2. 11.1.1 Describe the nature of standing (stationary) waves.Be sure to consider energy transfer, amplitude and phase. 11.1.2 Explain the formation of one-dimensional standing waves. Know what nodes and antinodes are, and how they form. 11.1.3 Discuss the modes of vibration of strings and air in open and in closed pipes.The lowest-frequency mode is known either as the fundamental or as the first harmonic. The term overtone will not be used. 11.1.4 Compare standing waves and traveling waves. 11.1.5 Solve problems involving standing waves. Topic 11: Wave phenomena 11.1 Standing (stationary) waves FYI  Topic 11.1 is an extension of Topic 4.5. Review it!

3. Describe the nature of standing (stationary) waves.  The principle of superposition yields a surprising resultant for two identical waves traveling in opposite directions.  Snapshots of the blue and the green waves and their red resultant follow:  Because the resultant red wave appears to not be traveling it is called a standing wave. Topic 11: Wave phenomena 11.1 Standing (stationary) waves

4. Explain the formation of standing waves. Know what nodes and antinodes are, and how they form.  The standing wave has two important properties.  It does not travel to the left or the right as the blue and the green wave do.  Its “lobes” grow and shrink and reverse, but do not go to the left or the right.  Any points where the standing wave has no displacement is called a node (N).  The lobes that grow and shrink and reverse are called antinodes (A). Topic 11: Wave phenomena 11.1 Standing (stationary) waves N N NN N N N AAA A A A

5. Discuss the modes of vibration of strings and air in open and in closed pipes.  You may be wondering how a situation could ever develop in which two identical waves come from opposite directions. Well, wonder no more.  When you pluck a stringed instrument, waves travel to the ends of the string and reflect at each end, and return to interfere under precisely the conditions needed for a standing wave.  Note that there are two nodes and one antinode.  Why must there be a node at each end of the string? Topic 11: Wave phenomena 11.1 Standing (stationary) waves L N N A Because it is fixed at each end.

6. Discuss the modes of vibration of strings and air in open and in closed pipes.  Observe that precisely half a wavelength fits along the length of the string.  Thus we see that = 2L.  Since v = f we see that f = v/(2L) for a string.  This is the lowest frequency you can possibly get from this string configuration, so we call it the fundamental frequency f 1.  The fundamental frequency of any system is called the first harmonic. Topic 11: Wave phenomena 11.1 Standing (stationary) waves L N N A fundamental frequency f 1 = v/(2L) 1 st harmonic

7. Discuss the modes of vibration of strings and air in open and in closed pipes.  The next higher frequency has another node and another antinode.  We now see that = L.  Since v = f we see that f = v/L.  This is the second lowest frequency you can possibly get and since we called the fundamental frequency f 1, we’ll name this one f 2.  This frequency is also called the second harmonic. Topic 11: Wave phenomena 11.1 Standing (stationary) waves L N N A f 2 = v/L A N 2 nd harmonic

8. PRACTICE: Complete the table below with both sketch and formula.  Remember that there are always nodes on each end of a string.  Add a new well-spaced node each time.  Decide the relationship between and L.  We see that = (2/3)L.  Since v = f we see that f = v/(2/3)L = 3v/2L. Solve problems involving standing waves. Topic 11: Wave phenomena 11.1 Standing (stationary) waves f 2 = v/L f 1 = v/2L f 3 = 3v/2L

9. Discuss the modes of vibration of strings and air in open and in closed pipes.  We can also set up standing waves in pipes.  In the case of pipes, longitudinal waves are created (instead of translational waves), and these waves are reflected from the ends of the pipe.  Consider a closed pipe of length L which gets its wave energy from a mouthpiece on the left side.  Why must the mouthpiece end be an antinode?  Why must the closed end be a node? Topic 11: Wave phenomena 11.1 Standing (stationary) waves (1/4) 1 = L(3/4) 2 = L (5/4) 2 = L f 1 = v/4L f 2 = 3v/4L f 3 = 5v/4L Air can’t move. Source.

10. Discuss the modes of vibration of strings and air in open and in closed pipes.  In an open-ended pipe you have an antinode at the open end because the medium can vibrate there (and, of course, at the mouthpiece). Topic 11: Wave phenomena 11.1 Standing (stationary) waves (1/2) 1 = L 2 = L (3/2) 2 = L f 1 = v/2L f 2 = 2v/2L f 3 = 3v/2L FYI  The IBO requires you to be able to make sketches of string and pipe harmonics (both open and closed) and find wavelengths and frequencies.

