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Title: HNC HND in Mechanical Electrical Electronic Engineering Unit 1 - Analytical Methods for Engineers
Description: HNC HND in Mechanical Electrical Electronic Engineering Unit 1 - Analytical Methods for Engineers
Description: HNC HND in Mechanical Electrical Electronic Engineering Unit 1 - Analytical Methods for Engineers
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Tom Obrien
HNC Level 4
Unit Number and title: Unit 1
Analytical Methods for Engineers
Assignment Title: Trigonometric
Methods
Assignment Number: 2
Date: November 2016
1
Tom Obrien
HNC Level 4
Contents Page:
Page 1:
Page 2:
-
Title Page
-
Contents Page
Page 3 to 4:
-
Task 2
...
1
Question 2
Page 5:
Page 6 to 7:
-
Task 2
...
2
Question 1a
Question 1b
Question 1c
Question 1d
Question 1e
Question 1f
-
Task 2
...
3
Question 1c
2
Tom Obrien
HNC Level 4
3
Task 2
...
5
-4
...
5
t
t
V1
V2
V1+V2
V1-V2
180
π
0
4
...
5
-4
...
5
0
-2
...
5
V2 = 9 sin ωt −
30
π
6
2
...
5
2
...
33
-4
...
83
0
...
33
4
...
83
-0
...
79
12
...
79
-5
-7
...
79
2
...
33
9
13
...
67
300
5π
3
-4
...
33
4
...
5
7
...
29
-5
...
5
-7
...
29
5
...
5
-4
...
5
Graph inserted on following page as FIG 1
...
Tom Obrien
FIG: 1
HNC Level 4
4
Tom Obrien
HNC Level 4
Question 2:
The graph shows the equation drawn on a graph
...
56
...
56) + 3 sin(1
...
16
-2
...
36
30
π
6
3
...
5
3
...
57
-2
...
19
60
π
3
3
...
98
2
...
94
-2
...
98
90
π
2
3
...
44
2
...
28
-2
...
71
120
2π
3
3
...
86
1
...
38
-2
...
37
150
5π
6
2
...
22
0
...
04
-2
...
94
360
2π
-1
...
73
-4
...
This shows the plotted lines for V1, V2 and
V1+V2
...
2
Question 1a:
V = 200 sin(50πt − 0
...
2
t
t
V
180
π
83
...
5
210
7π
6
60
π
3
86
...
4
-154
...
7
270
3π
2
-190
...
99
300
5π
3
-193
...
99 from calculation
Periodic time t = 2π/ = 2π/50π = 0
...
04 = 25Hz
Phase Angle = -0
...
Question 1b:
t = 0s so V = -126
...
01s so V = 200sin (50πX0
...
683) = 487
...
683)
100
= sin(50πt − 0
...
5 = sin(50πt − 0
...
683 = sin (0
...
52
150
5π
6
168
...
5
360
2π
-36
...
52 + 0
...
65 × 10 or 0
...
683)
−58
= sin(50πt − 0
...
29 = sin(50πt − 0
...
683 = sin (−0
...
29
t=
−0
...
683
= 2
...
0025s
50π
Question 1f:
200 = 200 sin(50πt − 0
...
683)
sin
=
= 50πt − 0
...
57 + 0
...
57 + 0
...
0144
50
9
Tom Obrien
HNC Level 4
10
Task 2
Title: HNC HND in Mechanical Electrical Electronic Engineering Unit 1 - Analytical Methods for Engineers
Description: HNC HND in Mechanical Electrical Electronic Engineering Unit 1 - Analytical Methods for Engineers
Description: HNC HND in Mechanical Electrical Electronic Engineering Unit 1 - Analytical Methods for Engineers