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Unformatted text preview: Lecture 20: Sound 1 REVIEW: Superposition and Standing Waves When two waves traveling in the same direction, with the same amplitude A , the same angular frequency ω , and the same wavelength λ = 2 π/k , but separated by a phase difference φ meet at the same place, they will add algebraically (superposition) y 1 ( x, t ) = A sin ( kx ωt ) and y 2 ( x, t ) = A sin ( kx ωt φ ) y 3 ≡ y 1 + y 2 = A sin ( kx ωt ) + A sin ( kx ωt φ ) y 3 ( x, t ) = µ 2 A cos φ 2 ¶ sin ( kx ωt φ 2 ) The phase difference φ depends on the location x relative to the source of the two waves. For certain values of x it is possible that φ = 0 in which case the resultant wave has twice the amplitude of the original wave (constructive interference). In other cases, the value of φ could be π/ 2 or an odd half integer multiple of π/ 2 in which case the resultant wave will have zero amplitude (destructive interference). For any value of φ in this example, the resultant wave has the same frequency and the same wavelength as the original two waves. Another example of superposition is to have a string fixed at both ends such that there are waves traveling in opposite directions along the string from reflections at either end. In this case one would have two waves of the form y 1 ( x, t ) = A cos ( kx ωt ) and y 2 ( x, t ) = A cos ( kx + ωt ) y 3 = y 1 + y 2 = A (cos ( kx ωt ) cos ( kx + ωt )) = µ 2 A sin kx ¶ sin ωt This is called a standing wave which looks like just a sine function of time but with an amplitude according to the position x . In fact, at certain positions called nodes , the amplitude will always be zero (no motion). The node positions are given by kx = nπ where n is any integer kx = nπ = ⇒ x = nπ k = nπ 2 π/λ = nλ 2 Since the end of a fixed string at x = L must also be a node, this sets a condition on the wavelengths and frequencies of standing waves therein f n = v λ n = v 2 L/n = q F/μ 2 L/n = n 2 L v u u t F μ Lecture 20: Sound 2 Standing Wave Problems Two waves in a long string are given by y 1 ( x, t ) = 0 . 015 cos ( x 2 40 t ) and y 2 ( x, t ) = . 015 cos ( x 2 + 40 t ) where x , y 1 , and y 2 are in meters, and t is in seconds. Determine the posi tions of the nodes of the resulting standing wave, and what is the maximum displacement of the standing wave at the position x = 0 . 4 meters ?...
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 Spring '08
 Maguire
 Physics, Frequency, Wavelength

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