WAVES. Wave Equation. Waves Chap 16. So far this quarter. An example of Dynamics Conservation of Energy. Conservation theories. mass energy.

Transcription

1 Waes Chap 16 An example of Dynamics Conseration of Energy Conceptual starting point Forces Energy WAVES So far this quarter Conseration theories mass energy momentum angular momentum m E p L All conserations theories: Conseration of X scalar scalar ector ector X system f X system i X input X output X system X transfer So far we hae focused on the properties of the system i.e. How does the system energy change? Nowwebeginamoredetailedstudyof Energy Transfer Mechanisms How can energy be transferred into or out of a system? How can energy be transferred from one system to another? Mechanisms of energy transfer across a distance Objects carry it off Waes transport energy across a distance without transporting matter across that distance. Examples: Wae on a rope The pulse moes with speed but no object, or part of an object, moes with that speed after elastic collision before elastic collision No object carries it off Drop object in pool of water 0 Where does energy go? Parts of the rope moe up and down, the wae moes horizontally Transerse wae Easiest to isualize Most examples Example compression wae in a spring: Parts of the spring moe back and forth, the pulse moes horizontally Longitudinal Wae Wae Equation ψ To ψ ρo z y F y Notation in the textbook µ x Solution What function? Second deriatie with respect to time gies same function as second deriatie with respect to position. Since position and time are different quantities, choose each function so that the second deriatie is the same as the function. You will explore the properties of both types of waes in the laboratory.

2 y F y µ x try y A cos( ax + b) cos( ct + d) y Acos( ax + b) csin( ct+ d) y Acos( ax + b) c cos( ct+ d) y Aasinax ( + b) cos( ct + d) x y Aa cos( ax + b) cos( ct + d) x F A cos( ax + b) c cos( ct + d) Aa cos( ax + b) cos( ct + d) µ F c a µ c F a µ Need another way to find either a or c. y A cos ( ax + b ) cos ( ct + d ) At what position does this repeat? Waelength y 1 A cos( ax1 + b) cos( ct + d) y y A cos ( ax + b) cos ( ct + d) 1 y A cos( ax1 + b) cos( ct + d) A cos( ax + b ) cos ( ct + d) cos( ax1 + b) cos( ax + b) ax1 + b + π ax + b π ax ax 1 π a( x x1 ) a π a At what time does this repeat? At what time does this repeat? Period y 1 A cos( ax + b ) cos( ct1 + d) y A cos ( ax + b) cos ( ct + d) A cos( ax + b) cos( ct1 + d) A cos( ax + b) cos( ct + d) cos( ct1 + d) cos( ct + d) ct1 + d + π ct + d π ct ct 1 π c( t t 1 ) cτ π c τ y 1 y y F y µ x y A cos( ax + b) cos( ct + d) c F a µ π a π τ F π µ ( ) F ( τ) µ π c τ τ F µ Waes Energy transport through a medium TRAVELING WAVES Our normal obserations are in both space and time The wae pattern repeats oer a distance is called the waelength A x 1 x x -x 1 x 3 +x z0 If the wae takes a time τ to trael a distance τ is called the period y0 A Its speed is / τ x 1 x x 3 +x τ

3 Consider two different simple obserations Obsere at fixed time (snapshot) A y0 +x x 1 x x -x 1 Obsere at a fixed position τ y0 A Wae elocity Determined by properties of medium Restoring force Mass density Waelength Distance of pulse repetition Period τ For a wae Time for pulse repetition τ or f1/τ f t t 1 t +t t -t 1 τ Material is moing in oscillations Spring forces in material F-k x Velocity of material t Perpendicular to wae motion transerse wae Parallel (and anti-parallel) to wae motion longitudinal wae Oscillating motion Wae amplitude A Maximum displacement of medium Angular frequency πf ω For a spring (oscillating system) How does the wae speed depend on the properties of the system (spring)? Some intuitie reasoning which could be wrong (check it with nature in the lab) Frequency depends on mass and spring constant frequency frequency For wae motion f k m restoring force strength mass term Frequency is determined by wae elocity It is reasonable that wae elocity goes like restoring force strength mass term restoring force strength mass term want units to be elocity k m units force / length 1 mass time to get units of elocity multiply by length length time force length / length mass Wae phenomena Determined by boundary conditions An important example: standing waes Waes that seem to be fixed in space length time Guess force mass / length T µ µ mass / length

