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Opus in profectus

# Periodic Waves

## Practice

### practice problem 1

The graph below is a record of sea level heights recorded at Hanimaadhoo, Maldives during the tsunami of 26 December 2004. The data were filtered to eliminate the normal tidal fluctuations, so what you are seeing is the increase in sea level due to the tsunami. (Data source: University of Hawaii Sea Level Center) Natural phenomena are normally noisy (in the statistical sense) but from 10:50 to 12:05 local time the changes in sea level at Hanimaadhoo were most nearly periodic. During this time interval determine the tsunami's mean…
1. amplitude
2. period
The speed of a tsunami varies with depth. In the open ocean they normally move as fast as a commercial jet airplane (about 250 m/s or 900 kph) but slow down to the speed of a car on a neighborhood street when they reach the shallow waters of the shore (about 15 m/s or 55 kph). Given these speeds, determine the mean wavelength of the segment of the tsunami that arrived in Hanimaadhoo between 10:50 and 12:05 when they were…
1. in deep water
2. near the shore
One final question.
1. How would the amplitude of a tsunami near shore compare to the amplitude of the same wave in the open ocean? Explain your reasoning.

#### solution

1. The amplitude is open to some interpretation. The crests seem to average around +90 cm between 10:50 and 12:05, so 90 cm would be an acceptable answer. The highest crest appears to be +100 cm, so 100 cm would also be an acceptable answer. The troughs seem to average −60 cm. If we split the difference between +90 cm and −60 cm we'd get 75 cm for an amplitude. The deepest trough looks like its −70 cm. If we split the difference between +100 cm and −70 cm we'd get 87.5 cm for an amplitude. Any one of these answers is reasonable as far as I'm concerned. Recall that the problem said that real physical data is statistically noisy. This makes it impossible to point to an exact answer. Exact isn't a scientific concept, anyway. It's a word that really belongs to mathematics.

2. There are 3.5 wave cycles from 10:50 to 12:05 (75 minutes or 4500 seconds). Divide time by number to get period — the time for one cycle. We'll do it once with each unit just because neither unit seems better than the other right now.

 T = t = 75 min = 21.4 min n 3.5 T = t = 4500 s = 1290 s n 3.5
3. Start with the wave speed equation (wave speed in terms of period) and solve it for wavelength.

 λ = vT ⇐ v = λ T

Put numbers in. Get answer out. I think I'll use the SI values (meters and seconds, not kilometers or hours or minutes).

 λ = vTλ = (250 m/s)(1290 s)λ = 321,000 m

Tsunamis have long periods and long wavelengths. This would be hard to perceive in the open ocean. The changes would be too gradual to see or feel.

4. Repeat the previous problem with one number different — the slower wavespeed near shore.

 λ = vTλ = (15 m/s)(1290 s)λ = 19,300 m

Near shore wavelength are much shorter, but they're still pretty long. Note that the period hasn't changed. That value is set by the source of the wave — an earthquake in this case.

5. The amplitude and wavelength of a wave in the ocean together describe something like a volume. This volume can't change. Squash the tsunami horizontally and it grows vertically. The amplitude increases as the wave moves toward the shore.

### practice problem 2

Each of the six strings on an acoustic guitar is 80 cm long and generates a wave that's twice the length of the string when picked or strummed. Determine the wave speed of each string given the following tuning.
1. 082.41 Hz
2. 110.00 Hz
3. 146.83 Hz
4. 196.00 Hz
5. 246.94 Hz
6. 329.63 Hz

#### solution

Use the equation v = fλ over and over again. Double the string length to get the wavelength. This problem is probably best solved using the SI units of meters and seconds and their various combinations.

 f λ = v E. (082.41 Hz) (1.60 m) = 132 m/s A. (110.00 Hz) (1.60 m) = 176 m/s D. (146.83 Hz) (1.60 m) = 235 m/s G. (196.00 Hz) (1.60 m) = 314 m/s B. (246.94 Hz) (1.60 m) = 395 m/s E. (329.63 Hz) (1.60 m) = 527 m/s

The highest frequency string has the fastest wave speed. Frequency is directly proportional to wave speed when wavelength is constant as it is in this case.

