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# Physics Learning Object 6

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# Physics Learning Object 6

This presentation helps explain the concepts surrounding standing waves and gives interactive examples

This presentation helps explain the concepts surrounding standing waves and gives interactive examples

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### Physics Learning Object 6

1. 1. Standing Waves Rachel Sunderland
2. 2.  A standing wave is two waves that are moving in opposite directions with the same amplitude, wavelength, and frequency.  They result from the superstition of waves when a wave reflects from a surface and interferes with the wave travelling in the original direction What is a Standing Wave?
3. 3.  D1(x,t) = A sin(kx + ωt) <- occurs at increasing x  D2(x,t) = A sin(kx - ωt) <- occurs at decreasing x  Together, the equation of a standing wave becomes:  D(x,t) = D1(x,t) + D2(x,t)  = A sin(kx – ωt) + A sin(kx + ωt)  = A sin(kx – wt) + sin(kx – wt)  = 2A sin(kx) cos (ωt) Standing Wave Equations
4. 4.  In this photo, the wave on the top is moving to the right and the wave to the bottom is moving to the left  They combine together in order to form one wave with an amplitude of 2A at the anti-node and 0 at the node Standing Waves Visual Example
5. 5.  A node is the point where the amplitude is zero and therefore, there is no displacement from equilibrium  The antinode is the point where the maximum displacement occurs from equilibrium and thus, the amplitude is also at its maximum point  Since a standing wave is two waves moving in opposite directions, the maximum amplitude of a standing wave is twice the amplitude of one wave component of the travelling wave What is a Node?
6. 6. Picture of Nodes/Anti-nodes  A node is where the displacement is zero and an anti- node is where the displacement is 2A as seen in the photograph below
7. 7.  Nodes occur when the displacement is zero  Therefore, sin (2πx/𝜆) = 0  Since is zero at every pi interval so we can substitute it in for 0  2πx/𝜆 = mπ where m = positive or negative (1,2,3…)  In order to simplify, we bring x to one side, and cancel out the π’s in order to get the equation:  X = m𝜆/2 Calculating the Position of a Node
8. 8.  At what position does the wave function: 0.75m sin(17.2x) have its first node?  We use the equation: X = m𝜆/2  1st we need to solve for the wavelength  We know that k = 2π/𝜆  Therefore, 17.2 = 2π/𝜆 and 𝜆 = 0.365  We can then plug it into the equation x = 1(0.365/2) = 0.1825 Example
9. 9.  Anti-Nodes occur when the displacement is (+/-) 2A  Therefore, sin (2πx/𝜆) = (+/-) 1  Since (+/-) occurs at every pi/2 interval so we can substitute it in for (+/-) 1  2πx/𝜆 = (m+ 1/2)π where m = positive or negative (1,2,3…) which also equals (m+1/2) *𝜆/2  In order to simplify, we bring x to one side in order to get the equation:  X = (+/-) 𝜆/4, 3𝜆/4, 5𝜆/4 … Calculating the Position of an Anti-Node
10. 10.  At what position does the wave function: 0.50m sin(5.9x) have its first anti-node?  We use the equation: X = (+/-) 𝜆/4, 3𝜆/4  1st we need to solve for the wavelength  We know that k = 2π/𝜆  Therefore, 5.9 = 2π/𝜆 and 𝜆 = 1.07  We can then plug it into the equation x = 1(1.07/2) = 0.535 Example
11. 11.  What is the amplitude of a wave when its at the location x = 0.35m, and its wavelength is 2.13m  1st we need to determine what the value of K is  We can use the equation K = 2π/𝜆  Therefore, k = 2π/2.13 = 2.95 rad/m  We can then use this information to complete the formula Integrated Example
12. 12.  A(x) = 2A sin (2π/𝜆)  Plugging in values gives us  0.35m = 2(x) sin (2.95)  0.35m = 2x (0.0514)  6.81 = 2x  x = 3.41  Therefore, at the location x = 0.35m and with a wavelength of 2.13m, the amplitude of the wave is 3.41m Integrated Example Part 2
13. 13.  "Standing Waves." Mr Magarey's Official Year 11 Physics Class Wikispace -. NHS, n.d. Web. 09 Mar. 2015.  "S-cool, the Revision Website." Standing Waves. N.p., n.d. Web. 09 Mar. 2015. Sources Cited