1206 - Concepts in Physics Friday, November 26th 2009
Notes There is a new point on the webpage things to look at for the final exam So far you have the two midterms there More things will be posted over the weekend FINAL EXAM, Thursday, December 10th at 9:00 am til NOON --- GYM B
Notes Format of the final exam Multiple choice and short questions (some of them a little longer) will make up 40% of the possible points Problems will be 60% of the final points, you will have additional points in the problems again There won t be twice as many problems - so you have more time Topics are everything we have done...
Principle of linear Superposition Often, two or more sound waves are present at the same place at the same time, as it is the case when everyone is talking at a party of when music plays from speakers of a stereo system. Let s take a closer look at what happens when several waves pass simultaneously through one region. Two pulses beginning to move towards each other, when they are at the same spot the resulting pulse is the sum of the two individual pulses. After they have passed, they continue to travel in the same form as before. If the two pulses were opposite - one up one down, they would cancel out at the point where they meet. But also continue as individual pulses afterwards. THE PRINCIPLE OF LINEAR SUPERPOSITION When two or more waves are present simultaneously at the same place, the resultant disturbance is the sum of the disturbances from the individual waves
Constructive and Destructive Interference of Sound Waves Suppose that the sounds from two speakers overlap in the middle of a listening area, and that each speaker produces a sound wave of the same amplitude and frequency. When two waves always meet condensationto-condensation and rarefaction-to-rarefaction (or crest-to-crest and trough-trough), they are said to be exactly in phase and to exhibit constructive interference. When two waves always meet condensationto-rarefaction (or crest-to-trough), they are said to be exactly out of phase and to exhibit destructive interference. Both cases described above mean that the wave patterns do not shift relative to one another as time passes. Sources that produce waves in this fashion are called coherent sources. For two wave sources vibrating in phase, a difference in path lengths that is zero or an integer number (1,2,3,...) of wavelength leads to constructive interference; a difference in path length that is a half-integer number (0.5, 1.5, 2.5,...) of wavelength leads to destructive interference.
Noise canceling headphones On a 1978 flight to Europe, Amar Bose, the founder of Bose Corporation, put on a pair of airline-supplied headphones, only to find that the roar of the jet engines prevented him from enjoying the audio. He started making calculations right there on the plane to see if it was possible to use the headphones themselves as a noise-reducing agent. Bose introduced the first noise-canceling headphones a decade later. Unfortunately for music lovers, many types of ambient sounds can interfere with or even block the sounds coming through their headphones. If you have ever tried to listen to a CD or MP3 player on a plane, then you know the problem well: The roar of the engines makes it difficult to hear what's being piped through the speakers -- even when those speakers are situated in or on your ear. Noise-canceling headphones come in either active or passive types. Technically speaking, any type of headphone can provide some passive noise reduction. That's because the materials of the headphones themselves block out some sound waves, especially those at higher frequencies. The best passive noise-canceling headphones, however, are circum-aural types that are specially constructed to maximize noisefiltering properties. That means they are packed with layers of high-density foam or other sound-absorbing material, which makes them heavier than normal headphones. The tradeoff of all that extra weight is a reduction in noise of about 15 to 20 decibels (db). But considering jet engines create 75 to 80 db of noise inside the aircraft cabin, passive models have some serious limitations. That's where active noise-canceling headphones come in.
Active noise-canceling headphones can do everything that passive ones can do -- their very structure creates a barrier that blocks high-frequency sound waves. They also add an extra level of noise reduction by actively erasing lower-frequency sound waves. How do noise-canceling headphones accomplish this? They actually create their own sound waves that mimic the incoming noise in every respect except one: the headphone's sound waves are 180 degrees out of phase with the intruding waves. If you look at the illustration below, you can see how this works. Notice that the two waves -- the one coming from the noise-canceling headphone and the one associated with the ambient noise -- have the same amplitude and frequency, but their crests and troughs (compressions and rarefactions) are arranged so that the crests (compressions) of one wave line up with the troughs (rarefactions) of the other wave and vice versa. In essence, the two waves cancel each other out, a phenomenon known as destructive interference. The result: the listener can focus on the sounds he wants to hear. noise-canceling headphones are able to provide an additional reduction in noise of 20 decibels. That means about 70 percent of ambient noise is effectively blocked, making noisecanceling headphones ideal for airline and train travel, open office environments or any other location with a high level of background noise.
Example: Two in-phase loudspeakers, A and B, are separated by 3.20 m. A listener is stationed at point C, which is 2.40 m in from of speaker B. The triangle ABC is a right triangle. Both speakers are playing identical 214-Hz tones, and the speed of sound is 343 m/s. Does the listener hear a loud sound or no sound? A 3.20 m 2.40 m B C The listener will hear either a loud sound or no sound, depending on whether the interference occurring at point C is constructive or destructive. To determine which it is, we need to find the difference in the distances traveled by the two sound waves that reach point C and see whether the difference is an integer or half-integer number of wavelengths. In either event, the wavelength can be found from the relation λ = v/f. Since the triangle ABC is a right triangle, the distance AC is given by the Pythagorean theorem: sqrt{(3.30 m)2 + (2.40 m) 2 } = 4.00 m. The distance BC is given as 2.40 m. Thus, the difference in the travel distances for the waves is 4.00 m - 2.40 m = 1.60 m. The wavelength of the sound is λ = v/f = (343 m/s)/(214 Hz) = 1.60 m. Since the difference in the distances is one wavelength, constructive interference occurs at point C, and the listener hears a loud sound.
