justify the previous discussion dealing with only one spatial dimension. First, with a source at one point in space and a receiver at another, there are two points, and they can be connected by a straight line. Arguably that is the most interesting path for a wave, and that path is one dimensional. Thus, a situation that really has three dimensions of space has been reduced to a situation that has only one. A second justification is that a one-dimensional treatment is a helpful simplification. It enables us to draw a graph like Fig.5.7. A third justification for the one-dimensional treatment is that facts about the speed of sound, the concept of wavelength, and the formula vDf are the same whether the number of spatial dimensions is one, two, or three.
48 5 Sound Waves Fig. 5.8 There is a source of
sound at the origin where the x and y axes meet. The solid circles indicate the peaks in air pressure as seen in a snapshot taken at timet1. Successive peaks are separated by one wavelength, as expected. Slightly later, at timet2, the wave has moved outward, and the peaks are now at the positions indicated by the dashed circles
Two-Dimensions Next, we ought to think about more spatial dimensions. Suppose there are two spatial dimensions. Then the sound waves are analogous to the ripples of water waves on the surface of a pool. A surface has two spatial dimensions. We can show a wave traveling on the surface by plotting only the peaks of the waveform at two different instants in time, timet1and timet2a moment later. In the interval between these two times the peaks of the waveform will move, as shown in Fig.5.8.
Three-Dimensions From two spatial dimensions it is not hard to make the jump to three. Look at Fig.5.8 and imagine that the concentric circles are turned into concentric hollow shells. As time goes on the shells expand like the surface of a spherical balloon as it is blown up. The shells expand at the speed of sound. That gives you the image of how the peaks of a sound wave move in three dimensions.
5.5 Unit Analysis
The term “Unit analysis” (also known as “Dimensional analysis”) refers to the fact that equations must make sense in terms of their physical units. This logical fact can help you avoid errors in applying equations. It can help you remember an equation that you have forgotten. It can even enable you to make a good guess about an equation that you never learned!
The best way to learn about unit analysis is with a few examples.
You know that the distance you travel (d) is equal to your speed (v) multiplied by the duration of your travel (t), ord D vt. In terms of units, this might be stated as, “(miles) equals (miles per hour) multiplied by (hours).” As an equation this is:
milesD miles
hour hours:
Exercises 49
We have used the fact that the word “per” means “divided by.” The key point is that the units of hours cancel in the numerator and denominator, leading to the logical conclusion that miles equal miles. Unit analysis enforces that kind of logic.
Notice that unit analysis can prevent you from making an error such as multiplying a speed in miles per hour by the number of minutes you travel.
Unit analysis applies to every equation we study. For instance, Eq. (5.2) says vDf . In terms of units
meters per secondD cycles
secondmeters cycle :
The units of cycles cancel, and this equation reduces to a logical result:
meters per secondD meters second:
Notice that unit analysis helps you avoid an error in writing the equation. For instance, vD=f is a nonsense formula. It says that a speed (v) is equal to a length () multiplied by a time (1=f), but you know that a speed is equal to a length divided by a time.
As another example, Chap. 2 says that frequency,f, has units of inverse time.
Therefore,1=f has units of seconds. Formally we say that period,T, is given byT D1=f, or,
seconds per cycleD 1 cycles=second:
You can improve your life by applying unit analysis to every practical calculation you do.
Exercises
Exercise 1, Constant pressure
The pressure of the sound shown in Fig.5.2is more than 101,000 Pa. Bozo says that such a big pressure is bound to be painful. Set Bozo straight on the role of this constant part of the pressure.
Exercise 2, Check those powers of ten
The first page of this chapter says that 100,000 N/m2is equivalent to 10 N/cm2. Is that really right? Consider that 1 m is 100 cm. Why isn’t 100,000 N/m2equivalent to 1,000 N/cm2?
50 5 Sound Waves
Exercise 3, Pressure conversion
Your goal is to show that 14.7 pounds per square inch is equivalent to about 101,000 N/m2. You begin by realizing that pressure is force per unit area. Therefore, you need English-metric conversions for force and for area. For force: 4.45 N/pound.
For distance: 0.0254 m/in., therefore, 0.000645 m2/in.2. Can you work it from here?
Exercise 4, Extreme weather
On a really cold day the temperature is18ıC (0ıF). On a really hot day it is C40(104ıF). What is the speed of sound at18? AtC40?
