Chapter 6, Problem 1P

Question

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Answer 1

Question

Chapter 6, Problem 1P

Expert Solution & Answer

To determine

To plot:

The x -component and y -component as a function of time and determine the amplitude, frequency, period, phase angle and time shift.

Answer to Problem 1P

The graph for x -componentas a function of timeis plotted is as shown in Figure 1.

The graph for y -component as a function of time is plotted is as shown in Figure 2.

The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

Explanation of Solution

Given:

The tip of a one-link robot is initially located at θ=0 attime t=0 s. The given figure is shown below,

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 1

Time taken for the robot to move form θ=0 to θ=2π rad is 1 s.

A one-link robot is length of 5 in.

Concept used:

Write the expression for the linear frequency.

f=1T ....... (1)

Here, f is the linear frequency and T is the time period.

Write the expression for the angular frequency.

ω=2πT ....... (2)

Here, ω is the angular frequency.

Write the expression for the time period.

T=2πω ....... (3)

Write the expression for x -component at any time,

x(t)=lcos(θ) ....... (4)

Here, x(t) is the expression for x -component at any time, l is the length of one-link robot, θ is the angle.

Write the expression for x -component at any time.

y(t)=lsin(θ) ....... (5)

Here, y(t) is the expression for y -component at any time.

Write the expression for the time shift.

t0=ϕω ....... (6)

Here, t0 is the expression for the time shift.

Calculation:

The one-link robot completes one revolution in 1 s.

Substitute 1 s for T in equation (2).

ω=2π1 s=2π rad/s

Substitute 2π rad/s for ω in equation (3).

T=2π2π s=1 s

Therefore, the period of the tip of a one-link robot is 1 s.

Substitute 1 s for T in equation (1).

f=11 s=1 Hz

Therefore, the frequency is 1 Hz.

Since the one-link robot initially start form θ=0 and the phase angle is 0 rad.

Substitute ωt+ϕ for θ in equation (4).

x(t)=lcos(ωt+ϕ) ....... (7)

Here, t is any given time and ϕ is the phase angle.

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (7).

x(t)=(5)cos(( 2π)t+0)=5cos2πt

The plot for the x -component as a function of time is shown in Figure 1.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 2

Substitute ωt+ϕ for θ in equation (5).

y(t)=lsin(ωt+ϕ) ....... (8)

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (8).

x(t)=(5)sin(( 2π)t+0)=5sin(2πt+0)=5sin(2πt)

The plot for the y -component as a function of time is shown in Figure 2.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 3

Figure 2

Substitute 0 rad for ϕ and 2π rad/s for ω in equation (6).

t0 =0 rad2π rad/s=0 s

Therefore, the time shift of the one-link robot is 0 s.

The amplitude of sine and cosine function is 5.

Conclusion:

Thus, the graph for x -component is plotted in Figure 1 and y -component as a function of time is plotted in Figure 2. The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

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Answer 2

Question

Chapter 6, Problem 1P

Expert Solution & Answer

To determine

To plot:

The x -component and y -component as a function of time and determine the amplitude, frequency, period, phase angle and time shift.

Answer to Problem 1P

The graph for x -componentas a function of timeis plotted is as shown in Figure 1.

The graph for y -component as a function of time is plotted is as shown in Figure 2.

The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

Explanation of Solution

Given:

The tip of a one-link robot is initially located at θ=0 attime t=0 s. The given figure is shown below,

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 1

Time taken for the robot to move form θ=0 to θ=2π rad is 1 s.

A one-link robot is length of 5 in.

Concept used:

Write the expression for the linear frequency.

f=1T ....... (1)

Here, f is the linear frequency and T is the time period.

Write the expression for the angular frequency.

ω=2πT ....... (2)

Here, ω is the angular frequency.

Write the expression for the time period.

T=2πω ....... (3)

Write the expression for x -component at any time,

x(t)=lcos(θ) ....... (4)

Here, x(t) is the expression for x -component at any time, l is the length of one-link robot, θ is the angle.

Write the expression for x -component at any time.

y(t)=lsin(θ) ....... (5)

Here, y(t) is the expression for y -component at any time.

