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# Acoustics PHYS 170

DU

GPA 3.82

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This 35 page Class Notes was uploaded by Milton O'Hara on Sunday October 11, 2015. The Class Notes belongs to PHYS 170 at Duquesne University taught by Staff in Fall. Since its upload, it has received 52 views. For similar materials see /class/221278/phys-170-duquesne-university in Physics 2 at Duquesne University.

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Date Created: 10/11/15

Chapter 4 Homework Questions for Thought and Discussion 24 Exercises 3 4 5 68 Chapter 4 Resonance Resonance is an effect when a system is driven at a natural frequency When an object or system is driven at a natural frequency the resulting amplitude of the vibration dramatically increases The simplest example of resonance and one thatI hope you ve all experienced is that of the swing A swing is essentially a pendulum and from Chapter 2 the natural frequency of the swing is determined by the length of the rope Resonance occurs when the person in the swing drives the system at or near the natural frequency 1 fi 5 271 l The resulting effect the person in the swing increases the amplitude of the motion gains gravitational potential energy by gaining height The longer the swinger drives the system at the resonant frequency the higher he or she will go 41 Mass Spring System Another system that is useful for demonstrating resonance is the familiar mass spring system If you grab the end of the spring the end not attached to the mass D and move that hand up and down you can observe resonance as well as the conditions for resonance I there will be a range of frequencies that increase the amplitude of the mass spring system I the frequency that produces the largest increase in amplitude is the resonant frequency I as you sweep through the frequencies you will also see a change in the phase between your hand and the motion of the mass At low f 1 f 1 39 below your hand and the mass will move in the same direction They will move in phase with each other At frequencies above resonance your hand will move in the opposite direction of the mass They will move out of phase with each other A mm A f The graph in Figure 42 is What you would get if you graphed the amplitude of the mass as you changed the applied frequency of the system that of your hand The resonantfrequencyfo is found from the peak of the curve Notice that there are other frequencies that increase the amplitude of the mass but the resonant frequency gives the maximum amplitude This is also a Vibration spectrum as discussed in Chapter 2 There is only one line or one resonant frequency present However as is normally the case there are frequencies close to resonance The linewidth ofthe peak is denoted by Afand is related to the term Qifactor which is a term used to discuss the sharpness of the resonance f0 9 3 A max The smaller the linewidth Af the sharper the resonance or the higher the Qifactor Basically the higher the Qifactor the closer you have to get to resonance to see an effect While resonance is an important acoustical phenornenon resonant effects are not always desired As the book also mentions the Qifactor is also related to the damping constant of the spring mass system The higher the Qifactor the smaller the damping of the system ie The slower the Vibration dies out and conversely the lower the Qifactor the larger the damping constant ie The quicker the Vibration dies out Hi gher Q factor Lower Q factor rm I m not really sure Why this is brought up but the tech majors among you might make use of this fact 42 Phase ofDn39ven Vibrau39ons quotIfwecarefully L L 439 ofnnon39on of L 39 39 39 39 frequencies farbelow o nee 39 439 39 39 hnwe enmeymovein 439 39 We describe this phenomenon by using the term phase which may be thought ofas a specification ofthe starting point nfn vibration At low frequencies L 39 quot 39 39 ofthecrank A L 39 L 439 stretc es m1 39 of L 39 39 ho e er 39 39 quot 39 39 39 39 39 39 quot quot 439 39 I quot bythecrankAtresonancethemassis onerfounh 39 39 quot quot 39 quot quot quot 39 39 39 39 Value L 1 39 39 llfunherthe L quot 39 nun nally L 39 L 39r 39 39 quot 39 39 39 thatis 39 39 39 39 opposite direenons quot Crank39s Manon Mass39s Mou39on 2 3 43 Standing Waves on a String Now we will get back to some of the terminology from Chapter 2 and maybe clear things up as well and put them with ideas presented in Chapter 3 Ideas from Ch 3 Terminology from Ch 2 Standing waves Re ection of waves The fundamental frequency Harmonics Putting this all together Standing waves are created by waves that have the same frequency and amplitude but are traveling in opposite directions The best example of this is take a string or spring with a decently high K that has one end attached to something that won t move As the free end is moved up and down the wave travels down the string or spring at a speed given by When the wave hits the end of the string it will be re ected and retum back to the free end The only catch is that a standing wave will be created only at certain frequencies So in a sense standing waves can also be thought of as a resonant phenomena Standing waves have areas of maximum amplitude and occur at special frequencies will new build fonnulas to figure out what these special frequencies are from the terrninology frorn Chapter2 and the speed equations from Chapter3 Our goal is to relate the frequency of vibration to the length ofthe string by looking a what are called boundary conditionsquot We start