Wave and Vector Functions

Wave and Vector Functions

TABLE OF CONTENTS
SCENARIO 1...................................................................................................................................1
SCENARIO 2...................................................................................................................................5

SCENARIO 1
X1 = 3.80 sin [100π t + (2π /9)]
X2 = 4.62 sin [100π t - (2π /5)]
1.)
X1 = 3.80 sin [100π t + (2π /9)]
Amplitude: 3.80
Phase: 2π /9 leading
Frequency: 50 Hz
Periodic time: 0.02 seconds
X2 = 4.62 sin [100π t - (2π /5)]
Amplitude: 4.62
Phase: - (2π /5) lagging
Frequency: 50 Hz
Periodic time: 0.02 seconds
2.)
The maximum displacement can be found by differentiating x1 and x2 with respect to y.
x1 = 3.80 sin [100π t + (2π /9)]
dx1 /dt = 3.80* 100π cos [100π t + (2π /9)]
For finding maximum displacement equate derivative to zero:
dx1 /dt = 0
100π t + (2π /9) = π/2
t1 = 0.0027 seconds
Similarly,
X2 = 4.62 sin [100π t - (2π /5)]
dx2 /dt = 4.62 * 100π cos [100π t - (2π /5)]
dx1 /dt = 0
100π t - (2π /5) = π/2
t = 0.009 seconds
1

3 .)
When x1 = -2 mm time can be calculated as follows:
-2 = 3.80 sin [100π t + (2π /9)]
sin [100π t + (2π /9)] = -0.0052
[100π t + (2π /9)] = -0.553
t = 0.0039 seconds
When x2 = -2
-2 = 4.62 Sin [100π t - (2π /5)]
t = 0.002 seconds
4.) Compound angle formula
x1 = 3.80 sin [100π t + (2π /9)]
Sin (A+B) = Sin A Cos B + Cos A Sin B
x1 =3.80 {[Sin 100πt * Cos (2π /9)] + [Cos 100πt * Sin (2π /9)]
x1 = 2.91 Sin 100πt + 2.43 Cos 100πt
Thus, from the equation of x1:
A = 2.91 B= 2.43
Similarly, for x2:
X2 = 4.62 sin [100π t - (2π /5)]
x2 =4.62 {[Sin 100πt * Cos (-2π /5)] - [Cos 100πt * Sin (-2π /5)]
x2= 1.38 Sin 100πt + 4.38 Cos 100πt
Thus, from the equation of x1:
A = 1.38 B= 4.38
5.)
From the above simplified expression x1 and x2 can be written as :
x1 = 2.91 Sin 100πt + 2.43 Cos 100πt
x2= 1.38 Sin 100πt + 4.38 Cos 100πt
2