1. 今日課程內容
CH15 波 干涉 駐波 繞射 折射
CH16聲音

2. 15-8 Interference(干涉)
The superposition principle says that when two waves pass through the same point, the displacement is the arithmetic sum of the individual displacements.
In the figure below, (a) exhibits destructive interference and (b) exhibits constructive interference.
Figure Two wave pulses pass each other. Where they overlap, interference occurs: (a) destructive, and (b) constructive.

3. 15-8 Interference
These graphs show the sum of two waves. In (a) they add constructively; in (b) they add destructively; and in (c) they add partially destructively.
Figure Graphs showing two identical waves, and their sum, as a function of time at three locations. In (a) the two waves interfere constructively, in (b) destructively, and in (c) partially destructively.
in phase同相
建設性干涉
out of phase反相
破壞性干涉

4. 15-9 Standing Waves(駐波); Resonance(共振)
Standing waves occur when both ends of a string are fixed. In that case, only waves which are motionless at the ends of the string can persist. There are nodes(節點), where the amplitude is always zero, and antinodes, where the amplitude varies from zero to the maximum value.
Figure Standing waves corresponding to three resonant frequencies.

5. 15-9 Standing Waves; Resonance
The frequencies of the standing waves on a particular string are called resonant frequencies.
They are also referred to as the fundamental and harmonics(諧音).
Figure (a) A string is plucked. (b) Only standing waves corresponding to resonant frequencies persist for long.

6. 15-9 Standing Waves; Resonance
The wavelengths and frequencies of standing waves are:
and

7. 15-9 Standing Waves; Resonance
Example 15-8: Piano string.
A piano string is 1.10 m long and has a mass of 9.00 g. (a) How much tension must the string be under if it is to vibrate at a fundamental frequency of 131 Hz? (b) What are the frequencies of the first four harmonics?
Solution: a. First we find the wave speed; v = 288 m/s. This gives a tension force of 679 N.
b. 131 Hz (fundamental), 262 Hz, 393 Hz, 524 Hz.

8. 15-9 Standing Waves; Resonance
Example 15-9: Wave forms.
Two waves traveling in opposite directions on a string fixed at x = 0 are described by the functions
D1 = (0.20 m)sin(2.0x – 4.0t) and D2 = (0.20m)sin(2.0x + 4.0t)
(where x is in m, t is in s), and they produce a standing wave pattern. Determine (a) the function for the standing wave, (b) the maximum amplitude at x = 0.45 m, (c) where the other end is fixed (x > 0), (d) the maximum amplitude, and where it occurs.
Solution: a. The sum is a standing wave: D = (0.40m)sin(2.0x)cos(4.0t)
b. The maximum amplitude is 0.31m.
c. Nodes occur every half wavelength, which is 1.57 m. The string must be an integral multiple of 1.57 m long.
d. The maximum amplitude is 0.40 m, and occurs at the antinodes, halfway between the nodes.

9. 15-10 Refraction(折射)
If the wave enters a medium where the wave speed is different, it will be refracted—its wave fronts and rays will change direction.
We can calculate the angle of refraction, which depends on both wave speeds:
Figure Refraction of waves passing a boundary.

10. 15-10 Refraction
The law of refraction works both ways—a wave going from a slower medium to a faster one would follow the red line in the other direction.
Figure 15-30b. Law of refraction for waves.

11. 15-10 Refraction Example 15-10: Refraction of an earthquake wave.
An earthquake P wave passes across a boundary in rock where its velocity increases from 6.5 km/s to 8.0 km/s. If it strikes this boundary at 30°, what is the angle of refraction?
Solution: Applying the law of refraction gives the angle as 38°.

12. 15-11 Diffraction(繞射)
When waves encounter an obstacle, they bend around it, leaving a “shadow region.” This is called diffraction.
Figure Wave diffraction. The waves are coming from the upper left. Note how the waves, as they pass the obstacle, bend around it, into the “shadow region” behind the obstacle.

13. 15-11 Diffraction
The amount of diffraction depends on the size of the obstacle compared to the wavelength. If the obstacle is much smaller than the wavelength, the wave is barely affected (a). If the object is comparable to, or larger than, the wavelength, diffraction is much more significant (b, c, d).
Figure Water waves passing objects of various sizes. Note that the longer the wavelength compared to the size of the object, the more diffraction there is into the “shadow region.”

14. Chapter 16 Sound
Chapter opener. “If music be the food of physics, play on.” [See Shakespeare, Twelfth Night, line 1.] Stringed instruments depend on transverse standing waves on strings to produce their harmonious sounds. The sound of wind instruments originates in longitudinal standing waves of an air column. Percussion instruments create more complicated standing waves. Besides examining sources of sound, we also study the decibel scale of sound level, sound wave interference and beats, the Doppler effect, shock waves and sonic booms, and ultrasound imaging.