Write the expression for the time shift.

t0=ϕω ....... (6)

Here, t0 is the expression for the time shift.

Calculation:

The one-link robot completes one revolution in 1 s.

Substitute 1 s for T in equation (2).

ω=2π1 s=2π rad/s

Substitute 2π rad/s for ω in equation (3).

T=2π2π s=1 s

Therefore, the period of the tip of a one-link robot is 1 s.

Substitute 1 s for T in equation (1).

f=11 s=1 Hz

Therefore, the frequency is 1 Hz.

Since the one-link robot initially start form θ=0 and the phase angle is 0 rad.

Substitute ωt+ϕ for θ in equation (4).

x(t)=lcos(ωt+ϕ) ....... (7)

Here, t is any given time and ϕ is the phase angle.

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (7).

x(t)=(5)cos(( 2π)t+0)=5cos2πt

The plot for the x -component as a function of time is shown in Figure 1.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 2

Substitute ωt+ϕ for θ in equation (5).

y(t)=lsin(ωt+ϕ) ....... (8)

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (8).

x(t)=(5)sin(( 2π)t+0)=5sin(2πt+0)=5sin(2πt)

The plot for the y -component as a function of time is shown in Figure 2.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 3

Figure 2

Substitute 0 rad for ϕ and 2π rad/s for ω in equation (6).

t0 =0 rad2π rad/s=0 s

Therefore, the time shift of the one-link robot is 0 s.

The amplitude of sine and cosine function is 5.

Conclusion:

Thus, the graph for x -component is plotted in Figure 1 and y -component as a function of time is plotted in Figure 2. The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

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Answer 3

Question

Chapter 6, Problem 1P

Expert Solution & Answer

To determine

To plot:

The x -component and y -component as a function of time and determine the amplitude, frequency, period, phase angle and time shift.

Answer to Problem 1P

The graph for x -componentas a function of timeis plotted is as shown in Figure 1.

The graph for y -component as a function of time is plotted is as shown in Figure 2.

The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

Explanation of Solution

Given:

The tip of a one-link robot is initially located at θ=0 attime t=0 s. The given figure is shown below,

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 1

Time taken for the robot to move form θ=0 to θ=2π rad is 1 s.

A one-link robot is length of 5 in.

Concept used:

Write the expression for the linear frequency.

f=1T ....... (1)

Here, f is the linear frequency and T is the time period.

Write the expression for the angular frequency.

ω=2πT ....... (2)

Here, ω is the angular frequency.

Write the expression for the time period.

T=2πω ....... (3)

Write the expression for x -component at any time,

x(t)=lcos(θ) ....... (4)

Here, x(t) is the expression for x -component at any time, l is the length of one-link robot, θ is the angle.

Write the expression for x -component at any time.

y(t)=lsin(θ) ....... (5)

Here, y(t) is the expression for y -component at any time.

Write the expression for the time shift.

t0=ϕω ....... (6)

Here, t0 is the expression for the time shift.

Calculation:

The one-link robot completes one revolution in 1 s.

Substitute 1 s for T in equation (2).

ω=2π1 s=2π rad/s

Substitute 2π rad/s for ω in equation (3).

T=2π2π s=1 s

Therefore, the period of the tip of a one-link robot is 1 s.

Substitute 1 s for T in equation (1).

f=11 s=1 Hz

Therefore, the frequency is 1 Hz.

Since the one-link robot initially start form θ=0 and the phase angle is 0 rad.

Substitute ωt+ϕ for θ in equation (4).

x(t)=lcos(ωt+ϕ) ....... (7)

Here, t is any given time and ϕ is the phase angle.

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (7).

x(t)=(5)cos(( 2π)t+0)=5cos2πt

The plot for the x -component as a function of time is shown in Figure 1.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 2

Substitute ωt+ϕ for θ in equation (5).

y(t)=lsin(ωt+ϕ) ....... (8)

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (8).

x(t)=(5)sin(( 2π)t+0)=5sin(2πt+0)=5sin(2πt)

The plot for the y -component as a function of time is shown in Figure 2.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 3

Figure 2

Substitute 0 rad for ϕ and 2π rad/s for ω in equation (6).

t0 =0 rad2π rad/s=0 s

Therefore, the time shift of the one-link robot is 0 s.

