2. The acceleration of a simple harmonic oscillator is zero whenever the oscillating object is at the equilibrium position.

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1 CHAPTER : Vibrations and Waes Answers to Questions The acceleration o a siple haronic oscillator is zero wheneer the oscillating object is at the equilibriu position 5 The iu speed is gien by = A k Various cobinations o changing A, k, and/or can result in a doubling o the iu speed For exaple, i k and are kept constant, then doubling the aplitude will double the iu speed Or, i A and k are kept constant, then reducing the ass to one-ourth its original alue will double the iu speed Note that changing either k or will also change the requency o the oscillator, since k = π 7 The period o a pendulu clock is inersely proportional to the square root o g, by Equation -a, T = π L g When taken to high altitude, the alue o g will decrease (by a sall aount), which eans the period will increase I the period is too long, the clock is running slow and so will lose tie 0 Soe exaples o resonance: Pushing a child on a playground swing you always push at the requency o the swing Seeing a stop sign oscillating back and orth on a windy day When singing in the shower, certain notes will sound uch louder than others Utility lines along the roadside can hae a large aplitude due to the wind Rubbing your inger on a wineglass and aking it sing Blowing across the top o a bottle A rattle in a car (see Question ) A rattle in a car is ery oten a resonance phenoenon The car itsel ibrates in any pieces, because there are any periodic otions occurring in the car wheels rotating, pistons oing up and down, ales opening and closing, transission gears spinning, drieshat spinning, etc There are also ibrations caused by irregularities in the road surace as the car is drien, such as hitting a hole in the road I there is a loose part, and its natural requency is close to one o the requencies already occurring in the car s noral operation, then that part will hae a larger than usual aplitude o oscillation, and it will rattle This is why soe rattles only occur at certain speeds when driing The requency o a siple periodic wae is equal to the requency o its source The wae is created by the source oing the wae ediu that is in contact with the source I you hae one end o a taut string in your hand, and you oe your hand with a requency o Hz, then the end o the string in your hand will be oing at Hz, because it is in contact with your hand Then those parts o the ediu that you are oing exert orces on adjacent parts o the ediu and cause the to oscillate Since those two portions o the ediu stay in contact with each other, they also ust be oing with the sae requency That can be repeated all along the ediu, and so the entire wae throughout the ediu has the sae requency as the source 005 Pearson Education, Inc, Upper Saddle Rier, NJ All rights resered This aterial is protected under all copyright laws as they 69

2 Chapter Vibrations and Waes The speed o the transerse wae is easuring how ast the wae disturbance oes along the cord For a unior cord, that speed is constant, and depends on the tension in the cord and the ass density o the cord The speed o a tiny piece o the cord is easuring how ast the piece o cord oes perpendicularly to the cord, as the disturbance passes by That speed is not constant i a sinusoidal wae is traeling on the cord, the speed o each piece o the cord will be gien by the speed relationship o a siple haronic oscillator (Equation -9), which depends on the aplitude o the wae, the requency o the wae, and the speciic tie o obseration 7 (a) Siilar to the discussion in section -9 or spherical waes, as a circular wae expands, the circuerence o the wae increases For the energy in the wae to be consered, as the circuerence increases, the intensity has to decrease The intensity o the wae is proportional to the square o the aplitude (b) The water waes will decrease in aplitude due to dissipation o energy ro iscosity in the water (dissipatie or rictional energy loss) 8 Assuing the two waes are in the sae ediu, then they will both hae the sae speed Since = λ, the wae with the saller waelength will hae twice the requency o the other wae Fro Equation -8, the intensity o wae is proportional to the square o the requency o the wae Thus the wae with the shorter waelength will transit 4 ties as uch energy as the other wae 9 The requency ust stay the sae because the edia is continuous the end o one section o cord is physically tied to the other section o cord I the end o the irst section o cord is ibrating up and down with a gien requency, then since it is attached to the other section o cord, the other section ust ibrate at the sae requency I the two pieces o cord did not oe at the sae requency, they would not stay connected, and then the waes would not pass ro one section to another The energy o a wae is not localized at one point, because the wae is not localized at one point, and so to talk about the energy at a node being zero is not really a eaningul stateent Due to the intererence o the waes the total energy o the ediu particles at the nodes points is zero, but the energy o the ediu is not zero at points o the ediu that are not nodes In act, the anti-node points hae ore energy than they would hae i only one o the two waes were present Solutions to Probles node node The particle would trael our ties the aplitude: ro x = A to x = 0 to x = A to x = 0 to x = A So the total distance = 4A = 4( 08 ) = 07 The spring constant is the ratio o applied orce to displaceent F 80 N 75 N 05 N k = = = = 5 0 N x The spring constant is ound ro the ratio o applied orce to displaceent 005 Pearson Education, Inc, Upper Saddle Rier, NJ All rights resered This aterial is protected under all copyright laws as they 70

