Sketch for the condition y → = ( y 0 cos 2 π t T ) j ^ .

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Answer 1

Question

Chapter 2, Problem 18PQ

(a)

To determine

Sketch for the condition y→=(y0cos2πtT)j^.

(a)

Expert Solution

Answer to Problem 18PQ

The sketch for the condition y→=(y0cos2πtT)j^ is given in figure 1.

Explanation of Solution

Write an expression for the condition.

y→=(y0cos2πtT)j^

Here, y→ is the instantaneous position, y0 is the initial position, t is the time and T is the time period.

Here, y axis points upwards. The time starts at t=0. there will not be any negative value for time. At time t=0 the particle is at +y0.

Thus, the sketch for the condition y→=(y0cos2πtT)j^ is given in figure 1.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 2, Problem 18PQ

(b)

To determine

The position of the particle at t=0,12T,T,32T,2T and 52T.

(b)

Expert Solution

Answer to Problem 18PQ

The position of the particle at t=0, 12T, T,32T, 2T and 52T are y0j^, −y0j^, y0j^, −y0j^, y0j^ and −y0j^ respectively and the positions are marked in figure 1.

Explanation of Solution

Write an expression for the condition.

y→=(y0cos2πtT)j^ (I)

Conclusion:

For t=0:

Substitute 0 for t in equation (I).to find y→.

y→=(y0cos2π(0)T)j^=(y0cos(0))j^=y0j^

For t=T2:

Substitute T/2 for t in equation (I).to find y→.

y→=(y0cos2π(T2)T)j^=(y0cos(π))j^=−y0j^

For t=T:

Substitute T for t in equation (I).to find y→.

y→=(y0cos2π(T)T)j^=(y0cos(2π))j^=y0j^

For t=3T2:

Substitute 3T/2 for t in equation (I).to find y→.

y→=(y0cos2π(3T2)T)j^=(y0cos(3π))j^=−y0j^

For t=2T:

Substitute 2T for t in equation (I).to find y→.

y→=(y0cos2π(2T)T)j^=(y0cos(4π))j^=y0j^

For t=5T2:

Substitute 5T/2 for t in equation (I).to find y→.

y→=(y0cos2π(5T2)T)j^=(y0cos(5π))j^=−y0j^

Thus, the particle can be seen at y0j^, −y0j^, y0j^, −y0j^, y0j^ and −y0j^ when the time is t=0, 12T, T,32T, 2T and 52T respectively and the positions are marked in figure 1.

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Answer 2

Question

Chapter 2, Problem 18PQ

(a)

To determine

Sketch for the condition y→=(y0cos2πtT)j^.

(a)

Expert Solution

Answer to Problem 18PQ

The sketch for the condition y→=(y0cos2πtT)j^ is given in figure 1.

Explanation of Solution

Write an expression for the condition.

y→=(y0cos2πtT)j^

Here, y→ is the instantaneous position, y0 is the initial position, t is the time and T is the time period.

Here, y axis points upwards. The time starts at t=0. there will not be any negative value for time. At time t=0 the particle is at +y0.

Thus, the sketch for the condition y→=(y0cos2πtT)j^ is given in figure 1.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 2, Problem 18PQ

(b)

To determine

The position of the particle at t=0,12T,T,32T,2T and 52T.

(b)

Expert Solution

Answer to Problem 18PQ

The position of the particle at t=0, 12T, T,32T, 2T and 52T are y0j^, −y0j^, y0j^, −y0j^, y0j^ and −y0j^ respectively and the positions are marked in figure 1.

Explanation of Solution

Write an expression for the condition.

y→=(y0cos2πtT)j^ (I)

Conclusion:

For t=0:

Substitute 0 for t in equation (I).to find y→.

y→=(y0cos2π(0)T)j^=(y0cos(0))j^=y0j^

For t=T2:

Substitute T/2 for t in equation (I).to find y→.

y→=(y0cos2π(T2)T)j^=(y0cos(π))j^=−y0j^

For t=T:

Substitute T for t in equation (I).to find y→.

y→=(y0cos2π(T)T)j^=(y0cos(2π))j^=y0j^

For t=3T2:

Substitute 3T/2 for t in equation (I).to find y→.