Exercise 5, Americanize the equation
Our formula Eq. (5.1) for the speed of sound as a function of temperature assumes that temperature is given on the Celsius (also known as Centigrade) scale. On this scale water freezes at zero and water boils at 100ı. Most Americans are more familiar with the Fahrenheit scale where water freezes at 32ı and boils at 212ı. A temperature on the Fahrenheit scaleTF can be converted to a temperature on the Celsius scaleTC with the formula
TC D 5
9.TF32/: (5.6)
(a) Show that the speed of sound, in meters per second, is given by vD321C1
3TF: (5.7)
(b) Room temperature on the Fahrenheit scale is about 69ı. Show that formula (5.7) leads to a speed of sound of 344 m/s at this temperature as you would expect.
(c) To further Americanize the equation, show that the speed of sound in feet per second is given by
vD1053C1:1 TF: (5.8)
Use the fact that 1 ft is exactly 0.3048 m.
Exercise 6, A fair start
(a) How long does it take for sound to travel 10 m? (b) The runner in the far lane is 10 m farther away from the starter than the runner in the near lane. When the starter fires the starting pistol, the near runner hears it first. Could this make a significant difference?
Exercise 7, Stormy weather
Thunder and lightning are produced simultaneously. If you see the lighting 2 s before you hear the thunder, how far away did lightning strike?
Exercise 8, Mach 1
The text says that the speed of sound at room temperature is 1,128 feet per second or 769 miles per hour. Use the fact that there are 5,280 ft in a mile to show that this equivalence is correct.
Exercises 51
Exercise 9, Sound vs light
The section called “Sound vs light” gives the range of audible sound frequencies and the range of visible light frequencies. Which range is broader?
Exercise 10, Longest and shortest
Nominally, the limits of human hearing are 20 and 20,000 Hz. Find the wave- lengths of those two waves in air. Here, and in exercises below, assume that the air is at room temperature.
Exercise 11, Medium long
What is the frequency of a tone with a wavelength of 1 m?
Exercise 12, Frequency and wavelength
Equations (5.2), (5.3), and (5.5) are all variations on the same equation. Is one more fundamental than the others? Knowing what you do about waves that travel from one medium into another, can you make the case that Eq. (5.5) [Dv=f] is the most fundamental?
Exercise 13, Making waves
Clap your hands four times with a perfectly regular rhythm so that there is 1/2 s between each clap. Think about the hand claps as a periodic signal. It is not a sine wave signal, but it is periodic—at least for a few seconds. Now look at Fig.5.8.
Could this apply to the hand clapping exercise? What is the period, the frequency, the wavelength?
Exercise 14, Unit analysis
By law, the maximum water usage of a new American toilet is 1.6 gpf. What do you think is meant by “gpf.” Write the equation that describes how many gallons of water the toilet uses in a day. Your equation should include all the units, which is the main point of this exercise.
Exercise 15, Dispersion
The speed of a sound wave in open air does not depend on the frequency of the wave. We say that the propagation in air is non-dispersive. Some physical systems do have dispersion. For instance, when a light wave travels through a glass prism, the speed of light in the glass depends on the color (frequency) of the light. The dispersion of the prism separates the different colors that compose white light. Suppose that sound propagation were dispersive so that low-frequency sounds travel 10 % faster than high-frequency sounds. What would be the consequences for communication?
Exercise 16, Bending water?
Can transverse waves propagate in a fluid like water?
Exercise 17, Breaking waves
Breaking waves that you see coming up the beach are complicated. They are more complicated than the sound waves that we describe as longitudinal or transverse. (a) Make the case that the polarization of breaking waves is longitudinal.
(b) Make the case that the polarization of breaking waves is transverse.
52 5 Sound Waves
Exercise 18, Helium gas
Helium gas is much lighter than ordinary air, i.e. it is less dense. That is why it is used as a lifting gas for blimps. Because helium is so light, the speed of sound in helium is high, about 1000 m/s.
(a) Show that the speed of sound in helium is about three times faster than the speed of sound in air.
(b) Show that the wavelength of a 1000-Hz tone in helium gas is about 1 meter.
(c) Explain why the frequency of a 1000-Hz tone in helium gas is still 1000 Hz.
Chapter 6
Wave Properties
This chapter describes some properties of waves—properties like interference, beats, reflection, and refraction. Naturally, the focus will be on sound waves, but the principles apply equally to light waves. Because of the unifying principles of wave physics, you get two for one—once you’ve learned the acoustics, you automatically know the optics.