3 Giancoli Physics: Principles with Applications, 6 th Edition ( 68 kg)( 98 s ) F g 5 k = = = = 0 N x x 5 0 The requency o oscillation is ound ro the total ass and the spring constant 5 k 0 N 467 Hz 5 Hz = = = π π 568 kg 9 (a) At equilibriu, the elocity is its iu k = A = ω A = π A = π ( Hz)( 0 ) = 450 s 5 s (b) Fro Equation (-5), we ind the elocity at any position ( 00 ) ( 0 ) x = ± = ± ( 45 s) = ± 565 s ± 6 s A (c) E total = = = 060 kg 45 s 80J 8 J (d) Since the object has a iu displaceent at t = 0, the position will be described by the cosine unction ( 0 ) cos ( π ( 0 Hz ) ) ( 0 ) cos( 60π ) x = t x = t 0 The relationship between the elocity and the position o a SHO is gien by Equation (-5) Set that expression equal to hal the iu speed, and sole or the displaceent = ± x A = ± x A = x A = x A = x = ± A 0866 A 005 Pearson Education, Inc, Upper Saddle Rier, NJ All rights resered This aterial is protected under all copyright laws as they Since F = kx = a or an object attached to a spring, the acceleration is proportional to the displaceent (although in the opposite direction), as a = x k Thus the acceleration will hae hal its iu alue where the displaceent has hal its iu alue, at ± x 0 5 (a) The work done to copress a spring is stored as potential energy W ( 0 J) W = kx k = = = 467 N 4 0 N x 0 (b) The distance that the spring was copressed becoes the aplitude o its otion The k iu acceleration is gien by a = A Sole this or the ass k k N a = A = A = 0 = kg kg a 5 s 6 The general or o the otion is x = Acosωt = 045cos 640t (a) The aplitude is A = x = s (b) The requency is ound by ω = π = 640 s = = 09 Hz 0 Hz π

4 Chapter Vibrations and Waes (c) The total energy is gien by ( ω ) E = = A = 060 kg 640 s 045 = 488J 5 J total (d) The potential energy is gien by = = ω = 060 kg 640 s 00 = J J potential E kx x The kinetic energy is gien by E = E E = 488 J J = 77 J 4 J kinetic total potential 7 I the energy o the SHO is hal potential and hal kinetic, then the potential energy is hal the total energy The total energy is the potential energy when the displaceent has the alue o the aplitude E = E kx = pot tot ( ka ) x = ± A ± 0707 A 80 s 7 The period o the juper s otion is T = = 475 s The spring constant can then be ound 8 cycles ro the period and the juper s ass ( 475 s) 4π 4π 650 kg T = π k = = = 7N 4 N k T The stretch o the bungee cord needs to proide a orce equal to the weight o the juper when he is at the equilibriu point ( 650 kg)( 980 s ) g k x = g x = = = 560 k 7N Thus the unstretched bungee cord ust be = 94 6 The wae speed is gien by = λ The period is 0 seconds, and the waelength is 65 = λ = λ T = 65 0 s = s 7 The distance between wae crests is the waelength o the wae λ = = 4 s 6 Hz = 8 To ind the waelength, use λ = AM: FM: s 00 0 s λ = = = 545 λ = = = 88 AM: 90 to Hz Hz s 00 0 s λ = = = 4 λ = = = 78 FM: 78 to Hz 08 0 Hz Pearson Education, Inc, Upper Saddle Rier, NJ All rights resered This aterial is protected under all copyright laws as they 7

5 Giancoli Physics: Principles with Applications, 6 th Edition 44 (a) Both waes trael the sae distance, so x = t = t We let the saller speed be, and the larger speed be The slower wae will take longer to arrie, and so t is ore than t t = t + 0in = t + 0 s t + 0 s = t t = = = 55k s 0 s 0 s 0 s 85k s 55k s x = t = = 85 k s 0 s 9 0 k (b) This is not enough inoration to deterine the epicenter All that is known is the distance o the epicenter ro the seisic station The direction is not known, so the epicenter lies on a circle o radius 9 0 k ro the seisic station Readings ro at least two other seisic stations are needed to deterine the epicenter s position 48 Fro Equation (-8), i the speed, ediu density, and requency o the two waes are the sae, then the intensity is proportional to the square o the aplitude I I = E E = A A = A A = = 4 The ore energetic wae has the larger aplitude 50 The bug oes in SHM as the wae passes The iu KE o a particle in SHM is the total energy, which is gien by E = ka Copare the two KE ia total KE ka A 5 c = = 056 = = KE ka A 0 c 5 (a) (b) (c) The energy is all kinetic energy at the oent when the string has no displaceent There is no elastic potential energy at that oent Each piece o the string has speed but no displaceent 5 The requencies o the haronics o a string that is ixed at both ends are gien by = n n, and so the irst our haronics are = 440 Hz, = 880 Hz, = 0 Hz, = 760 Hz 4 5 The undaental requency o the ull string is gien by = = 94 Hz uningered I the length is L reduced to / o its current alue, and the elocity o waes on the string is not changed, then the new requency will be ingered = = = uningered = 94 Hz = 44 Hz L L Pearson Education, Inc, Upper Saddle Rier, NJ All rights resered This aterial is protected under all copyright laws as they 7

