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Opus in profectus

Standing Waves

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Practice

practice problem 1

Tuning forks
note pitch (Hz)
C4 256
D4 288
E4 320
F4 341
G4 384
A4 427
B4 480
C5 512
D5 576
E5 640
F5 682
G5 768
A 1.00 m vertical tube is partially submerged in water. The height of the tube above the water can be adjusted to any value from 0 m (the tube is completely submerged) to 1 m (the bottom of the tube is just touching the top of the water). A vibrating tuning fork is held just over the open top end of the tube. Pick one of the notes from the table on the right and answer the following questions.
  1. Determine the wavelength of the sound emitted from the tuning fork in air at room temperature (vsound = 343 m/s).
  2. How long is ¼, ½, ¾, 1¼, 1½, 1¾ of the wavelength you calculated in part a.?
  3. At what heights will resonance occur? (Just highlight the answers from part b. that satisfy this condition.)

To see a similar experiment using a shorter tube and, more importantly, to hear what resonance sounds like watch the video below.

youtu.be/YpCA9CVsVos

solution

The entire set of solutions is summarized in the table below the numbered list. The problem said to pick one of the notes.

  1. Wavelengths are calculated using the formula below. Replace v with 343 m/s (the speed of sound in air at room temperature). Compute with whatever value of f you chose.

    λ =  v  ⇐  v = fλ
    f

  2. Multiply the results of part a. by ¼, ½, ¾, 1¼, 1½, 1¾. A boring, but simple task.

  3. A tube submerged in water is a half open, half closed system. Resonance will only occur when an odd number of quarter wavelengths fit inside the tube. Since the tube is only one meter long, only heights that are less than one meter will fit at all. The answers that satisfy both of these conditions are highlighted in the table below.

Tube lengths that will resonate for various tuning forks

note		 pitch
(Hz)		 ¼ λ
(m)		 ½ λ
(m)		 ¾ λ
(m)		 1 λ
(m)		 1¼ λ
(m)		 1½ λ
(m)		 1¾ λ
(m)
C4 256 0.335 0.670 1.005 1.340 1.675 2.010 2.345
D4 288 0.298 0.595 0.893 1.191 1.489 1.786 2.084
E4 320 0.268 0.536 0.804 1.072 1.340 1.608 1.876
F4 341 0.251 0.503 0.754 1.006 1.257 1.509 1.760
G4 384 0.223 0.447 0.670 0.893 1.117 1.340 1.563
A4 427 0.201 0.402 0.602 0.803 1.004 1.205 1.406
B4 480 0.179 0.357 0.536 0.715 0.893 1.072 1.251
C5 512 0.167 0.335 0.502 0.670 0.837 1.005 1.172
D5 576 0.149 0.298 0.447 0.595 0.744 0.893 1.042
E5 640 0.134 0.268 0.402 0.536 0.670 0.804 0.938
F5 682 0.126 0.251 0.377 0.503 0.629 0.754 0.880
G5 768 0.112 0.223 0.335 0.447 0.558 0.670 0.782

practice problem 2

In contemporary Western music, the most common type of clarinet is the soprano clarinet. The most common type of flute is the concert flute. Some characteristics of the two instruments are summarised in the table below. The two instruments have almost the same length, but noticeably different frequency ranges. The purpose of this exercise is to explain why that is the case.
Woodwind instruments compared Source: Vienna Symphonic Library
soprano
clarinet		 concert
flute
key B♭ C
highest
concert pitch		 G6
1567.98 Hz		 D7
2349.32 Hz
lowest
concert pitch		 D3
146.83 Hz		 B3
246.94 Hz
length 66 cm 68 cm
inner diameter
(bore)		 12.7 mm 19 mm
  1. How does the
  2. How is this problem supposed to work? I forgot.

solution

Answer it.

practice problem 3

Schumann Resonances
The ionosphere is a layer in the Earth's upper atmosphere where a large portion of the atoms and molecules have been ionized by exposure to the ultraviolet radiation of the Sun. With so many charged particles free to roam around, the ionosphere is a reasonably good conductor of electricity. The surface of the Earth is also a reasonably good conductor. This should be somewhat obvious since 70% of the Earth's surface is covered in saltwater, which will short out electrical equipment as everyone knows, and the remaining 30% is exposed rock or soil, the stuff that electric circuits are grounded to. The layer of atmosphere in between these two conductors is ordinary, non ionized air, which is transparent to radio waves. For extremely low frequency (ELF) radiation, the gap between the Earth and its ionosphere acts as a spherical wave guide — a kind of racetrack for radio waves. Lightning and other natural phenomena generate ELF waves at all sorts of different frequencies. Those frequencies that are just right will travel around the Earth, meet themselves in phase, and form standing waves. The set of frequencies that will do this are known as the Schumann resonances in honor of Winfried Schumann (1888–1974), the scientist who predicted their existence in 1952.
  1. Complete the following table…
    Schumann resonances
    harmonic λ (km) fpredicted (Hz) fobserved (Hz) ∆f/fobserved (%)
    first     7.8  
    second     14  
    third     20  
    fourth     26  
    fifth     33  
    sixth     39  
    seventh     45  
  2. Do the predicted Schumann resonances agree with the observed values to a reasonable degree? Account for any significant discrepancies.