4 STANDING WAVES Waes as oscillations Standing Waes fixed supports TRAVELING WAVES snapshot y0 A x 1 x +x To keep the same shape oer the whole rope Each piece of the rope must oscillate up and down at the same rate. STANDING WAVES y0 Each part of wae has same time dependence x 1 x +x node anti-node node A standing wae is described by y f(x) g(t) / / 3/ 4/ nodes Standing wae has zero amplitude anti-nodes Standing wae has maximum amplitude Ln/ for a standing wae Time dependence of eery position of rope is the same. Eachpointonropeismoingupand down. Harmonic motion Periodic in time Shape of wae is periodic. Periodic in space y 0 at x 0, y 0 at xl, y 0 at nodes Try yasin(kx)cos(ω t) k π/ ω π/τ Where does 0 When x n/ anode Example: While watching a concert you wonder about the physics of playing a iolin. One of the iolin strings sounds an A note (440 Hz) when played without fingering. The string is 30 cm long and has a mass of.0 g. Where must one put one s finger to play acnote(58hz)withthesamestring? A finger placed on a string becomes a node. Geometry of standing wae gies waelength. Frequency related to waelength by speed Same string, same wae speed. L No fingering L 1 / f Hz d Finger at a node f 58 Hz d / Find distance of finger from end of string d Use: τ τ 1/f

5 d / need τ /f d/f need From string without fingering 1 /f 1 1 f 1 d 1 f 1 /f need 1 L 1 / Example: You are helping a friend design a high tech electric iolin. She wants one steel string to ibrate at a frequency of 51 Hz in its lowest mode. The next string should be made of the same material and hae a lowest mode of 56 Hz. If the tensions of the two strings are different, they will warp the instrument. To decide on the thickness of the string, she asks You to calculate the ratio of the mass per unit length of the two strings so that they hae the same tension. dlf 1 /f d (30 cm) (440 Hz)/(58 Hz) d5cm Units are correct for a length Finger position is reasonable, it is less than the length of the string. L 1 What is µ /µ 1? f 1 51 Hz f 56 Hz string tension T 1 T T mass per unit length µ 1, µ Use: Wae properties τ τ 1/f Tension mass density Vibration of strings is standing wae. Lowest mode - nodes only at ends. Mass per unit length and tension gie wae speed Waelength and frequency gies speed Geometry gies waelength for standing wae L T µ For the two strings with the same tension µ 1 T 1 µ T L/ string 1 1 string µ 1 need µ 1 1, µ 1 µ 1 1 τ 1 /τ 1 1 τ /τ 1 µ τ 1 µ 1 τ 1 1 τ τ µ f 1 µ 1 f What is the mechanism for getting Standing Waes? Shake one end of a rope and keep shaking it. Get traeling waes Hit end and reflect ( ) ( ) µ 51Hz µ 1 ( 56Hz) 56Hz ( 56Hz) 4 Units are correct since a ratio has no units The lower frequency string has a larger mass so it ibrates slower at the same tension.

6 But waes are still coming in Combine and How the waes match depends on Their Reflection Properties Two cases Fixed end Free end Fixed end 3rd Law Pairs F rw force of wall on rope At each alue of x, add the y from the incoming wae and the y from the outgoing wae. Superposition force of rope on wall F wr Result depends on how the waes match Reflections from a fixed end are inerted Free end Question At t0, your friend claps hands, with 1.00 second between each clap. You are standing 550 ft away from your friend. The speed of sound in air is 1100 ft/s. When do you hear the first clap? What is the time interal you hear between the first clap and the second clap? Reflections from a free end are not inerted At t0, your friend claps hands, with 1.00 second between each clap. You are 550 ft away from your friend and running away with a constant speed of ft/s. The speed of sound in air is 1100 ft/s. When do you hear the first clap? What is the time interal you hear between the first clap and the second clap? At t0, your friend claps hands, with 1.00 second between each clap. You are standing 550 ft away from your friend. The speed of sound in air is 1100 ft/s. When do you hear the first clap? Time for sound to trael 550 ft s d t 550 ft t 1100 ft / s 0.500s At t0, your friend claps hands, with 1.00 second between each clap. You are 550 ft away from your friend and running away with a constant speed of ft/s. The speed of sound in air is 1100 ft/s. When do you hear the first clap? s to 0 s y y xo 0 xa 550ft xo to s x 1 t 1 s t 1 x 1 xa to y x 1 x a t 1 x1 t1 unknowns, equations t1? x1? s 1100 ft/s y 3ft/s What is the time interal you hear between the first clap and the second clap? Time interal 1.00 s y st 1 x a t 1 x a ( s y )t 1 x a ( s y ) t ft 1097 ft / s t s t1

