Sound Production o Sound Sound is produced by a vibrating object. A loudspeaker has a membrane or diaphragm that is made to vibrate by electrical currents. Musical instruments such as gongs or cymbals are some examples o vibrating sources o sound The human voice is a result o vibrations o the vocal cords, two membranes located in the throat. 1
Brass instruments produce their sound as a result o vibration o the lips o the perormer (similar to the voice). Reed instruments have a thin wooden strip that vibrated as a result o air blown across it. In lutes, recorders, organ pipes or whistles, air is blown across an opening in a pipe. Stringed instruments have a wire string that is set into vibration. Detection o Sound The ear is an amazing sound detector. Not only can it detect sound waves over a very wide range o requencies, it is also sensitive to an enormous range o sound intensities. The ear is a very complex organ. The interpretation o sounds by the brain is even more complex and not totally understood. 2
Anatomy o the Ear http://www.wisc-online.com/objects/ index_tj.asp?objid=ap1502 3
The ear is not equally sensitive to all requencies. Most people cannot hear requencies below 20 Hz or above 16000 Hz. Most are most sensitive to requencies between 1000 and 5000 Hz. lder people are less sensitive to requencies above 10000 Hz. Relative Intensity Intensity is measured with acoustical instruments and does not depend on the hearing o the listener. At 1000 Hz, the intensity o the aintest audible sound (the threshold o hearing) is 10-12 W/m 2. (I 0 ) We compare the intensity o sound to this threshold o hearing by setting up ratios. The ratio o the intensities yield an expression called relative intensity. 4
Relative intensity is given by the equation: β = 10 log β (relative intensity) is measured in decibels (db). I 0 is the intensity o the threshold o hearing (10-12 W/m 2 ). I is the intensity o the sound in question. Decibels is a dimensionless unit. I I 0 A decibel is actually 0.1 o a bel. A term that is named ater Alexander Graham Bell (inventor o the telephone). 5
Example 2: Relative Intensity A small source uniormly emits sound energy at a rate o 2.0 W. Calculate the relative intensity at 34 m rom the source. First, we need to ind the intensity o the sound being emitted. For that we use the equation: I=P/A. I=intensity, P=power, A=area. Sound is emitted in a spherical pattern in all directions rom the emission. A=4πr 2. I I = = P A 2.0W 4π (34) 2 = 6
β = 10 log 1.376772864 10 β = 10 log 12 10 W/m β = 10 log13767728.64 β = 81dB I I 0 4 2 W/m 2 Practice Problem A jet plane is ound to have a relative intensity upon takeo o 110 db. Calculate the intensity o the sound the jet makes. 7
Human Perception A change in the sound power level o 1 db is just barely perceptible. However this 1 db change represents a change in intensity o 26%. The next slide includes a graph o requency compared to intensity and loudness. Eect o temperature on speed o sound The speed o sound is aected by the temperature o the air. @ 0ºC, the velocity o sound is 331.5 m/s. For every degree increase you will ADD 0.6m/s/ºC. For every degree decrease you will SUBTRACT 0.6m/s/ºC. Ex.: What is the speed o sound @ 7.0ºC? 331.5 0.6m 7. C ( ) 0 m + C 8
The Doppler Eect In previous discussions, we assumed the source o emission and the listener were stationary. In that case the pitch o the sound heard is characteristic o the requency emitted. When there is relative motion between the source and the listener, the pitch o the sound heard is not the same as when they are both stationary. Two situations are common: the source is moving, the listener is stationary or the listener is moving, the source is stationary Doppler Demos http:// www.kettering.edu/ ~drussell/demos/ doppler/doppler.html 9
Calculations o perceived requency The observed requency ( ) is related to the requency o the source ( S ) and the relative speeds o the source (v s ), observer (v o ), and the speed (v) o waves in the medium @ that temp. by = = observer, S= source. Towards use top sign Away use bottom sign S v ± v v v S Example 3: The Doppler Eect A sound source is moving at 15.0 m/s towards an observer. The air temperature is 15.0ºC. The requency o the sound being emitted is 300. Hz. What is the apparent requency as the sound approaches the observer? What is the apparent requency as the sound goes past the observer? What is the change in requency that is observed? 10
