3/2/15 Oregon State University PH 212, Class #251 Other Effects of Interference We have considered interference between waves of the same frequency. But of course, that’s not always the situation…. Beats The phenomenon of sound beats is a simple, common example. This is an alternating rise and fall in the observed amplitude (which we hear as loudness) of the combined (superimposed) sound of two sources that are close, but not identical, in frequency. The lows and highs of the two functions occur together only periodically—just a few cycles per second—so that we can hear distinctly each combined high (constructive interference) or low (destructive interference).

3/2/15 Oregon State University PH 212, Class #252 As it turns out: f pitch = ( f 1 + f 2 )/2 f beats =  f 1 – f 2  Musicians use this phenomenon regularly to tune their instruments: When trying to match a given pitch, they try to eliminate the beats; so long as they can still hear beats, they’re not yet in tune. Example: The lead oboe sounds an “A” note of exactly 440 Hz. When the second oboe then joins in with her “A,” it produces beats that pulse with a frequency of 1 Hz. If the second oboe could correct this by lengthening her instrument slightly (lowering her pitch), what frequency (pitch) was your ear hearing while you heard the 1 Hz beats? (see After class 25.)

Standing Waves Standing waves correspond to another simple kind of wave interference. A standing wave is the combination of two or more identical continuous traveling waves moving (in one direction or another) along the same portion of the same medium. The two waves are identical except for the direction in which they are traveling. They have the same amplitude, wavelength, frequency and speed. 3/2/15 3Oregon State University PH 212, Class #25

3/2/15 Oregon State University PH 212, Class #254 Standing Waves Musical instruments use standing waves in solid materials or in air to generate musical tones of very reliable frequency (pitch). How? A standing wave occurs when many waves overlap due to reflections of the same disturbance in the same available space. The interference of all these reflected waves tend to cancel each other out except for waves that “fit the available space.” The reflections that don’t cancel each other out are those that are completing a cycle or half-cycle when they encounter the end of the available space. They reflect back in a “harmonized” way, thus reinforcing each other (i.e. forming constructive interference), increasing the amplitude of their sum. We hear this as resonance.

3/2/15 Oregon State University PH 212, Class #255 How does the end of the available space govern the standing wave? It fixes the value of the wave function. For example, a guitar string is fixed at both ends. Its function (a transverse displacement) is thus forced to be zero at each end. We call such zero displacement a node in the standing wave. The guitar string is a good example of a transverse standing wave. What about a longitudinal standing wave—such as sound?

3/2/15 Oregon State University PH 212, Class #256 In a longitudinal pressure wave (sound), the air particles are being displaced back and forth (along the axis of the wave travel)—and they are undergoing pressure variations, too. Both of these—displacement and pressure—vary sinusoidally, but it turns out that they “peak” and “dip” at different times. So when we analyze how the “end of the space” affects a longitudinal wave, we have to be careful about what wave we’re talking about—either the pressure (what we hear and measure as sound) or the displacement of the particles back and forth. Here’s how it works.…

3/2/15 Oregon State University PH 212, Class #257 A pipe open at both ends fixes the air pressure at each end to be atmospheric pressure. This means that it can have no standing pressure variation at those ends. A pipe closed at one end fixes the air displacement to be zero at the closed end (and the standing pressure variation must still be zero at the open end). Now let’s summarize the requirements for resonance—for strings and for two kinds of pipes.…

3/2/15 Oregon State University PH 212, Class #258 With two fixed ends, a string will have standing waves of displacement at certain frequencies, according to this formula: L = n( /2) or f n = nv/2L where n = 1, 2, 3... Each such resonant frequency is called a harmonic. The first harmonic (the fundamental frequency) is when n = 1; the second harmonic is when n = 2; etc. Nodes (points of zero displacement) will, of course, appear at the fixed ends. They will also appear at spaced intervals of /2 from those ends. Antinodes (points of maximum displacement) appear halfway between the nodes.

3/2/15 Oregon State University PH 212, Class #259 With both ends open, a cylindrical pipe will have standing longitudinal waves of pressure and displacement at certain frequencies, according to the same formula: L = n( /2) or f n = nv/2L where n = 1, 2, 3... The nodes and antinodes? It depends which disturbance you’re looking at; displacement and pressure peak at different points. Displacement antinodes (points of maximum displacement) appear at the open ends and at spaced intervals of /2 from those ends. Displacement nodes (points of zero displacement) appear halfway between those antinodes. Pressure nodes (points of no pressure variation) appear at the open ends and at spaced intervals of /2 from those ends. Pressure antinodes (points of maximum pressure variation) appear halfway between those nodes.

3/2/15 Oregon State University PH 212, Class #2510 With just one end open, a cylindrical pipe will have standing waves of pressure and displacement at certain frequencies, with a different constraint: L = n( /4) or f n = nv/4L where n = 1, 3, 5... A pressure node (point of minimum pressure variation) will appear at the open end and at spaced intervals of /2 from that end. Pressure antinodes (points of maximum pressure variation) appear at the closed end and at spaced intervals of /2 from that end. A displacement antinode (point of maximum displacement) appears at the open end and at spaced intervals of /2 from that end. Displacement nodes (points of minimum displacement) appear at the closed end and at spaced intervals of /2 from that end.

A string is clamped at both ends and plucked so it vibrates in a standing mode between two extreme positions a and b. Let upward motion correspond to positive velocities. When the string is in position c, the instantaneous velocity of points along the string… 1. is zero everywhere. 2. is positive everywhere. 3. is negative everywhere. 4. depends on location. 3/2/15 11Oregon State University PH 212, Class #25

3/2/15 Oregon State University PH 212, Class #2512 Example: A standing wave of the 4th harmonic is set up on a guitar string. The frequency is 440 Hz. (a) How many loops are in this standing wave? (b) How many nodes does this wave have? (c) Is the string’s midpoint a node or antinode? (d) Find the frequency of the string’s 1st harmonic. Example: The “G” string on a guitar has a fundamental frequency of 196 Hz and a length of 0.62 m. When you press this string against the proper fret, you can produce a “C” note (262 Hz). What is the new, shortened length of the vibrating string? (In other words, what is the distance from that fret to the bridge of the instrument?)

3/2/15 Oregon State University PH 212, Class #2513 Example: When you blow across the top of a partially-filled bottle of water, you are setting up a standing wave in a chamber that is open only at one end. (a) Is there an air pressure node or antinode at the top of the bottle? How about at the water level? (b) Is there an air displacement node or antinode at the top of the bottle? How about at the water? (c) If the standing wave is vibrating at its fundamental frequency, how far is it from the top to the water level? (d) If you pour out some water, then blow again, will the fundamental frequency then be higher or lower?

3/2/15 Oregon State University PH 212, Class #2514 Example: The auditory canal in the human ear has an average length of 2.9 cm, and it functions approximately as a tube with one open end. For a speed of sound of 343 m/s, what is the fundamental frequency of the auditory canal? Example: For each of the following systems, the fundamental frequency is 400 Hz. Find the three lowest frequencies at which other standing waves will occur: (a) a string fixed at both ends; (b) a cylindrical pipe with both ends open; (c) a cylindrical pipe with only one end open.