The amplitude of sine and cosine function is 5.

Conclusion:

Thus, the graph for x -component is plotted in Figure 1 and y -component as a function of time is plotted in Figure 2. The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

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Answer 4

Question

Chapter 6, Problem 1P

Expert Solution & Answer

To determine

To plot:

The x -component and y -component as a function of time and determine the amplitude, frequency, period, phase angle and time shift.

Answer to Problem 1P

The graph for x -componentas a function of timeis plotted is as shown in Figure 1.

The graph for y -component as a function of time is plotted is as shown in Figure 2.

The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

Explanation of Solution

Given:

The tip of a one-link robot is initially located at θ=0 attime t=0 s. The given figure is shown below,

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 1

Time taken for the robot to move form θ=0 to θ=2π rad is 1 s.

A one-link robot is length of 5 in.

Concept used:

Write the expression for the linear frequency.

f=1T ....... (1)

Here, f is the linear frequency and T is the time period.

Write the expression for the angular frequency.

ω=2πT ....... (2)

Here, ω is the angular frequency.

Write the expression for the time period.

T=2πω ....... (3)

Write the expression for x -component at any time,

x(t)=lcos(θ) ....... (4)

Here, x(t) is the expression for x -component at any time, l is the length of one-link robot, θ is the angle.

Write the expression for x -component at any time.

y(t)=lsin(θ) ....... (5)

Here, y(t) is the expression for y -component at any time.

Write the expression for the time shift.

t0=ϕω ....... (6)

Here, t0 is the expression for the time shift.

Calculation:

The one-link robot completes one revolution in 1 s.

Substitute 1 s for T in equation (2).

ω=2π1 s=2π rad/s

Substitute 2π rad/s for ω in equation (3).

T=2π2π s=1 s

Therefore, the period of the tip of a one-link robot is 1 s.

Substitute 1 s for T in equation (1).

f=11 s=1 Hz

Therefore, the frequency is 1 Hz.

Since the one-link robot initially start form θ=0 and the phase angle is 0 rad.

Substitute ωt+ϕ for θ in equation (4).

x(t)=lcos(ωt+ϕ) ....... (7)

Here, t is any given time and ϕ is the phase angle.

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (7).

x(t)=(5)cos(( 2π)t+0)=5cos2πt

The plot for the x -component as a function of time is shown in Figure 1.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 2

Substitute ωt+ϕ for θ in equation (5).

y(t)=lsin(ωt+ϕ) ....... (8)

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (8).

x(t)=(5)sin(( 2π)t+0)=5sin(2πt+0)=5sin(2πt)

The plot for the y -component as a function of time is shown in Figure 2.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 3

Figure 2

Substitute 0 rad for ϕ and 2π rad/s for ω in equation (6).

t0 =0 rad2π rad/s=0 s

Therefore, the time shift of the one-link robot is 0 s.

The amplitude of sine and cosine function is 5.

Conclusion:

Thus, the graph for x -component is plotted in Figure 1 and y -component as a function of time is plotted in Figure 2. The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

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Answer 5

Expert Solution & Answer

To determine

To plot:

The x -component and y -component as a function of time and determine the amplitude, frequency, period, phase angle and time shift.

Answer to Problem 1P

The graph for x -componentas a function of timeis plotted is as shown in Figure 1.

The graph for y -component as a function of time is plotted is as shown in Figure 2.

The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

Explanation of Solution

Given:

The tip of a one-link robot is initially located at θ=0 attime t=0 s. The given figure is shown below,

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 1

Time taken for the robot to move form θ=0 to θ=2π rad is 1 s.

A one-link robot is length of 5 in.

Concept used:

Write the expression for the linear frequency.

f=1T ....... (1)

Here, f is the linear frequency and T is the time period.

Write the expression for the angular frequency.