v T = 15.0 C v = 340.5m v S S = 15.0m = 300. Hz = 0m v =.5 + C 331 / ( 0.6 m s 15.0 C) = 340.5m 340.5m 0m 340.5m 0m = 300. Hz( ) ( ) 340.5m + 15.0m = 300. Hz 340.5m + 15.0m 340.5m 340.5m = 300. Hz( ) = 300. Hz( ) = 313.82... Hz = 314Hz 325.5m = 287.37... Hz = 287Hz 355.5m Δ = 314 Hz 287Hz = 27Hz Resonance In sound applications, a resonant requency is a natural requency o vibration determined by the physical parameters o the vibrating object. This same basic idea o physically determined natural requencies applies throughout physics in mechanics, electricity and magnetism, and even throughout the realm o modern physics. Some o the implications o resonant requencies are: 11
It is easy to get an object to vibrate at its resonant requencies, hard to get it to vibrate at other requencies A vibrating object will pick out its resonant requencies rom a complex excitation and vibrate at those requencies, essentially "iltering out" other requencies present in the excitation. Most vibrating objects have multiple resonant requencies. Ease o Excitation at Resonance It is easy to get an object to vibrate at its resonant requencies, hard at other requencies. A child's playground swing is an example o a pendulum, a resonant system with only one resonant requency. With a tiny push on the swing each time it comes back to you, you can continue to build up the amplitude o swing. I you try to orce it to swing a twice that requency, you will ind it very diicult, and might even lose teeth in the process! Swinging a child in a playground swing is an easy job because you are helped by its natural requency. But can you swing it at some other requency? 12
Picking out resonant requencies A vibrating object will pick out its resonant requencies rom a complex excitation and vibrate at those requencies, essentially "iltering out" other requencies present in the excitation. I you just whack a mass on a spring with a stick, the initial motion may be complex, but the main response will be to bob up and down at its natural requency. The blow with the stick is a complex excitation with many requency components (as could be shown by Fourier analysis), but the spring picks out its natural requency and responds to that. A resonating tube with one end closed is called a closed-pipe resonator. A tube with both ends open is an open tube resonator. 13
How does resonance occur? The vibrating tuning ork produces a sound wave. This wave o alternate high-and lowpressure variations moves down the air column. When the wave hits the end o the tube, it is relected back. I the relected high-pressure wave reaches the ork at the same time that the ork produces another high-pressure wave then the leaving and returning waves reinorce each other. This reinorcement produces a standing wave, and resonance occurs. 14
Resonance Frequencies in a Closed Pipe The shortest column o air that can have an antinode at the closed end and a node at the open end is ¼ o a wavelength long. As the requency is increased, additional resonance lengths are ound at quarter wavelength intervals. So tube lengths o λ/4, 3 λ/4, 5 λ/4, 7 λ/4 and so on will be in resonance. Resonance in an open pipe The shortest column o air that can have nodes at both ends is ½ wavelength long. As requency is increased, additional resonances are ound at hal-wavelength intervals. So tube lengths o λ/2, 3 λ/2, 2 λ and so on will be in resonance. 15
Calculation o wavelength o a closed pipe λ = 4(l + 0.4d) A child s whistle is 15.0 cm long with a diameter o 1.25 cm. What is the requency o the sound that resonates through the whistle i the temperature is 20.0ºC? Hint: ind the wavelength, then plug that into the v= λ ormula. Calculation o wavelength o a open pipe λ = 2(l + 0.8d) An organ pipe open at both ends is 1.23 m long and has a diameter o 10.0 cm. What is the requency when the air is 15.0ºC? 16
Fundamentals, Pitch, Harmonics and Harmonic Series All periodic oscillations are either pure tones (sine waves) or are a set o pure tones all sounding together. The most prominent pure tone is known as the undamental and determines the undamental requency or pitch o the tone. The others are known as harmonics and have requencies related to the requency o the undamental by integer (whole number) actors. An harmonic series contains all the integral multiples o a pitch or as ar as the series extends. For example, a undamental requency F has harmonics o requencies 2F, 3F, 4F, 5F, 6F, etc. The undamental, F, is also reerred to as the irst harmonic, 2F the second harmonic and so on. 17