15. 16-1 Characteristics of Sound
Sound can travel through any kind of matter, but not through a vacuum.
The speed of sound is different in different materials; in general, it is slowest in gases, faster in liquids, and fastest in solids.
The speed depends somewhat on temperature, especially for gases.

16. 16-1 Characteristics of Sound
Loudness(響度): related to intensity of the sound wave
Pitch(音高): related to frequency
Audible range(可聽範圍): about 20 Hz to 20,000 Hz; upper limit decreases with age
Ultrasound(超音波): above 20,000 Hz; see ultrasonic camera focusing in following example

17. 16-1 Characteristics of Sound
Example 16-2: Autofocusing with sound waves.
Older autofocusing cameras determine the distance by emitting a pulse of very high frequency (ultrasonic) sound that travels to the object being photographed, and include a sensor that detects the returning reflected sound. To get an idea of the time sensitivity of the detector, calculate the travel time of the pulse for an object (a) 1.0 m away, and (b) 20 m away.
Figure Example 16–2. Autofocusing camera emits an ultrasonic pulse. Solid lines represent the wave front of the outgoing wave pulse moving to the right; dashed lines represent the wave front of the pulse reflected off the person’s face, returning to the camera. The time information allows the camera mechanism to adjust the lens to focus at the proper distance.
Solution: (remember the pulse makes a round trip)
5.8 ms
120 ms

18. 16-2 Mathematical Representation of Longitudinal Waves
Longitudinal waves are often called pressure waves(壓力波). The displacement is 90° out of phase with the pressure.
Figure Longitudinal sound wave traveling to the right, and its graphical representation in terms of pressure.
Figure Representation of a sound wave in space at a given instant in terms of (a) displacement, and (b) pressure.

19. 16-2 Mathematical Representation of Longitudinal Waves
By considering a small cylinder within the fluid, we see that the change in pressure is given by (B is the bulk modulus):
Figure Longitudinal wave in a fluid moves to the right. A thin layer of fluid, in a thin cylinder of area S and thickness Δx, changes in volume as a result of pressure variation as the wave passes. At the moment shown, the pressure will increase as the wave moves to the right, so the thickness of our layer will decrease, by an amount ΔD.

20. 16-2 Mathematical Representation of Longitudinal Waves
If the displacement is sinusoidal, we have
where

21. 16-3 Intensity of Sound: Decibels(分貝)
The intensity of a wave is the energy transported per unit time across a unit area.
The human ear can detect sounds with an intensity as low as W/m2 and as high as 1 W/m2.
Perceived loudness, however, is not proportional to the intensity.

22. 16-3 Intensity of Sound: Decibels
The loudness of a sound is much more closely related to the logarithm of the intensity.
Sound level is measured in decibels (dB) and is defined as:
I0 is taken to be the threshold of hearing:

23. 16-3 Intensity of Sound: Decibels
Example 16-3: Sound intensity on the street.
At a busy street corner, the sound level is 75 dB. What is the intensity of sound there?
Solution: Substitution gives I = 3.2 x 10-5 W/m2.

24. 16-3 Intensity of Sound: Decibels
Example 16-4: Loudspeaker response.
A high-quality loudspeaker is advertised to reproduce, at full volume, frequencies from 30 Hz to 18,000 Hz with uniform sound level ± 3 dB. That is, over this frequency range, the sound level output does not vary by more than 3 dB for a given input level. By what factor does the intensity change for the maximum change of 3 dB in output sound level?
Solution: Call I1 the average intensity and β1 the average sound level. Then maximum intensity, I2 corresponds to a level β2 = β1 + 3 dB. Then use Eq Substitution shows this corresponds to a doubling or halving of the intensity.

25. 16-3 Intensity of Sound: Decibels
Conceptual Example 16-5: Trumpet players.
A trumpeter plays at a sound level of 75 dB. Three equally loud trumpet players join in. What is the new sound level?
Solution: The intensity has gone up by a factor of 4; the new sound level is 81 dB.

26. 16-3 Intensity of Sound: Decibels
An increase in sound level of 3 dB, which is a doubling in intensity, is a very small change in loudness.
In open areas, the intensity of sound diminishes with distance:
However, in enclosed spaces this is complicated by reflections, and if sound travels through air, the higher frequencies get preferentially absorbed.

27. 16-3 Intensity of Sound: Decibels
Example 16-6: Airplane roar.
The sound level measured 30 m from a jet plane is 140 dB. What is the sound level at 300 m? (Ignore reflections from the ground.)
Solution: Assuming the intensity decreases as the square of the distance, the sound level at 300 m is 120 dB.