solution

Start with a picture. Here's what the 5th harmonic looks like as an example…

Cartoon representation of standing waves wrapped around a sphere

  1. For the wavelength, divide the circumference of the Earth by the number of the harmonic.

    λ =  C  =  2πr
    n n

    Use the wave speed equation to get the frequency.

    f =  c  ⇐  c = fλ
    λ

    Compute the relative uncertainty with the given equation.

    relative uncertainty  =  ∆f
    fobserved
    relative uncertainty  =  |fpredicted − fobserved|
    fobserved

    Repeat the procedure over and over again. I suggest letting a computer do the work for you. Here's a suggested method using the 5th harmonic as an example…

    (2*pi*radius of earth)/5 = 8014.95684 km
    speed of light/(8014.95684 km) = 37.4041263 Hz
    (37.4041263Hz-33 Hz)/(33 Hz) = 0.133458373

    Compile your results. (Please report only a reasonable number of significant digits.) You should get something like this …

    Schumann resonances
    harmonic λ (km) fpredicted (Hz) fobserved (Hz) ∆f/fobserved (%)
    first 40,100 7.48 7.8 4.1
    second 20,000 15.0 14 6.9
    third 13,400 22.4 20 12
    fourth 10,000 29.9 26 15
    fifth 08,010 37.4 33 13
    sixth 06,680 44.9 39 15
    seventh 05,720 52.4 45 16

  2. The disagreement between theory and observation is significant (more than 10% in most cases). The flaw lies in our use of a one-dimensional wave model. The earth is a sphere, an object whose surface is a two-dimensional object. A proper solution to this problem is beyond the scope of this book (and the skills of the author).

practice problem 4

A seiche (pronounced "saysh") is a type of standing wave that occurs in bays, lakes, and seas. Seiches typically form when strong winds push a surge of water toward one side of a large body of water. This raises the water level on one side and lowers it on another. When the wind stops, the surge sloshes back until it hits the other side where it is reflected. The seiche may then continue to oscillate back and forth for several hours or days, depending on the size of the body of water. Seiches are sometimes confused with tides, but those are caused by the gravitational interaction of the Earth with the Moon (and to a lesser extent, the Earth with the Sun). The animation below shows a cross-sectional representation of a seiche oscillating at its fundamental frequency in a cartoon body of water.

Cartoon animation of a seiche

Lake Erie on the border between the US and Canada is a convenient place to observe seiches. The long axis of the lake is nearly aligned with the direction of the prevailing winds, which makes seiches relatively common, and the lake is surrounded by monitoring stations, which makes accessing water level data relatively easy. The map below shows the location of seven of these monitoring stations on the American shore of Lake Erie.

Map of Lake Erie with water level monitoring stations labeled

On Monday 31 December 2018 (New Year's Eve), a noticeable seiche formed in Lake Erie. Water level displacement data from the seven stations shown on the map above were plotted on the graph below. Although the seiche lasted for six days, the graph below only shows data for two whole days and a few hours before and after. The data after Thursday morning are a little too messy for an introductory physics textbook.

Water level displacement (mm) vs. time (days, hours)

Answer the following questions about the seiche described in the text, animation, map, and graph above. Start by answering some questions that are about standing waves.

  1. Which two monitoring stations are closest to the one node of the seiche? Explain your reasoning.
  2. Which two monitoring stations are closest to each of the two antinodes of the seiche? Explain your reasoning.
  3. When it comes to seiches, is Lake Erie more like a system that is fixed at both ends, free at both ends, or fixed at one end and free at the other. Explain your reasoning.
  4. How long is one wavelength of a seiche oscillating in its fundamental mode in Lake Erie compared to the length of the lake itself?

Determine some temporal characteristics of the seiche. State all times to the nearest hour (or day and hour).

  1. At what time did the seiche start? How did you identify this time?
  2. How many half-cycles of the seiche are shown in the graph of water level displacement?
  3. At what time did the last half-cycle you counted in the previous question end? How did you identify this time?
  4. What is the period of the seiche?
  5. If the seiche dissipated after six days, how many complete cycles were there?

