Wave and Vector Functions

Added on - 21 Feb 2021

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Wave and Vector Functions
TABLE OF CONTENTSSCENARIO 1...................................................................................................................................1SCENARIO 2...................................................................................................................................5
SCENARIO 1X1¿3.80sin¿+¿9]X2= 4.62 sin [100π t - (2π5)]1.)X1¿3.80sin¿+¿9]Amplitude:3.80Phase:2π9leadingFrequency:50 HzPeriodic time:0.02 secondsX2= 4.62 sin [100π t - (2π5)]Amplitude:4.62Phase:-2π5laggingFrequency:50 HzPeriodic time:0.02 seconds2.)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 seconds1
Similarly,X2= 4.62 sin [100π t - (2π5)]dX2dt= 4.62 * 100π cos [100π t -2π5]dX1dt= 0100π t -2π5=π2t = 0.009 seconds3 .)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] = -0.553t = 0.0039 secondsCHECKX1 = 3.80 sin [100π t +2π9]Putting t=0.0039X1 = -2Thus proved RHS = LHSWhen X2= -2-2 = 4.62 Sin[100π t -2π5]t = 0.002 secondsCHECK2
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