Chapter 23: Superposition, Interference, and Standing Waves

Transcription

1 Chapter 3: Superposition, Intererence, and Standing Waves Previously, we considered the motion o a single wave in space and time What i there are two waves present simultaneously in the same place and time Let the irst wave have λ and T, while the second wave has λ and T The two waves (or more) can e added to give a resultant wave this is the Principle o Linear Superposition Consider the simplest example: λ λ

2 Since oth waves travel in the same medium, the wave speeds are the same, then T T We make the additional condition, that the waves have the same phase i.e. they start at the same time Constructive Intererence The waves have A and A. Here the sum o the amplitudes A sum A +A 3 (yy +y ) A Sum A

3 I the waves (λ λ and T T ) are exactly out o phase, i.e. one starts a hal cycle later than the other Destructive Intererence I A A, we have complete cancellation: A sum 0 yy +y 0 A sum A These are special cases. Waves may have dierent wavelengths, periods, and amplitudes and may have some ractional phase dierence.

4 Here are a ew more examples: exactly out o phase (π), ut dierent amplitudes Same amplitudes, ut out o phase y (π/)

5 Example Prolem Speakers A and B are virating in phase. They are directed acing each other, are 7.80 m apart, and are each playing a 73.0-Hz tone. The speed o sound is 343 m/s. On a line etween the speakers there are three points where constructive intererence occurs. What are the distances o these three points rom speaker A? Solution: Given: A B 73.0 Hz, L7.80 m, v343 m/s

6 λ vt v 343 m/s 73.0 Hz 4.70 m L x λ L + x λ x is the distance to the irst constructive intererence point The next point (node) is hal a wave-length away. x L n λ Where n0,,,3, or all nodes n 0 : x m

7 n : x.55 m n : x m Behind speaker A n : x ( ) 6.5 m Speaker A A A! " sum B x x λ

8 Beats Dierent waves usually don t have the same requency. The requencies may e much dierent or only slightly dierent. I the requencies are only slightly dierent, an interesting eect results the eat requency. Useul or tuning musical instruments. I a guitar and piano, oth play the same note (same requency, ) constructive intererence I and are only slightly dierent, constructive and destructive intererence occurs

9 The eat requency is T as T T, or In terms o periods The requencies ecome ``tuned Example Prolem 0 When a guitar string is sounded along with a 440- Hz tuning ork, a eat requency o 5 Hz is heard. When the same string is sounded along with a 436- Hz tuning ork, the eat requency is 9 Hz. What is the requency o the string?

10 Solution: Given: T 440 Hz, T 436 Hz, 5 Hz, 9 Hz But we don t know i requency o the string, s, is greater than T and/or T. Assume it is. 445 Hz Hz and T s T s T s T s I we chose s smaller 47 Hz Hz and T s T s s T s T

11 Standing Waves A standing wave is an intererence eect due to two overlapping waves - transverse wave on guitar string, violin, - longitudinal sound wave in a lute, pipe organ, other wind instruments, The length (dictated y some physical constraint) o the wave is some multiple o the wavelength You saw this in la last semester Consider a transverse wave (, T ) on a string o length L ixed at oth ends.

12 I the speed o the wave is v (not the speed o sound in air), the time or the wave to travel rom one end to the other and ack is I this time is equal to the period o the wave, T, then the wave is a standing wave L / L v v T λ L v L λ Thereore the length o the wave is hal o a wavelength or a hal-cycle is contained etween the end points We can also have a ull cycle contained etween end points v v λ L λ L v

13 Or three hal-cycles λ 3 3 Or n hal-cycles v v 3v L 3 λ L L Some notation: n & v # n$!, n % L ",, 3, 4, For a string ixed at oth ends... The zero amplitude points are called nodes; the maximum amplitude points are the antinodes st harmonic or undamental nd st overtone 3 3 3rd nd overtone 4 4th 3rd overtone 4

14 Longitudinal Standing Waves Consider a tue with oth ends opened I we produce a sound o requency at one end, the air molecules at that end are ree to virate and they virate with The amplitude o the wave is the amplitude o the virational motion (SHM) o the air molecule changes in air density Similar to the transverse wave on a string, a standing wave occurs i the length o the tue is a hal-multiple o the wavelength o the wave

15 For the irst harmonic (undamental), only hal o a cycle is contained in the tue Following the same reasoning as or the transverse standing wave, all o the harmonic requencies are n & v # n$!, n % L " L Identical to transverse wave, except numer o nodes is dierent v,, 3,... Open-open tue # nodes n # nodes n string Open-open tue

16 An example is a lute. It is a tue which is open at oth ends. a v L v L a, < a x x L L a mouthpiece We can also have a tue which is closed at one end and opened at the other (open-closed) At the closed end, the air molecules can not virate the closed end must e a ``node The open end must e an anti-node

17 The ``distance etween a node and the next adjacent anti-node is /4 o a wavelength. Thereore the undamental requency o the openclosed tue is The next harmonic does not occur or / o a wavelength, ut 3/4 o a wavelength. The next is at 5/4 o a wavelength every odd /4 wavelength n v since L λ / 4 or λ 4L 4L & v # n$!, n % 4L ",3,5,... Note that the even harmonics are missing. Also, # nodes n Open-closed

18 Complex (Real) Sound Waves Most sounds that we hear are not pure tones (single requency like the undamental o a standing wave) But are superpositions o many requencies with various amplitudes For example, when a note (tone, requency) is played on a musical instrument, we actually hear all o the harmonics (,, 3, ), ut usually the amplitudes are decreased or the higher harmonics This is what gives each instrument it s unique sound

19 For example, the sound o a piano is dominated y the st harmonic while or the violin, the amplitudes o the st, nd, and 5 th harmonic are nearly equal gives it a rich sound Violin wave orm n & v # n$!, n % L " Summary String ixed at oth ends and the open-open tue,, 3,... n Open-closed tue & v # n$!, n % 4L ",3,5,...

20 Example Prolem A tue with a cap on one end, ut open at the other end, produces a standing wave whose undamental requency is 30.8 Hz. The speed o sound is 343 m/s. (a) I the cap is removed, what is the new undamental requency? () How long is the tue? Solution: Given: oc 30.8 Hz, n, v343 m/s oc n & n$ % v!" # 4L oo n & n$ % v L #! "

21 (a) We don t need to know v or L, since they are the same in oth cases. Solve each equation or v/l and set equal v L oo 4 oc, oc v L oo (30.8 Hz) 4 oc 6.6 Hz () Can solve or L rom either open-open or openclosed tues oc ' v $ % " & 4L # v 343 m/s L m oc 4 4(30.8 Hz) v 343 m/s L m oo (6.6 Hz) oo

22 Example prolem Two violinists, one directly ehind the other, play or a listener directly in ront o them. Both violinists sound concert A (440 Hz). (a) What is the smallest separation etween the violinists that will produce destructive intererence or the listener? () Does the smallest separation increase or decrease i the violinists produce a note with higher requency? (c) Show this or 540 Hz. Example prolem A pair o in-phase speakers are placed side y side, 0.85 m apart. You stand directly in rom o one speaker,. m rom the other speaker. What is the lowest requency that will produce constructive intererence at your location?