ω=2πT ....... (2)

Here, ω is the angular frequency.

Write the expression for the time period.

T=2πω ....... (3)

Write the expression for x -component at any time,

x(t)=lcos(θ) ....... (4)

Here, x(t) is the expression for x -component at any time, l is the length of one-link robot, θ is the angle.

Write the expression for x -component at any time.

y(t)=lsin(θ) ....... (5)

Here, y(t) is the expression for y -component at any time.

Write the expression for the time shift.

t0=ϕω ....... (6)

Here, t0 is the expression for the time shift.

Calculation:

The one-link robot completes one revolution in 1 s.

Substitute 1 s for T in equation (2).

ω=2π1 s=2π rad/s

Substitute 2π rad/s for ω in equation (3).

T=2π2π s=1 s

Therefore, the period of the tip of a one-link robot is 1 s.

Substitute 1 s for T in equation (1).

f=11 s=1 Hz

Therefore, the frequency is 1 Hz.

Since the one-link robot initially start form θ=0 and the phase angle is 0 rad.

Substitute ωt+ϕ for θ in equation (4).

x(t)=lcos(ωt+ϕ) ....... (7)

Here, t is any given time and ϕ is the phase angle.

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (7).

x(t)=(5)cos(( 2π)t+0)=5cos2πt

The plot for the x -component as a function of time is shown in Figure 1.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 2

Substitute ωt+ϕ for θ in equation (5).

y(t)=lsin(ωt+ϕ) ....... (8)

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (8).

x(t)=(5)sin(( 2π)t+0)=5sin(2πt+0)=5sin(2πt)

The plot for the y -component as a function of time is shown in Figure 2.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 3

Figure 2

Substitute 0 rad for ϕ and 2π rad/s for ω in equation (6).

t0 =0 rad2π rad/s=0 s

Therefore, the time shift of the one-link robot is 0 s.

The amplitude of sine and cosine function is 5.

Conclusion:

Thus, the graph for x -component is plotted in Figure 1 and y -component as a function of time is plotted in Figure 2. The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

Answer 6

Expert Solution & Answer

To determine

To plot:

The x -component and y -component as a function of time and determine the amplitude, frequency, period, phase angle and time shift.

Answer to Problem 1P

The graph for x -componentas a function of timeis plotted is as shown in Figure 1.

The graph for y -component as a function of time is plotted is as shown in Figure 2.

The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

Explanation of Solution

Given:

The tip of a one-link robot is initially located at θ=0 attime t=0 s. The given figure is shown below,

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 1

Time taken for the robot to move form θ=0 to θ=2π rad is 1 s.

A one-link robot is length of 5 in.

Concept used:

Write the expression for the linear frequency.

f=1T ....... (1)

Here, f is the linear frequency and T is the time period.

Write the expression for the angular frequency.

ω=2πT ....... (2)

Here, ω is the angular frequency.

Write the expression for the time period.

T=2πω ....... (3)

Write the expression for x -component at any time,

x(t)=lcos(θ) ....... (4)

Here, x(t) is the expression for x -component at any time, l is the length of one-link robot, θ is the angle.

Write the expression for x -component at any time.

y(t)=lsin(θ) ....... (5)

Here, y(t) is the expression for y -component at any time.

Write the expression for the time shift.

t0=ϕω ....... (6)

Here, t0 is the expression for the time shift.

Calculation:

The one-link robot completes one revolution in 1 s.

Substitute 1 s for T in equation (2).

ω=2π1 s=2π rad/s

Substitute 2π rad/s for ω in equation (3).

T=2π2π s=1 s

Therefore, the period of the tip of a one-link robot is 1 s.

Substitute 1 s for T in equation (1).

f=11 s=1 Hz

Therefore, the frequency is 1 Hz.

Since the one-link robot initially start form θ=0 and the phase angle is 0 rad.

Substitute ωt+ϕ for θ in equation (4).

x(t)=lcos(ωt+ϕ) ....... (7)

Here, t is any given time and ϕ is the phase angle.