with the lowest possible mode of vibration for a string of len th L thatis xed at both ends The boundary condlhons require there to be antinodes at the ends of the string which are attached to an immovable objectlike a wall This mode will look like Figure 2 m It will always be true that the lowest mode of vibration for a standing wave on a string will have the We can i i i i of iA i i i ii in U H mm Remember lii mndin a i atdbyna iii a androina liii wavelengthandfrequencyarerelated VAf so that the frequency is given by and with A 2L fL 2L And for a string under tension we know how the speed is related to the properties of the string putting this into the last equation gives g fi 2L which can be rewritten into a neater form 1 T ZLM and since this would be the frequency of the lowest mode of Vibration it s called the fundamental frequency or the rst harmonic Let s call this frequency So that 1 T fli 2L M VI mtjon will look like Now the question becomes is there a rebitionship between the frequency and 1ength of the stnng for all othe standing waves I guess we39ll nd out Again h o niidni quot 39 39 39 A L mum gm i i b W We can also relate the wavelength of this mode to the 1ength othe stn39ng m x L N and following the same steps we nd Lhatthe frequency of this mode is given by v f X The speed of a wave on a stnng is sti11 there v J However thisu39me AL M l T putting these pieces together gives us fj 7 lM Now if we are clever a panein starts to emerge le Z 1 Z 2 Z L M 2L M 2L M Let39s label this frequency f2 4a quot139 39 39har mnnirl If we continue this panem the next mode looks like 2 x L N N 3 and the frequency of this mode will be given by f 1 Z 3L 1 2L M 2L M Again we have a harmonic on our hands 3 Thisishow 39 39 1 39 vibmtjonfora 39 39 4 mm here A l 2 3 ande will be the frequency of the n hannonic the 15 harmonic is usually called the fundamental Fre uenc if 15tharmomc q y 7 I fundamental In summary 39 Antincdes For the vibrating string the number of the harmonic can be found by counting the number of antinodes in the mode 2nd harmonic F 39 requency 21 151 overtone The 15t harmonic or fundamental has one antinode The 2quotd harmonic has two antinodes Extra problems Standing Waves on a String Example 1 The A string on a bass is tuned to vibrate at a fundamental frequency of 550 Hz If the tension in the string were increased by a factor of four what would be the new fundamental frequency Example 2 A string of length 028 In is xed at both ends The string is plucked and a standing wave is set up that is vibrating at its second harmonic The traveling waves that make up the standing wave have a speed of 140 ms What is the frequency of vibration Example 3 A 41 cm length of Wire has a mass of 60 grams It is stretched between two xed supports and is under a tension of 160 N What is the fundamental frequency of this Wire Example 4 On a cello the string with the largest linear mass density 14 is the C string This string has M156 X 10 2 kgm and produces a fundamental frequency of 654 Hz If the length of the string is 0800 meters nd the tension in the string 44 Partials Harmonics and Overtones Vocabulary Time Harmonic modes of vibration of a system that are Whole number multiples of the fundamental mode and the sounds they generate This term may also be used for nearly Whole number multiples of the fundamental such as 3005 for example Fundamental the lowest frequency mode of vibration of a system The fundamental can also be referred to as the rst harmonic Overtone higher modes of vibration than the fundamental but not a Whole number multiple of the fundamental A harmonic is a special kind of overtone but there are many other overtones 15 times the fundamental is an overtone but not a harmonic The term overtone does not include the fundamental As a consequence the 1st overtone is the second harmonic The 2quot01 overtone is the 3rd harmonic Partial all modes of vibration the fundamental pulse overtones everything Even worse is the term upper partial which excludes the fundamental The upper partial is the same thing as the overtones 45 Open and Closed Pipes In way similar to section 43 we will now build formulas for the resonant frequencies of pipes with two open ends and pipes with one end closed Our boundary conditions are a bit different than in section 43 For a resonant sound there will be an antinode at any open end to a pipe Why Well the loudest sound is going to occur when the air molecules are moving the most This only happens at antinodes If the end of a pipe is closed there will be a node Even though we are dealing with pipes I always try to go back to the string as a visual reference I will include pictures of the pipe overlaid with a picture of the vibrating string for reference whenever possible I think this might help you to visualize what is going on in the pipe because nodes correspond to areas of low pressure antinodes to areas of high pressure The Pipe of Length L With Two Open Ends Ignon39ng edge effects the lowest vibrational mode will contain one node at the center of the pipe and antinodes at each end and would look like l E1 771 m 2 7 9L H e z e H L g p A x A quot vw solvingforfrequency f Like the ease of the stn39ng we an relate the 1ength of the pipe to the wavelength of this mode name1y A 2L Unlike the ease of the stn39ng we don39t have a formula forthe speed of sound in air unless you have really good memory there was one in ChapterZ The speed of sound in air is usua11y something that is given Using this re1ation with the equation for the speed ofthe wave Where I39ve used A 2L 39 