y→=(y0cos2π(3T2)T)j^=(y0cos(3π))j^=−y0j^

For t=2T:

Substitute 2T for t in equation (I).to find y→.

y→=(y0cos2π(2T)T)j^=(y0cos(4π))j^=y0j^

For t=5T2:

Substitute 5T/2 for t in equation (I).to find y→.

y→=(y0cos2π(5T2)T)j^=(y0cos(5π))j^=−y0j^

Thus, the particle can be seen at y0j^, −y0j^, y0j^, −y0j^, y0j^ and −y0j^ when the time is t=0, 12T, T,32T, 2T and 52T respectively and the positions are marked in figure 1.

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Answer 3

Question

Chapter 2, Problem 18PQ

(a)

To determine

Sketch for the condition y→=(y0cos2πtT)j^.

(a)

Expert Solution

Answer to Problem 18PQ

The sketch for the condition y→=(y0cos2πtT)j^ is given in figure 1.

Explanation of Solution

Write an expression for the condition.

y→=(y0cos2πtT)j^

Here, y→ is the instantaneous position, y0 is the initial position, t is the time and T is the time period.

Here, y axis points upwards. The time starts at t=0. there will not be any negative value for time. At time t=0 the particle is at +y0.

Thus, the sketch for the condition y→=(y0cos2πtT)j^ is given in figure 1.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 2, Problem 18PQ

(b)

To determine

The position of the particle at t=0,12T,T,32T,2T and 52T.

(b)

Expert Solution

Answer to Problem 18PQ

The position of the particle at t=0, 12T, T,32T, 2T and 52T are y0j^, −y0j^, y0j^, −y0j^, y0j^ and −y0j^ respectively and the positions are marked in figure 1.

Explanation of Solution

Write an expression for the condition.

y→=(y0cos2πtT)j^ (I)

Conclusion:

For t=0:

Substitute 0 for t in equation (I).to find y→.

y→=(y0cos2π(0)T)j^=(y0cos(0))j^=y0j^

For t=T2:

Substitute T/2 for t in equation (I).to find y→.

y→=(y0cos2π(T2)T)j^=(y0cos(π))j^=−y0j^

For t=T:

Substitute T for t in equation (I).to find y→.

y→=(y0cos2π(T)T)j^=(y0cos(2π))j^=y0j^

For t=3T2:

Substitute 3T/2 for t in equation (I).to find y→.

y→=(y0cos2π(3T2)T)j^=(y0cos(3π))j^=−y0j^

For t=2T:

Substitute 2T for t in equation (I).to find y→.

y→=(y0cos2π(2T)T)j^=(y0cos(4π))j^=y0j^

For t=5T2:

Substitute 5T/2 for t in equation (I).to find y→.

y→=(y0cos2π(5T2)T)j^=(y0cos(5π))j^=−y0j^

Thus, the particle can be seen at y0j^, −y0j^, y0j^, −y0j^, y0j^ and −y0j^ when the time is t=0, 12T, T,32T, 2T and 52T respectively and the positions are marked in figure 1.

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Answer 4

Question

Chapter 2, Problem 18PQ

(a)

To determine

Sketch for the condition y→=(y0cos2πtT)j^.

(a)

Expert Solution

Answer to Problem 18PQ

The sketch for the condition y→=(y0cos2πtT)j^ is given in figure 1.

Explanation of Solution

Write an expression for the condition.

y→=(y0cos2πtT)j^

Here, y→ is the instantaneous position, y0 is the initial position, t is the time and T is the time period.

Here, y axis points upwards. The time starts at t=0. there will not be any negative value for time. At time t=0 the particle is at +y0.

Thus, the sketch for the condition y→=(y0cos2πtT)j^ is given in figure 1.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 2, Problem 18PQ

(b)

To determine

The position of the particle at t=0,12T,T,32T,2T and 52T.

(b)

Expert Solution

Answer to Problem 18PQ

The position of the particle at t=0, 12T, T,32T, 2T and 52T are y0j^, −y0j^, y0j^, −y0j^, y0j^ and −y0j^ respectively and the positions are marked in figure 1.

