School

McGill University
Department

Mechanical Engineering

Course Code

MECH 262

Professor

Farhang Daneshmand

Study Guide

Final

5.1

CHAPTER 5

5.1 The ramp function appears as the following:

T = 0.02 sec

o = 2f

= 2(1/T)

= 2(1/0.02)

= 100 rad/sec

f(t) = 100t

0 001 t .

f(t) = 100t 2

001 002. . t

From Eq. 5.5,

T

on tdtntf

T

b0sin)(

2

Thus,

01.0

0

02.0

01.0

1)100sin()2100()100sin(100

02.02dtttdtttb

100 31831 10 31831 10

06366

3 3

. .

.

01.0

0

02.0

01.0

2)1002sin()2100()1002sin(100

02.02dtttdtttb

100 15916 10 15916 10

03183

3 3

. .

.

ao is the average over the period, Eq. 5.3

dttf

T

aT

o

0

1

Thus,

01.0

0

02.0

01.0

0)0cos()2100()0cos(100

02.01dttdtta

50 510 510

0

3 3

From Eq. 5.4,

5.2

T

on tdtntf

T

a0cos)(

2

01.0

0

02.0

01.0

1)100cos()2100()100cos(100

02.02dtttdttta

100 2023 10 2026 10

0

3 3

. .

0

)1002cos()2100()1002cos(100

02.0201.0

0

02.0

01.0

2

dtttdttta

Without actually evaluating the values for ao, a1, and a2, we could have found

that they were each equal to zero since f(t) is an odd function.

5.2 The ramp function appears as the following:

T = 0.04 sec

o = 2f

= 2(1/T)

= 2(1/0.04)

= 50 rad/sec

f(t) = 50t

02.00 t

f(t) = 50t 2

04.002.0 t

From Eq. 5.5,

T

on tdtntf

T

b0sin)(

2

Thus,

02.0

0

04.0

02.0

1)50sin()250()50sin(50

04.0

2dtttdtttb

6366.0

103662.6103662.650 33

02.0

0

04.0

02.0

2)502sin()250()502sin(50

04.0

2dtttdtttb

5.3

3183.0

101831.3101831.350 33

ao is the average over the period,Eq. 5.3

dttf

T

aT

o

0

1

Thus,

02.0

0

04.0

02.0

0)0cos()250()0cos(50

04.0

1dttdtta

0

01.001.025

From Eq. 5.4,

T

on tdtntf

T

a0cos)(

2

02.0

0

04.0

02.0

1)50cos()250()50cos(50

04.0

2dtttdttta

0

10053.410053.450 33

0

)502cos()250()502cos(50

04.0

202.0

0

04.0

02.0

2

dtttdttta

Without actually evaluating the values for ao, a1, and a2, we could have found

that they were each equal to zero since f(t) is an odd function.

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