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (7).

x(t)=(5)cos(( 2π)t+0)=5cos2πt

The plot for the x -component as a function of time is shown in Figure 1.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 2

Substitute ωt+ϕ for θ in equation (5).

y(t)=lsin(ωt+ϕ) ....... (8)

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (8).

x(t)=(5)sin(( 2π)t+0)=5sin(2πt+0)=5sin(2πt)

The plot for the y -component as a function of time is shown in Figure 2.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 3

Figure 2

Substitute 0 rad for ϕ and 2π rad/s for ω in equation (6).

t0 =0 rad2π rad/s=0 s

Therefore, the time shift of the one-link robot is 0 s.

The amplitude of sine and cosine function is 5.

Conclusion:

Thus, the graph for x -component is plotted in Figure 1 and y -component as a function of time is plotted in Figure 2. The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

Answer 7

Expert Solution & Answer

Answer 8

Expert Solution & Answer

Answer 9

To determine

To plot:

The x -component and y -component as a function of time and determine the amplitude, frequency, period, phase angle and time shift.

Answer 10

Answer to Problem 1P

The graph for x -componentas a function of timeis plotted is as shown in Figure 1.

The graph for y -component as a function of time is plotted is as shown in Figure 2.

The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

Answer 11

Explanation of Solution

Given:

The tip of a one-link robot is initially located at θ=0 attime t=0 s. The given figure is shown below,

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 1

Time taken for the robot to move form θ=0 to θ=2π rad is 1 s.

A one-link robot is length of 5 in.

Concept used:

Write the expression for the linear frequency.

f=1T ....... (1)

Here, f is the linear frequency and T is the time period.

Write the expression for the angular frequency.

ω=2πT ....... (2)

Here, ω is the angular frequency.

Write the expression for the time period.

T=2πω ....... (3)

Write the expression for x -component at any time,

x(t)=lcos(θ) ....... (4)

Here, x(t) is the expression for x -component at any time, l is the length of one-link robot, θ is the angle.

Write the expression for x -component at any time.

y(t)=lsin(θ) ....... (5)

Here, y(t) is the expression for y -component at any time.

Write the expression for the time shift.

t0=ϕω ....... (6)

Here, t0 is the expression for the time shift.

Calculation:

The one-link robot completes one revolution in 1 s.

Substitute 1 s for T in equation (2).

ω=2π1 s=2π rad/s

Substitute 2π rad/s for ω in equation (3).

T=2π2π s=1 s

Therefore, the period of the tip of a one-link robot is 1 s.

Substitute 1 s for T in equation (1).

f=11 s=1 Hz

Therefore, the frequency is 1 Hz.

Since the one-link robot initially start form θ=0 and the phase angle is 0 rad.

Substitute ωt+ϕ for θ in equation (4).

x(t)=lcos(ωt+ϕ) ....... (7)

Here, t is any given time and ϕ is the phase angle.

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (7).

x(t)=(5)cos(( 2π)t+0)=5cos2πt

The plot for the x -component as a function of time is shown in Figure 1.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 2

Substitute ωt+ϕ for θ in equation (5).

y(t)=lsin(ωt+ϕ) ....... (8)

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (8).

x(t)=(5)sin(( 2π)t+0)=5sin(2πt+0)=5sin(2πt)

The plot for the y -component as a function of time is shown in Figure 2.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip 3

Figure 2

Substitute 0 rad for ϕ and 2π rad/s for ω in equation (6).

t0 =0 rad2π rad/s=0 s

Therefore, the time shift of the one-link robot is 0 s.

The amplitude of sine and cosine function is 5.

Conclusion:

Thus, the graph for x -component is plotted in Figure 1 and y -component as a function of time is plotted in Figure 2. The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

Answer 12

Chapter 6, Problem 1P Chapter 6, Problem 2P Chapter 6, Problem 3P Chapter 6, Problem 4P Chapter 6, Problem 5P Chapter 6, Problem 6P Chapter 6, Problem 7P Chapter 6, Problem 8P Chapter 6, Problem 9P Chapter 6, Problem 10P

Chapter 6, Problem 1P

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Answer to Problem 1P

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