### practice problem 3

A viola string is 36 cm long and plays a certain note when the finger is not resting on the string. How far from the bridge should a finger be placed so as to produce a note that is…
1. an octave higher (double the original frequency)?
2. two octaves higher (quadruple the original frequency)?
3. a fifth higher (32 the original frequency)?
4. a fourth higher (43 the original frequency)? #### solution

This is a question about the equation…

v = fλ

but we won't be using the equation in the classic plug and chug fashion. This is really a question about a proportions.

Placing a finger on the fingerboard of a viola or any other similar stringed instrument changes the length of the segment that is allowed to vibrate. The wavelength of a sound is directly proportional to the size of the source of the sound.

λ ∝ L

Long segments of viola string of emit long wavelength sounds. Short segments emit short wavelengths.

Placing a finger on the fingerboard does not change the speed of the waves through the string, however. The speed of a wave in a string is determined by other factors. The equation…

v = fλ

says that if v is constant, then the product fλ is also constant or equivalently that wavelength and frequency are inversely proportional…

λ ∝ 1/f

Long wavelengths go with low frequencies and short wavelengths go with high frequencies.

Combining these two proportions into one gives us the solution. Take the frequency ratios given in the problem and invert them to get the string length ratios…

L ∝ 1/f

1. If an octave produces a note with double the original frequency (21), then the string length must be halved (12).

L = 12(36 cm) = 18 cm

2. If two octaves quadruples the frequency (41), then the new string length must be one-quarter the old one (14).

L = 14(36 cm) = 9 cm

3. If a fifth multiplies the frequency by three halves (32), then a fifth multiplies the string length by two thirds (23).

L = 23(36 cm) = 24 cm

4. If a fourth means 43 the original frequency, then it also means 34 the original string length.

L = 34(36 cm) = 27 cm

String lengths in this problem correspond to the distance from the bridge to the finger. The bridge is a thin, flat piece of wood that raises the strings above the soundboard. It is between the two ends of the strings, but is closer to the attachment to the soundboard (the tailpiece) than the attachment to the neck (the nut).

### practice problem 4

tsunamis.txt
The greatest recorded earthquake (magnitude of 9.5) occurred on 22 May 1960 in Chile. The second largest earthquake (magnitude 9.2) occurred on 27 March 1964 during the Christian Holiday of Good Friday, which is why it is also known as the Good Friday Earthquake. A large earthquake (magnitude 8.8) occurred in Chile on 27 February 2010 that grabbed my attention and motivated me to write this problem. All three earthquakes generated tsunamis for which I was able to find useful data.

Tsunamis are sometimes called "tidal waves" but this name is misleading. Tsunamis, which are seismic in origin, and tides, which are caused by the gravitational pull of the Moon and Sun, are completely unrelated. The word tsunami is derived from the Japanese phrase "harbor wave" (津波) since tsunamis are most intense in harbors where the underlying terrain focuses their energy. The term "tidal wave" is somewhat appropriate since the waves generated by earthquakes result in long period waves that sometimes look like the changes in water depth caused by the tides.

The accompanying tab-delimited text file provides the following data for the tsunamis associated with the three earthquakes described above. (Source: National Centers for Environmental Information)

1. Location of town, harbor, or facility
2. Region (state, province, or country)
3. Transit time in minutes after the earthquake began
4. Distance from the epicenter in kilometers measured along a great circle (the shortest path on the surface of a sphere)

Use this information to determine the speed of a tsunami in…

1. km/min
2. km/hr
3. m/s
4. and mph if you live in the United States

#### solution

Plot distance on the y axis and time on the x axis so that the slope of the best fit line will be the speed. It appears that the average speed of a tsunami in the pacific is…

1.  10.55 km/min + 11.85 km/min + 11.27 km/min = 11.22 km/min 3
2.  11.22 km × 60 min = 673.4 km/h 1 min 1 h
3.  11.22 km × 1 min × 1,000 m = 187.1 m/s 1 min 60 s 1 km
4.  11.22 km × 60 min × 1 mile = 418.5 mph 1 min 1 h 1.609 km