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Wave and Vector Functions in the Microwave Background

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Added on  2021-02-21

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X1 + ] = 3.80* 100 cos [100 t + ] For finding maximum displacement derivative to zero: = 0 100 t + = t1 = 0.0027 seconds Similarly, X2 = 4.62 sin [100 t - ()] = 4.62 * 100 cos [100 t - ] = 0 100 t - = t = 0.009 seconds 3 .) When X

Wave and Vector Functions in the Microwave Background

   Added on 2021-02-21

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Wave and Vector Functions
Wave and Vector Functions in the Microwave Background_1
TABLE OF CONTENTSSCENARIO 1...................................................................................................................................1SCENARIO 2...................................................................................................................................5
Wave and Vector Functions in the Microwave Background_2
SCENARIO 1X1 ¿3.80sin¿ + ¿9 ]X2 = 4.62 sin [100π t - (2π5)]1.)X1 ¿3.80sin¿ + ¿9 ]Amplitude: 3.80Phase: 2π9 leading Frequency: 50 Hz Periodic time: 0.02 seconds X2 = 4.62 sin [100π t - (2π5)]Amplitude: 4.62 Phase: -2π5laggingFrequency: 50 Hz Periodic time: 0.02 seconds 2.)The maximum displacement can be found by differentiating x1 and x2 with respect to y. X1 ¿3.80sin¿ + ¿9 ]dX1dt = 3.80* 100π cos [100π t + ¿9] For finding maximum displacement equate derivative to zero: dX1dt = 0100π t + ¿9= π2t1 = 0.0027 seconds 1
Wave and Vector Functions in the Microwave Background_3
Similarly, X2 = 4.62 sin [100π t - (2π5)]dX2dt = 4.62 * 100π cos [100π t - 2π5] dX1dt = 0 100π t - 2π5= π2t = 0.009 seconds 3 .) When X1 = -2 mm time can be calculated as follows: -2 = 3.80 sin [100π t + 2π9] sin [100π t + 2π9] = -0.0052[100π t + 2π9] = -0.553 t = 0.0039 seconds CHECK X1 = 3.80 sin [100π t + 2π9] Putting t=0.0039X1 = -2 Thus proved RHS = LHS When X2 = -2-2 = 4.62 Sin [100π t - 2π5] t = 0.002 seconds CHECK 2
Wave and Vector Functions in the Microwave Background_4

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