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This figure shows four one-dimensional standing waves. The waves are shown in a tan color and are composed of curves to represent standing waves that can be generated using string. The first image at the top of the figure shows a single long wave with no nodes, or points where the string appears to cross between the endpoints at the left and right sides of the figure. The second diagram just below shows a single node at the center of the wave, which divides the wave into two identical halves to the left and right. The third diagram shows two nodes, dividing the image into three identical parts to the left, center, and right. Similarly, the last image at the bottom of the figure shows three nodes, dividing the image into four identical parts.
A vibrating string shows some one-dimensional standing waves. Since the two end points of the string are held fixed, only waves having an integer number of half-wavelengths can form. The points on the string between the end points that are not moving are called the nodes.

An example of two-dimensional standing waves is shown in [link] , which shows the vibrational patterns on a flat surface. Although the vibrational amplitudes cannot be seen like they could in the vibrating string, the nodes have been made visible by sprinkling the drum surface with a powder that collects on the areas of the surface that have minimal displacement. For one-dimensional standing waves, the nodes were points on the line, but for two-dimensional standing waves, the nodes are lines on the surface (for three-dimensional standing waves, the nodes are two-dimensional surfaces within the three-dimensional volume). Because of the circular symmetry of the drum surface, its boundary conditions (the drum surface being tightly constrained to the circumference of the drum) result in two types of nodes: radial nodes that sweep out all angles at constant radii and, thus, are seen as circles about the center, and angular nodes that sweep out all radii at constant angles and, thus, are seen as lines passing through the center. The upper left image in [link] shows two radial nodes, while the image in the lower right shows the vibrational pattern associated with three radial nodes and two angular nodes.

This figure includes four images. In each image, a brown circular platform has been sprinkled with a tan powder, yellow wires connect to a cylindrical base beneath the platform. To the right of the platform is a white box with a blue front which is labeled, “5.289 k H z Function Generator.” The image in the top left shows three distinct rings formed from the tan powder evenly spaced from the center of the platform, with the first ring very close to the center of the platform. The box reads, “2.434 k H z.” The image in the top right is similar except that the rings are closer together and the central ring has a significantly greater radius than in the first diagram. In this photo, the box reads, “3.986 k H z.” The image at the lower left is similar to the image in the upper left except that more of the powder is present, and 8 evenly-spaced radii are formed from the tan powder on the platform, making a web-like image. In this photo, the box reads, “5.289 k H z.” In the lower right of the figure, the image is similar to what is shown in the upper right except that four evenly spaced radii are shown composed of the tan powder on the platform. In this photo, the box reads, “5.670 K H z.”
Two-dimensional standing waves can be visualized on a vibrating surface. The surface has been sprinkled with a powder that collects near the nodal lines. There are two types of nodes visible: radial nodes (circles) and angular nodes (radii).

Blackbody radiation and the ultraviolet catastrophe

The last few decades of the nineteenth century witnessed intense research activity in commercializing newly discovered electric lighting. This required obtaining a better understanding of the distributions of light emitted from various sources being considered. Artificial lighting is usually designed to mimic natural sunlight within the limitations of the underlying technology. Such lighting consists of a range of broadly distributed frequencies that form a continuous spectrum    . [link] shows the wavelength distribution for sunlight. The most intense radiation is in the visible region, with the intensity dropping off rapidly for shorter wavelength ultraviolet (UV) light, and more slowly for longer wavelength infrared (IR) light.

A graph is shown with a horizontal axis labeled, “Wavelength ( n m ),” and a vertical axis labeled, “Spectral irradiance ( W divided by m superscript 2 divided by n m ).” The horizontal axis begins at 250 and extends to 4000 with markings provided every 250 n m. Similarly, the vertical axis begins at 0.00 and extends to 2.00 with markings every 0.25 units. Two vertical dashed lines are drawn. The first appears at about 400 nanometers and the second at nearly 700 nanometers. To the left of the first of these lines, the label, “U V,” appears at the top of the graph. Between these lines, the label, “Visible,” appears at the top of the graph. To the right of the second of these lines, the label, “Infrared,” appears at the top of the graph. A grey curve begins on the vertical axis at about 0.10. This curve increases steeply to a maximum value between the two vertical line segments of approximately 1.75 at about 625 nanometers. This curve decreases rapidly at first, then tapers off to reach a value of about 0 at the far right end of the graph. A golden colored curve traces along the same path as the grey curve, but shows a significant degree of variation in the region of the peak of the graph. In this general region, the gold curve is jagged and somewhat erratic. This curve reaches a maximum over 2.00 at around 475 nanometers. A key provided in the open space of the graph shows that the gold graph represents sunlight at the top of the atmosphere, and the grey curve represents the 5250 degrees C Blackbody spectrum.
The spectral distribution (light intensity vs. wavelength) of sunlight reaches the Earth's atmosphere as UV light, visible light, and IR light. The unabsorbed sunlight at the top of the atmosphere has a distribution that approximately matches the theoretical distribution of a blackbody at 5250 °C, represented by the blue curve. (credit: modification of work by American Society for Testing and Materials (ASTM) Terrestrial Reference Spectra for Photovoltaic Performance Evaluation)

Questions & Answers

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
what is the solution
bill
how did you arrive at this answer?
bill
-24m+3+3mÁ^2
Susan
i really want to learn
Amira
I only got 42 the rest i don't know how to solve it. Please i need help from anyone to help me improve my solving mathematics please
Amira
Hw did u arrive to this answer.
Aphelele
hi
Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
complete the table of valuesfor each given equatio then graph. 1.x+2y=3
Jovelyn Reply
x=3-2y
Salma
y=x+3/2
Salma
Hi
Enock
given that (7x-5):(2+4x)=8:7find the value of x
Nandala
3x-12y=18
Kelvin
please why isn't that the 0is in ten thousand place
Grace Reply
please why is it that the 0is in the place of ten thousand
Grace
Send the example to me here and let me see
Stephen
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
Marry Reply
how far
Abubakar
cool u
Enock
state in which quadrant or on which axis each of the following angles given measure. in standard position would lie 89°
Abegail Reply
hello
BenJay
hi
Method
I am eliacin, I need your help in maths
Rood
how can I help
Sir
hmm can we speak here?
Amoon
however, may I ask you some questions about Algarba?
Amoon
hi
Enock
what the last part of the problem mean?
Roger
The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
mahnoor Reply
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
Munster
difference between rational and irrational numbers
Arundhati Reply
When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
Jakoiya Reply
how to reduced echelon form
Solomon Reply
Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
Zack Reply
d=r×t the equation would be 8/r+24/r+4=3 worked out
Sheirtina
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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