Lecture 7 Notes - Recap of last lecture The wave equation...

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Recap of last lecture The wave equation for a periodic wave is: 2 2 2 2 2 t y v 1 x y = This relates the curvature of the wave to the transverse acceleration. The velocity of a wave in string is given by Where F is the tension and μ is its linear mass density (mass per unit length) μ F v = Speed of mechanical = wave m equilibriu to return the opposing inertia m equilibriu to system the returning force restoring
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Problem 15.18 Without ignoring the weight of the rope find the velocity and wavelength of this transverse wave at the a) bottom, b) middle and c) top of the rope. One end of a nylon rope which weighs 2 kg is tied to a stationary support at the top of a vertical mine shaft that is 80.0 m deep. The rope is stretched taut by a box of mineral that weighs 20kg. The geologist at The bottom of the shaft jerks the cable sideways 2 times every seconds to signal to the colleague at the top.
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Problem 15.18 a) At the bottom the tension in the rope is just due to the 20 kg box. F = 20x9.8 = 196 N Mass of rope per unit length μ = 2/80 =0.025 kg/m Therefore v = sqrt(F/ μ29= 88.5 m/s Wavelength λ = v/f = 88.5/2 = 44.3m b) At the middle the tension includes half the weight of the rope, therefore F = 21x9.8 = 205.8 N Therefore v = 90.7 m/s and wavelength λ = 45.8 m c) At the top F = 22kg Therefore v = 92.9 m/s and λ = 46.5m
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Energy transferred by waves The slope at any point is = -F y /F t) FkAsin(kx x y F t) (x, F y ϖ - = - = Instantaneous power = F y v y = Fk ϖ A 2 sin 2 (kx- ϖ t) Using k = ϖ /v and μ F v = ωt) (kx sin A ω μF 2 2 2 - = Max Power 2 2 A ω μF = Average power = ½ Max power
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Wave intensity Go beyond the wave on a string and visualize, say … a sound wave spreading from a speaker. That wave has intensity dropping as 1/ r 2 Intensity = Power per unit area W/m 2 I = P/4 π r 2 I 1 /I 2 = r 2 2 /r 1 2
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An ice fisher is cutting a hole in the ice sheet with a saw on a frozen lake. A fish is swimming in the deep
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