Out-of-phase Speakers To make a speaker operate, two wires must be connected between the speaker and the receiver (amplifier). To ensure that the diaphragms of two speakers vibrate in phase, it is necessary to make these connections in exactly the same way. If the wires for one speaker are not connected just as they are for the other speaker, the two diaphragms will vibrate out of phase. Whenever one diaphragm move outward, the other will move inward, and vice versa. Assume that the connection are made so, that the speaker diaphragms vibrate out of phase. What kind of interference would result at the overlap point? Since the diaphragms are vibration out of phase, one of them os now moving exactly opposite to the way it was moving originally. The effect of this change is that every condensation originating from speaker becomes a rarefaction and every rarefaction becomes a becomes a condensation - therefore we will have destructive interference.
Diffraction When a wave encounters an obstacle or the edges of an opening, it bends around them. For instance, as sound wave produced by a stereo system bends around the edges of an open doorway. If such bending did not occur, sound could be heard outside the room only at locations directly in from of the doorway - assuming no sounds is transmitter directly through the walls. The bending of a wave around an obstacle or the edges of an opening is called diffraction. All kinds of waves exhibit diffraction. When the sound wave reaches the doorway, he air in the doorway is set into longitudinal vibration. In effect every air molecule in the doorway becomes a source of a sound wave in its own right, and, for purposes of illustration, we will just follow two of them.
D Θ Θ The angle Θ defines the location of the first minimum intensity point on either side of the center and is can be expressed in dependence of the wavelength λ and the width of the doorway D (treat the doorway as a slit - height large compared to width): sinθ = λ/d Note! The diffraction of sound by a circular opening, such as that in a loudspeaker. In this case the first minimum is found here: sinθ = 1.22 λ/d An important point to remember is that the extent of the diffraction depends on the ration of the wavelength to the size of the opening. If the ratio of λ/d is small, then Θ is small and little diffraction occurs narrow dispersion. Accordingly for larger angle Θ we have wide dispersion.
Notes of the Overtone Series
Designing a Loudspeaker with wide dispersion A 1500-Hz sound and a 8500-Hz sound each emerges from a loudspeaker through a circular opening whose diameter is 0.30 m. Assuming that the speed of sound in air is 343 m/s, fine the diffraction angle Θ for each sound. The diffraction angle Θ for each sound wave is given by sin Θ = 1.22(λ/D). However, it will first be necessary to calculate the wavelengths of the sounds from λ = v/f. The wavelength of the two sounds are λ1500 = (343 m/s)/1500 Hz = 0.23 m and λ8500 = (343 m/s)/1500 Hz = 0.040 m The diffraction angles can now be determined: 1550-Hz-sound sinθ = 1.22 (λ1500/d) = 1.22 (0.23 m/0.30 m) = 0.94 Θ = sin -1 (0.94) = 70 8500-Hz-sound sinθ = 1.22 (λ8500/d) = 1.22 (0.23 m/0.30 m) = 0.94 Θ = sin -1 (0.94) = 9.2 With a 0.30-m opening, the dispersion of the higher-frequency sound is limited to only 9.2. To increase the dispersion, a smaller opening is needed. This is the reason why loudspeaker designers use a small-diameter speaker (tweeter) to generate highfrequency sound.
Transverse standing waves A standing wave is another interference effect that can occur when two waves overlap. Standing waves can arise with transverse waves, such as those on a guitar string, and also with longitudinal sound waves, such as those in a flute. In any case, the principle of linear superposition provides an explanation of the effect, just as it does for diffraction. One end is fixed to the wall, while the other is vibrated. Regions of the string move fast while others don t move at all. Each of the patterns is called transverse standing wave. The nodes are places that don t vibrate at all, and the antinodes are places where maximum vibration occurs. Each standing wave pattern is produced at a unique frequency of vibration. These frequencies form a series, the smallest f1 corresponding to one-loop.
