Phase Angle of the Gravitational Waves V 3

Question 1
Part A
V 1= 30sin ( 4000 πt ) V
V 2= 140sin (4000 πt + π
2 ) V
Phasor angle for V 1 = 0
Phasor angle for V 2 = π
2 radians
Convering to degrees: π
2 * 180
π = 90 degrees
The resultant combination, V 3 of V 1 and V 2 calculated from:
Horizontal component
30cos 0 + 140cos 90 = 30
Vertical component
30sin 0 + 140 sin 90 = 140
Magnitude of V 3:

V 3=√302 +1402 = 143.18
To get the phase angle of V 3:
∅ = tan−1 vertical component
horizontal component
∅ = tan−1 140
30 = 77.9
Converting to radians we multiply the angle in degrees by π
180
77.9 * π
180 = 1.36 rad
Hence the expression of V 3 is:
V 3= 143.18sin ( 4000 πt+1.36 ) V
Part B
Angular velocity is expressed as:
ω=2 πf
Making f (frequency) the subject:

f = ω
2 π
Substituting the values:
f = 4000 π
2 π = 2000
f = 2000 Hz

Question 2
Parameters of the sample:
Thickness: 6mm
Width: 15mm
Original length: 65mm
New length: 72mm
Force: 27kN
Part A
Stress at point of failure, σ :
σ = Force
Cross−sectional area
σ = 27 k
6 mm∗15 mm = 27 k
9∗10−5 = 300 * 106 Pa
σ = 300.00 MPa
Part B

Fracture strain, :
ε = change∈length
originallength
ε = ∆ l
lo
ε = 72−65
65 = 7
65 = 0.1076
ε=0.11
Part C
Modulus of elasticity of the sample, E:
E= σ
ε
E=300 M Pa
0.1076
E=2.7857∗109
E= 2.79 GPa