Physics Assignment: Detailed Analysis of Wave and Vector Functions

This physics assignment solution covers two scenarios involving wave and vector functions. Scenario 1 analyzes two wave functions, calculating their amplitudes, phases, frequencies, periodic times, and maximum displacements. It also determines times for specific displacements, simplifies the functions using compound angle formulas, and discusses their superposition and interference. Scenario 2 focuses on vector functions, determining slope lengths, angles between vectors, and calculating vector components. The assignment provides detailed step-by-step solutions to illustrate the concepts and problem-solving techniques for a thorough understanding of wave and vector analysis.

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
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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

x1 +x2 = 4.29 Sin 100πt + 6.81 Cos 100πt
The equation represents the form A sin Ф +B cos Ф
A = 4.29 B= 6.81
Tan α = b/a
so α = 57.67
R = √ (A ² + B²)
R = √ (4.29 ² + 6.81²)
R= 8.04
A sin Ф +B cos Ф = R Sin (Ф+ α)
8.04 Sin (100π t + 57.67)
6.)
7.)
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8.)
From the graph it can be concluded that despite having different phase differences both
x1 and x2 are characterised by the constructive interference. It means the amplitudes of wave
forms x1 and x2 reinforce each others and thus the amplitude of resultant wave is higher than the
individual waves.
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SCENARIO 2
1.)
A = (0, -40, 0)
B = (40, 0, -20)
C = (a, b, 0)
Slope length AB = √ [(40)² + (40)² + (-20)²]
Slope length AB = 60
2.)
Lenght AB = B-A = (40, 40, -20)
BC is drilled in the direction of 3i +4j +k
Angle between AB and BC is given by the following formula:
Cos Ф = (AB.BC) / (magnitude AB*magnitude BC)
Cos Ф = (40*3 + 40*4 -20*1) / [√[(20)² + (20)² + (10)²] √ [(3)² + (4)² + (1)²]]
Cos Ф = 0.3333
Ф = 70.53 degree
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3.)
vector C = vector B + rho (3i +4j+ k)
(a,b,0) = (40, 0, -20) + rho (3, 4 ,1)
rho = 20
A = 100
B= 80
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