7 What is the time interal you hear between the first clap and the second clap? s s y y xo tb x s t t b xa to y x x a t unknows, equations s ( t t b ) x y s( t t b ) x a t y t + x a s ( t t b ) x t to0 tb1s xo0 xa550ft t? x? s 1100 ft/s y3ft/s x a + s t b ( s y ) t time interal between the first & second clap x a tt -t1 ( s y ) t 1 x a + s t b x a ( s y ) ( s y ) t s ( s y ) t b t 1100 ft / s 1097 ft / s t s t x a + s t b ( s y )t Or using waes Emitted wae (claps) s Ttb T Receied wae (heard claps) ' ' T t T' s y s t b t s Example: While trying to get home for spring break, you hae a flat tire on the freeway. As you change your tire a truck approaches you blowing his air horn which has a frequency of 50 Hz. Your friend who is helping you has perfect pitch and claims the frequency that you both hear is 80 Hz. You wonder if the truck is speeding. The speed limit is 65 mi/hr. You remember that the speed of sound is about 1100 ft/s which is 750 mph. s s y t b t same as before t Diagram of sound wae for a truck at rest. 750 mph f 80 Hz Distance between peaks of intensity is the same eerywhere Want to find t We know: f1/τ 50 Hz 750 mph You hear f 1/τ 80 Hz speed limit 65 mph peaks of intensity Approach: Use kinematics Use the relationship of frequency to wae elocity. truck X You Speed of wae does not depend on the speed of the object emitting it. τ f

8 Diagram of sound waes for truck moing with speed t Distance between peaks of intensity is not the same eerywhere. Wae speed does not depend on speed of source. peaks of intensity t t τ t τ +x Wae peaks are closer together in direction truckismoing. Truck moes toward wae peaks If truck is emitting a sound with aperiodτ The distance between peaks is - t τ So the distance between repetitions is - t τ τ is time between emitted peaks. For any wae there is a relationship between waelength and period τ τ is the time between receied peaks Find t t τ τ' Find τ f' 1 τ' Find τ Find τ f 1 τ need, τ, τ 750 mi 50Hz 1 hr 80Hz t 80 mi hr units are correct, speed in mph The truck is speeding but 80 mph is possible for a truck f t f f ' f f ' t f f f ' t 1 f f ' t check units Many possible ariations Suppose you know the truck s (source) speed and want to determine the frequency you hear. Find f f' ' Find - s /f Find s /f f' ' - s τ s τ s τ +x since truck moes toward you a distance t τ - s /f know, s,andfwantf s is source elocity iswae elocity τ is time between emitted peaks. /f- s /f f' f s f f' f ( s ) as s gets larger, the frequency heard gets greater. Higher pitch

9 Ifthesourceisgoingawayfromyou s has the opposite sign f' f ( + s ) as s gets larger, the frequency heard gets less. For practice draw the diagram and show this is true Lower pitch s Example: While trying to get home for spring break, you are driing down the freeway. On the side of the road is a truck with a flat tire. Since the drier needs help, he sounds his air horn as you pass him. Your friend who is riding with you has perfect pitch and claims the frequency that you both hear is 80 Hz. You wonder what the frequency of the truck s horn is. You know you are driing right at thespeedlimitof65mph. Youremember that the speed of sound is about 1100 ft/s which is 750 mph. +x s τ s τ The truck is stopped Stationary source, moing receier τ truck You What frequency do you hear? Speed of wae depends only on the material it traels through. τ is the period of the wae emitted by the source. f r f 80 Hz 750 mi/hr r 65mi/hr f is the frequency of thewaeemittedby the source. Since you are moing away from the waes, you obsere them to hae a slower speed, - r The distance between the wae peaks is the same, so you obsere the same The frequency you hear is greater since the time between peaks is smaller than if you were standing still. Find f τ f Find f' ' ' f' Find - r 1 f f' ' into 1 f' ' need need into f f' r f ( 80Hz) 750 mi hr 307Hz mi 65 hr 750 mi hr Units are correct, frequency in Hz Answer is reasonable since the moing receier hears a lower frequency than emitted by the source. f' f r soling for f from aboe If the receier is moing away from the source -(- r )+ r replace r by - r moing source (toward receier) moing receier (toward source) f' f ( s ) f' f + r RECAP s τ Moing source Emitted Obsered < source toward receier > source away from receier - s τ motion of s τ f τ f Moing receier Emitted Obsered > receier toward source < receier away from source + r motion of r τ f τ f from wae equation from wae equation

10 Both moing source and receier For emitted wae f For the wae you hear f moing source - s τ - s /f Source moing in SAME direction as wae moing receier - r Receier moing in SAME direction as wae Know: f, s, r Find f f need, s f f' ( r ) need /f s f f' ( r ) ( s ) f' f ( r ) f' f r ( s )