6 Chapter Vibrations and Waes 54 Four loops is the standing wae pattern or the 4 th haronic, with a requency gien by = 4 = 80 Hz Thus = 70 Hz, = 40 Hz, = 0 Hz and = 50 Hz 5 are all other 4 resonant requencies 55 Adjacent nodes are separated by a hal-waelength, as exaination o Figure -40 will show 9 s λ = x = λ = = = 97 0 node 475 Hz 56 Since = n n, two successie oertones dier by the undaental requency, as shown below = = n + n = = 50 Hz 80 Hz = 70 Hz n+ n 57 The speed o waes on the string is gien by equation (-), = The resonant requencies L n o a string with both ends ixed are gien by equation (-9b), =, where L n ib is the length Lib o the portion that is actually ibrating Cobining these relationships allows the requencies to be calculated n F 50 N L L kg 090 T = = = n ib = = 5854 Hz = = 87Hz ( ) F T 9077 Hz So the three requencies are 90 Hz, 580 Hz, 870 Hz, to signiicant igures 6 Fro the description o the water s behaior, there is an anti-node at each end o the tub, and a node in the iddle Thus one waelength is twice the tube length = λ = L = Hz = s tub 6 The speed in the second ediu can be ound ro the law o reraction, Equation (-0) o sin 5 = = = ( 80k s) 6k s o = sin 47 6 The angle o reraction can be ound ro the law o reraction, Equation (-0) s = = = sin 4 = 049 θ = sin 049 = 5 8 s o o 64 The angle o reraction can be ound ro the law o reraction, Equation (-0) The relatie elocities can be ound ro the relationship gien in the proble + 060T o + 060( 0) o 5 = = = sin 5 = sin 5 = T θ = sin = 4 o 005 Pearson Education, Inc, Upper Saddle Rier, NJ All rights resered This aterial is protected under all copyright laws as they 74

7 Giancoli Physics: Principles with Applications, 6 th Edition 66 The error o o is allowed due to diraction o the waes I the waes are incident at the edge o the dish, they can still diract into the dish i the relationship θ λ L is satisied λ o π rad θ λ = Lθ = ( 05 ) = o L 80 I the waelength is longer than that, there will not be uch diraction, but shadowing instead 67 The unusual decrease o water corresponds to a trough in Figure -4 The crest or peak o the wae is then one-hal waelength distant The peak is 5 k away, traeling at 750 k/hr x 5 k 60 in x = t t = = = 0 in 750k hr hr 7 The requency at which the water is being shaken is about Hz The sloshing coee is in a standing wae ode, with anti-nodes at each edge o the cup The cup diaeter is thus a hal-waelength, or λ = 6 c The wae speed can be calculated ro the requency and the waelength = λ = 6 c Hz = 6c s 7 Relatie to the ixed needle position, the ripples are oing with a linear elocity gien by re in π ( 008 ) = = 07 s in 60 s re This speed is the speed o the ripple waes oing past the needle The requency o the waes is 07 s = = = 0 Hz λ The equation o otion is x = 0650cos 740 t = Acos ωt (a) The aplitude is A = 0650 (b) The requency is gien by ω = π = 740 rad s 740 rad s = = 77 Hz 8 Hz π rad (c) The total energy is gien by ω 00 kg 740 rad s J J total E = ka = A = = (d) The potential energy is ound by = = ω = = PE kx x 00 kg 740 rad s J 70 J The kinetic energy is ound by KE = E PE = 6 J 70 J = 94 J total 79 (a) The iu speed is gien by = π A = π 64 Hz 8 0 = 0 s (b) The iu acceleration is gien by a 4 A 4 64 Hz s = π = π = 005 Pearson Education, Inc, Upper Saddle Rier, NJ All rights resered This aterial is protected under all copyright laws as they 75