quot 39 mode I39ll The second mode of vibran39on would look like Hopefully it is not Lenibly of this mode gives harmonic of The third mode of vibration would look like 2113 of this mode gives v or f23 From these rst three modes a panein can be seen For any resonant frequency involving a pipe open atboth ends 123 Where n keeps track of the harmonic starts at one and increases by one each time ie all harmonics are present One short cut exists similar to the string The number of the harmonic is equal to the number ofn The Pipe ofLength L With One End Closed Closing one end othe pipe adds one more restriction to our pictures and a twist we39ll talk about soon The only thing to remember is the boundary conditions say there will be a node atthe closed end of the pipe The lowest resonant rnode would be Es 141 V solving for frequency give us A relating the wavelength to the length of the pipe fz l 39 39 like we did u 39 39 39 The next resonant mode of vibration would consist of a node at the closed and open ends of he pipe There39s a problem here A node is where e particles in the medium are not moving and as a result no tone would be heard remember this is an ideal case a ma So the second 39 mud nld be 4143 reamnging a bit The next hannonic is not going to be audible like the second hannonic Thus the next audible frequency would be Now the panein emexges For a pipe with one end closed the audible resonant frequencies are given by riv f 7 n 13s f n135 2 4L The result of closing one end is that every even harmonic disappears and there is a 11 instead of a 12 in the formula My hope in going over how these formulas are created is two fold 1 these exercises have reinforced the concepts amp conditions for resonance in the different systems 2 as a result you are able to create these formulas if you wantneed to I m not a big fan of memorizing formulas Extra problems Pipes Example 5 An organ pipe is open at both ends It is producing a sound at its third harmonic 262 Hz The speed of sound is 343 ms What is the length of the pipe Example 6 Sound enters the ear travels through the auditory canal and reaches the eardrum The auditory canal is approximately a tube open at only one end The other end is closed by the eardrum A typical length for the auditory canal in an adult is about 29 cm If the speed of sound is 343 ms What is the fundamental frequency of the canal Interestingly the fundamental frequency is in the frequency range Where human healing is most sensitive Example 7 A tube of air is open at only one end and has a length of 15 m This tube sustains a standing wave at its third haimonic What is the distance between one node and the adjacent antinode Example 8 The fundamental frequency of a vibrating system is 400 Hz For each of the following systems give the three lowest frequencies excluding the fundamental at which standing waves can occur a a string xed at both ends b a cylindrical pipe with both ends open and c a cylindrical pipe with only one end open 48 Singing Rods Longitudinal modes of vibration can be created in metal rods with free ends The fundamental mode will have an antinode at each end and a node at the center of the rod just like a tube with both ends open The formula for the frequencies of the audible modes looks like the formula of a pipe with both ends open or those of the string xed at both ends nv fni 12 3 2L 1 but here is Where you use the speed formula given in Chapter 3 v p thus getting the formula in the book fn n 12 3 FIGURE 411 Strokng an aluminum rod with the ngers to excite longitudinal resonances Apparently exiting resonant modes in a bar is only exciting for the bar Also nice caption 49 More Examples of Resonance Can a Singer Break a Wine Glass Not likely The easiest mode of vibration to excite in an object is the fundamental The fundamental also contains the least amount of energy To break the wine glass you would need a lot of ampli cation at the fundamental frequency or excite higher harmonics in the glass or both 410 Sympathetic Vibrations This section explains why you place the handle of a tuning fork against a wood plate or table top Most if not all of you knew this before I did The tuning fork gets the surface to vibrate at the forks frequency sympathetic vibrations Even though these frequencies are not likely to be at resonant frequencies of the wood the sound is ampli ed do to the larger area vibrating at that frequency Questions for Thought and Discussion 2 List as many examples of Helmholtz resonators as you can other than those given in Section 47 Are the resonances sharp or broad 4 Does the end correction given in Section 45 lower all harmonics of a pipe proportionally or does it result in the overtones going out of tune An exact expression for the end correction shows that is varies slightly with wavelength Does that change your answer Exercises 3 Determine the frequencies of the fundamental and rst overtone second partial for the following Neglect end corrections a A 16 ft open pipe b A 16 ft stopped organ pipe one open end one closed end 4 Extend Figures 47 and 48 to include two more modes each 5 Find the difference in the fundamental frequency calculated with and Without the end correction of an open organ pipe 2 In long and 10 cm in diameter 6 A nylon guitar string 65 cm long has as mass of 8310 4 kgm and the tension is 56 N Find the frequencies of the rst four par1ials 8 Determine the frequencies of the pipes in Problem 3 if helium is substituted for air The speed of sound in helium is given in Table 31

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