Thus, if f1 is 10 Hz, the frequency needed to establish the 2- loop pattern is f2 = 2f1 or 20 Hz, while that needed to create the 3-loop pattern is f3 = 3f1 or 30 Hz and so on. These frequencies fn (n = 1, 2, 3,...) are called harmonics. F1 is referred to as the first harmonic or fundamental. All other s are called overtones or n th overtone. Standing waves arise because identical waves travel on the string in opposite directions and combine following the principle of linear superposition. They are called standing because it does not travel in one direction or the other, we have nodes and antinodes that are at fixed locations. A standing wave is composed of many individual pulses. Let s follow one of them on its way to the wall. If we have a string attached to the wall and a pulse (upward pointing half-cycle) is reaching the wall, it causes the string to pull upward on the wall. Consistent with Newton s third law (action = reaction) the wall pulls down on the string, and a downward pointing half-cycle is sent back away from the wall. Thus, the wave reflects from the wall. When it arrives back at the point of origin, if will reflect again, this time from the hand vibrating the string. Repeated reflection at both ends of the string create a multitude of wave cycles traveling in both directions. (The hand is basically fixed for small amplitudes) Suppose the string has a length L and its left end is being vibrated at a frequency f1. The time required to create a new wave cycle is the period T of the wave, where F = 1/f1. On the other hand, the time needed for a cycle to travel from the hand to the wall and back, a distance of 2L is 2L/v, where v is the wave speed. Reinforcement between new and reflected cycles occurs if these two times are equal; that is 1/f1 = 2L/v or f1 = v/(2l)
Repeated reinforcement between newly created and reflected cycles causes a large amplitude standing wave to develop on the string. Thus, the motion of the string is a resonance effect. The frequency f1 at which resonance occurs is sometimes call a natural frequency of the string, similar to the frequency at which an object oscillates on a spring. There is a difference between the resonance of the string and the resonance of a spring system. An object on a spring has only a single natural frequency, whereas the string has a series of natural frequencies. The series arises because a reflected wave cycle need not return to its point of origin in time to reinforce every newly created cycle. Reinforcement can occur, for instance, on every other new cycle, as it does if the string is vibrated at twice the frequency f1 (f2 = 2f1). More general this is true for every integer n and we can write fn = n f1, where n = 1,2,3,... Sting fixed at both ends: fn = n(v/2l)
Playing a guitar The heaviest string on an electric guitar has a linear density of m/l = 5.28 x 10-3 kg/m and is stretched with a tension of F = 226 N. This string produces the musical note E when vibrating along its entire length in a standing wave at the fundamental frequency of 164.8 Hz. (a) Find the length L of the string between its two fixed ends. (b) A guitar player wants the string to vibrate at a fundamental frequency of 2 x 164.8 Hz, as it must if the musical note E is to be sounded one octave higher in pitch. To accomplish this, he presses the string against the proper fret and then plucks the string. Find the distance L between the fret and the bridge of the guitar. The fundamental frequency f1 is given with n = 1: f1 = v/(2l). Since f1 is known in both parts (a) and (b), the length L in each case can be calculated directly form this expression, one the speed v is known. The speed is related to the tension F and the linear density m/l. (a) The speed is v = sqrt{f/(m/l)} = (226 N)/(5.28 x 10-3 kg/m) = 207 m/s Therefore L = v/2f1 = (207 m/s)/(2*164.8 Hz) = 0.628 m (b) frequency f1 is now 2*164.8 Hz = 329.9 Hz L = v/2f1 = (207 m/s)/(2*329.9 Hz) = 0.314 m exactly half the length to double the frequency!
Longitudinal standing wave Open tube on only one end: fn = n (v/4l) with n = 1, 3, 5,... It should not come as a big surprise that standing wave patterns can also be formed by longitudinal waves - for example when sound reflects from a wall, the forwardand backward-going waves can produce a standing wave. Musical instruments, such as trumpet, flute, clarinet, pipe organ, etc. are modified tubes of columns of air. To determine the natural frequencies of the air columns for a tube with two open ends (different from picture) we need to look at the antinode - node distribution. Since the ends are open, there is no reflection/displacement. The distance between two successive antinodes is onehalf of a wavelength, so the length L of the tube must be an inter number n of halfwavelength, so the length L of the tube must be L = n(1/2λn) or λn=2l/n or fn = n(v/2l), n = 1,2,3, 4
Longitudinal Standing Waves
Air Columns End effects
End Effect Experiment
End at right open End at right closed
End configurations for tubes 1. Phase change at open and closed ends
Open Tubes
Open Tube Frequencies and Wavelengths f = v / λ N = 1 λ = 2L o f = v/2l o = f o N = 2 λ = L o f = v/l o = 2f o N = 3 N = 4 N = 5 N = 6 λ = 2L /3 f = 3v/2L = 3f o o o λ = 2L /4 f = 4v/2L = 4f o o o λ = 2L /5 f = 5v/2L = 5f o o o λ = 2L /6 f = 6v/2L = 6f o o o
Closed Tubes
Closed Tube Frequencies and Wavelengths f = v / λ N = 1 λ = 4L c f = v/4l c = f c N = 2 does not exist. N = 3 λ = 4L /3 f = 3v/4L = 3f c c c N = 4 does not exist. N = 5 λ = 4L c /5 f = 5v/4L c = 5f c N = 6 does not exist.
Open and Closed Tube Comparison
Chladni Plates
Standing waves on a string. The mechanical driver at the right is made to oscillate up and down at a resonant frequency of the string producing standing waves. Mapping of the "nodes," or stationary lines, on the surface of a "violin" made to oscillate at two different resonant frequencies
Violin Body Vibrations