Compare your results to those predicted by theory. The fundamental period of a seiche can be estimated using Merian's equation, named after the Swiss mathematician Johann Rudolf Merian (1797–1871). If you have been doing this problem right, the period computed from this equation and the value computed from the data shown in the graph should be equal to two significant digits. Merian's equation works quite well for Lake Erie.

T =  2ℓ
√gh

Where

T =  period of the fundamental mode of oscillation
ℓ =  effective length of the body of water
h =  average depth of the body of water
g =  acceleration due to gravity
  1. What does Professor Google say the values of ℓ, h, and g are for Lake Erie?
  2. What is the fundamental period of a seiche in Lake Erie according to Merian's equation and the values you found online?

Some final questions on waves in general and standing waves in particular.

  1. What is the wave speed of this seiche in Lake Erie?
  2. The water level graph shows evidence of higher harmonics in the lake. How does the graph show this?
  3. What are the periods of the 2nd, 3rd, and 4th harmonics?

solution

Solutions!

  1. Cleveland and Fairport are nearest the node. Their water level displacement graphs have the smallest amplitudes, which is the essential characteristic of a node. They are also nearest the middle of Lake Erie, which agrees with what the cartoon animation shows.

  2. Toledo and Buffalo are nearest the antinodes. Their water level displacement graphs have the largest amplitudes. They are also nearest the extreme ends of the lake and are out of phase with one another. (When the water level is below normal in Toledo it is above normal in Buffalo and vice versa.) This agrees with what the cartoon animation shows.

  3. Lake Erie is more like a system that is free at both ends. The antinodes are located at the extreme ends of the lake, and antinodes always form at free ends.

  4. The wavelength of the seiche is twice the length of Lake Erie. When a system is free at both ends it contains a whole number (n) of half wavelengths. When a system is oscillating in its fundamental mode, as this one is, n = 1.

    λ1 = 2L ⇐ 
    L = nλn
    2

  5. 11 PM on Saturday is when this seiche started. This is the first moment on the graph where the water level displacements are closest to zero as a group.

  6. There are 7 half-cycles in the time frame shown on the graph. Just count the number of "bumps" between "pinched" regions on the graph.

  7. 7:00 AM on Thursday is the end of the last obvious half-cycle of this seiche. This is last moment on the graph where the water level displacements are closest to zero as a group.

  8. A period is the time for one cycle, or on average…

    T = t
    n

    The time was 2 days plus 1 hour on Saturday night and 7 hours on Thursday morning

    t = (1 h) + (2 × 24 h) + (7 h)
    t = 56 h

    7 half-cycles is 3½ full cycles. Thus…

    T = 56 h = 16 h
    3.5

  9. Rearrange the equation we just used.

    n = T
    t

    6 days is 144 hours. Substitute and solve…

    n = 144 h = 9 cycles
    16 h

  10. "The Internet" says that Lake Erie has the following characteristics…

    ℓ = 388 km
    h = 19 m
    g = 9.80 m/s2

  11. Use Merian's equation with the values above in units that work well together.

    T = 2ℓ
    √gh
    T = 2(388,000 m)
    √[(9.80 m/s2)(19 m)]
    T = 56,869 s = 15.8 h 
     

    I'd call that 16 hours — same as the value we determined using the graph.

  12. Speed is the rate of change of distance with time, or in the case of a wave, the ratio of wavelength to period.

    v = ∆s = λ
    ∆tT

    Numbers in, answer out. Remember that wavelength is twice the length of the lake. (An equivalent way to think about it is that the wave has to go all the way down the lake and back to complete a cycle, so double it.) I recommend using kilometers and hours as the units. Keep it simple.

    v = 2(388 km) = 48.5 km/h
    16 h

    That's the speed of a rather sedate drive. Nothing like the jet airplane speeds of a tsunami. Seiches may sometimes act like small tsunamis, but they aren't. They aren't tidal waves. They aren't tides. They are their own thing.

  13. The curves on the graph are not perfect sine curves. The superposition of other sine curves with different periods on to the fundamental gives the curves some of their "wobbly" appearance. The rest of it comes from other factors. Lake Erie is noisy (in the statistical sense).

  14. When a system is free at both ends the frequency of the harmonics are whole number multiples of the fundamental.

    f1 = 1f1
    f2 = 2f1
    f3 = 3f1
    f4 = 4f1
    ⋮
    fn = nf1
    ⋮

    But the question asked about period, not frequency. No problem. Period is just the inverse of frequency. Replace every f with a T and every n with a 1n.

    T1 = 11T1
    T2 = 12T1
    T3 = 13T1
    T4 = 14T1
    ⋮
    Tn = 1nT1
    ⋮

    Using the numbers from this problem, we get…

    T1 = 16 h
    T2 = 12(16 h) = 8.0 h
    T3 = 13(16 h) = 5.3 h
    T4 = 14(16 h) = 4.0 h