28. 16-3 Intensity of Sound: Decibels
Example 16-7: How tiny the displacement is.
Calculate the displacement of air molecules for a sound having a frequency of 1000 Hz at the threshold of hearing.
(b) Determine the maximum pressure variation in such a sound wave.
Solution: a. Substitution gives A = 1.1 x m.
b. Substitution gives the maximum pressure variation as 3.1 x 10-5 Pa, or 3.1 x atm.

29. 16-3 Intensity of Sound: Decibels
The intensity can be written in terms of the maximum pressure variation. With some algebraic manipulation, we find:

30. 16-4 Sources of Sound: Vibrating Strings(弦) and Air Columns
Musical instruments produce sounds in various ways—vibrating strings, vibrating membranes, vibrating metal or wood shapes, vibrating air columns.
The vibration may be started by plucking, striking, bowing, or blowing. The vibrations are transmitted to the air and then to our ears.

31. 16-4 Sources of Sound: Vibrating Strings and Air Columns
This figure shows the first three standing waves, or harmonics, on a fixed string.
Figure Standing waves on a string—only the lowest three frequencies are shown.

32. 16-4 Sources of Sound: Vibrating Strings and Air Columns
The strings on a guitar can be effectively shortened by fingering, raising the fundamental pitch.
The pitch of a string of a given length can also be altered by using a string of different density.
Figure The wavelength of (a) an unfingered string is longer than that of (b) a fingered string. Hence, the frequency of the fingered string is higher. Only one string is shown on this guitar, and only the simplest standing wave, the fundamental, is shown.

33. 16-4 Sources of Sound: Vibrating Strings and Air Columns
Example 16-8: Piano strings.
The highest key on a piano corresponds to a frequency about 150 times that of the lowest key. If the string for the highest note is 5.0 cm long, how long would the string for the lowest note have to be if it had the same mass per unit length and was under the same tension?
Solution: The wave speed is determined by the tension and the mass per unit length of the string. If these are the same, then the string would have to be 150 times as long to give a frequency 150 times lower, or 7.5 m. Instead, piano strings increase in mass per unit length as the frequency decreases, so the tension can be kept close.

34. 16-4 Sources of Sound: Vibrating Strings and Air Columns
Example 16-9: Frequencies and wavelengths in the violin.
A 0.32-m-long violin string is tuned to play A above middle C at 440 Hz.
What is the wavelength of the fundamental string vibration, and
(b) What are the frequency and wavelength of the sound wave produced?
(c) Why is there a difference?
Solution: a. The wavelength is twice the length of the string, 0.64 m.
b. The sound wave has the same frequency as the wave on the string; using the speed of sound in air gives a wavelength of 0.78 m.
c. They are different because the wave speeds are different. (It's the frequency that will be the same).

35. 16-4 Sources of Sound: Vibrating Strings and Air Columns
A tube open at both ends (most wind instruments) has pressure nodes, and therefore displacement antinodes, at the ends.
Figure Graphs of the three simplest modes of vibration (standing waves) for a uniform tube open at both ends (“open tube”). These simplest modes of vibration are graphed in (a), on the left, in terms of the motion of the air (displacement), and in (b), on the right, in terms of air pressure. Each graph shows the wave format at two times, A and B, a half period apart. The actual motion of molecules for one case, the fundamental, is shown just below the tube at top left.

36. 16-4 Sources of Sound: Vibrating Strings and Air Columns
A tube closed at one end (some organ pipes) has a displacement node (and pressure antinode) at the closed end.
Figure Modes of vibration (standing waves) for a tube closed at one end (“closed tube”).

37. 16-4 Sources of Sound: Vibrating Strings and Air Columns
Example 16-10: Organ pipes.
What will be the fundamental frequency(基音) and first three overtones(泛音) for a 26-cm-long organ pipe at 20°C if it is (a) open and (b) closed?
Solution: a. The fundamental is 660 Hz; the overtones are integer multiples: 1320 Hz, 1980 Hz, 2640 Hz.
b. The fundamental is 330 Hz; only odd harmonics are present: 990 Hz, 1650 Hz, 2310 Hz.

38. 16-4 Sources of Sound: Vibrating Strings and Air Columns
Example 16-11: Flute.
A flute is designed to play middle C (262 Hz) as the fundamental frequency when all the holes are covered. Approximately how long should the distance be from the mouthpiece to the far end of the flute? (This is only approximate since the antinode does not occur precisely at the mouthpiece.) Assume the temperature is 20°C.
Solution: This is an open tube; to get a fundamental of 262 Hz requires a tube 0.66 m long.

39. 習題
Ch15: 42, 47, 54, 60, 67
Ch16: 13, 21, 28, 39, 44, 46