11. Compare standing waves and traveling waves.  Because a standing wave consists of two traveling waves carrying energy in opposite directions, the net energy flow through the wave is zero. Topic 11: Wave phenomena 11.1 Standing (stationary) waves Standing waveTraveling wave EnergyNot transferred but the wave has it in each antinode Transferred from source to receiver at the wave speed AmplitudeMaximum amplitude different for all points in medium between nodes Maximum amplitude same for all points in medium FrequencyAll vibrations are SHM and same frequency WavelengthSame as component wavesDistance between crests PhaseSame for each point in a lobe, but adjacent lobes are phase shifted by 180º Different for each point along a single wavelength Wave patternDoes not moveMoves

12. PRACTICE: A tube is filled with water and a vibrating tuning fork is held above the open end. As the water runs out of the tap at the bottom sound is loudest when the water level is a distance x from the top. The next loudest sound comes when the water level is at a distance y from the top. Which expression for is correct, if v is the speed of sound in air? A. = x B. = 2x C. = y-x D. = 2(y-x)  v = f and since v and f are constant, so is.  The first possible standing wave is sketched.  The sketch shows that = 4x, not a choice. Solve problems involving standing waves. Topic 11: Wave phenomena 11.1 Standing (stationary) waves

13. PRACTICE: A tube is filled with water and a vibrating tuning fork is held above the open end. As the water runs out of the tap at the bottom sound is loudest when the water level is a distance x from the top. The next loudest sound comes when the water level is at a distance y from the top. Which expression for is correct, if v is the speed of sound in air? A. = x B. = 2x C. = y-x D. = 2(y-x)  The second possible standing wave is sketched.  Notice that y – x is half a wavelength.  Thus the answer is = 2(y - x). Topic 11: Wave phenomena 11.1 Standing (stationary) waves y-x Solve problems involving standing waves.

14. Topic 11: Wave phenomena 11.1 Standing (stationary) waves PRACTICE: This drum head set to vibrating at different resonant frequencies has black sand on it, which reveals 2D standing waves.  Does the sand reveal nodes, or does it reveal antinodes?  Why does the edge have to be a node? Nodes, because there is no displacement to throw the sand off. The drumhead cannot vibrate at the edge. Solve problems involving standing waves.

15. Topic 11: Wave phenomena 11.1 Standing (stationary) waves Solve problems involving standing waves.  Alternate lobes have a 180º phase difference.  See Slides 3 and 10.

16. Topic 11: Wave phenomena 11.1 Standing (stationary) waves Solve problems involving standing waves.  Make a sketch. Then use v = f. antinode antinode L / 2 = L v = f = 2L f = v/ f = v/(2L)

17. Topic 11: Wave phenomena 11.1 Standing (stationary) waves Solve problems involving standing waves.  Reflection provides for two coherent waves traveling in opposite directions.  Superposition is just the adding of the two waves to produce the single stationary wave.

18. Topic 11: Wave phenomena 11.1 Standing (stationary) waves Solve problems involving standing waves.  A snapshot of Slide 3 shows the points between successive nodes.  For every point between the two nodes f is the same.  But the amplitudes are all different.  Therefore the energies are also different.

19. Topic 11: Wave phenomena 11.1 Standing (stationary) waves Solve problems involving standing waves.  Energy transfer via a vibrating medium without interruption.  The medium itself does not travel with the wave disturbance.  Speed at which the wave disturbance propagates.  Speed at which the wave front travels.  Speed at which the energy is transferred.

20. Topic 11: Wave phenomena 11.1 Standing (stationary) waves Solve problems involving standing waves.  Frequency is number of vibrations per unit time.  Distance between successive crests (or troughs).  Distance traveled by the wave in one oscillation of the source. FYI: IB frowns on you using particular units as in “Frequency is number of vibrations per second.” FYI: There will be lost points, people!

21. Topic 11: Wave phenomena 11.1 Standing (stationary) waves Solve problems involving standing waves.  Each of the waves traveling in opposite directions carry energy at same rate in both directions. Thus there is NO energy transfer.  The amplitude is always changing and reversing.

22. Topic 11: Wave phenomena 11.1 Standing (stationary) waves Solve problems involving standing waves. L P / 4 = L = 4L L Q / 2 = L = 2L v = f f = v/ f P = v/(4L) f Q = v/(2L) v = 4Lf P f Q = 4Lf P /(2L) f Q = 2f P

23. Topic 11: Wave phenomena 11.1 Standing (stationary) waves Solve problems involving standing waves.  The tuning fork is the driving oscillator (and is at the top).  The top is thus an antinode.  The bottom “wall” of water allows NO oscillation.  The bottom is thus a node.

24. Topic 11: Wave phenomena 11.1 Standing (stationary) waves Solve problems involving standing waves.  Sound is longitudinal in nature.  Small displacement at P, big at Q.

25. Topic 11: Wave phenomena 11.1 Standing (stationary) waves Solve problems involving standing waves.  If the lobe at T is going down, so is the node at U.

26. Topic 11: Wave phenomena 11.1 Standing (stationary) waves Solve problems involving standing waves.  Pattern 1 is a 1/2 wavelength.  Pattern 2 is a 3/2 wavelength.  Thus f 2 = 3f 1 so that f 1 / f 2 = 1/3.