Explanation of Solution

Write an expression for the condition.

y→=(y0cos2πtT)j^ (I)

Conclusion:

For t=0:

Substitute 0 for t in equation (I).to find y→.

y→=(y0cos2π(0)T)j^=(y0cos(0))j^=y0j^

For t=T2:

Substitute T/2 for t in equation (I).to find y→.

y→=(y0cos2π(T2)T)j^=(y0cos(π))j^=−y0j^

For t=T:

Substitute T for t in equation (I).to find y→.

y→=(y0cos2π(T)T)j^=(y0cos(2π))j^=y0j^

For t=3T2:

Substitute 3T/2 for t in equation (I).to find y→.

y→=(y0cos2π(3T2)T)j^=(y0cos(3π))j^=−y0j^

For t=2T:

Substitute 2T for t in equation (I).to find y→.

y→=(y0cos2π(2T)T)j^=(y0cos(4π))j^=y0j^

For t=5T2:

Substitute 5T/2 for t in equation (I).to find y→.

y→=(y0cos2π(5T2)T)j^=(y0cos(5π))j^=−y0j^

Thus, the particle can be seen at y0j^, −y0j^, y0j^, −y0j^, y0j^ and −y0j^ when the time is t=0, 12T, T,32T, 2T and 52T respectively and the positions are marked in figure 1.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 2, Problem 18PQ

(a)

To determine

Sketch for the condition y→=(y0cos2πtT)j^.

(a)

Expert Solution

Answer to Problem 18PQ

The sketch for the condition y→=(y0cos2πtT)j^ is given in figure 1.

Explanation of Solution

Write an expression for the condition.

y→=(y0cos2πtT)j^

Here, y→ is the instantaneous position, y0 is the initial position, t is the time and T is the time period.

Here, y axis points upwards. The time starts at t=0. there will not be any negative value for time. At time t=0 the particle is at +y0.

Thus, the sketch for the condition y→=(y0cos2πtT)j^ is given in figure 1.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 2, Problem 18PQ

(b)

To determine

The position of the particle at t=0,12T,T,32T,2T and 52T.

(b)

Expert Solution

Answer to Problem 18PQ

The position of the particle at t=0, 12T, T,32T, 2T and 52T are y0j^, −y0j^, y0j^, −y0j^, y0j^ and −y0j^ respectively and the positions are marked in figure 1.

Explanation of Solution

Write an expression for the condition.

y→=(y0cos2πtT)j^ (I)

Conclusion:

For t=0:

Substitute 0 for t in equation (I).to find y→.

y→=(y0cos2π(0)T)j^=(y0cos(0))j^=y0j^

For t=T2:

Substitute T/2 for t in equation (I).to find y→.

y→=(y0cos2π(T2)T)j^=(y0cos(π))j^=−y0j^

For t=T:

Substitute T for t in equation (I).to find y→.

y→=(y0cos2π(T)T)j^=(y0cos(2π))j^=y0j^

For t=3T2:

Substitute 3T/2 for t in equation (I).to find y→.

y→=(y0cos2π(3T2)T)j^=(y0cos(3π))j^=−y0j^

For t=2T:

Substitute 2T for t in equation (I).to find y→.

y→=(y0cos2π(2T)T)j^=(y0cos(4π))j^=y0j^

For t=5T2:

Substitute 5T/2 for t in equation (I).to find y→.

y→=(y0cos2π(5T2)T)j^=(y0cos(5π))j^=−y0j^

Thus, the particle can be seen at y0j^, −y0j^, y0j^, −y0j^, y0j^ and −y0j^ when the time is t=0, 12T, T,32T, 2T and 52T respectively and the positions are marked in figure 1.

Answer 6

Chapter 2, Problem 18PQ

(a)

To determine

Sketch for the condition y→=(y0cos2πtT)j^.

(a)

Expert Solution

Answer to Problem 18PQ

The sketch for the condition y→=(y0cos2πtT)j^ is given in figure 1.

Explanation of Solution

Write an expression for the condition.

y→=(y0cos2πtT)j^

Here, y→ is the instantaneous position, y0 is the initial position, t is the time and T is the time period.

Here, y axis points upwards. The time starts at t=0. there will not be any negative value for time. At time t=0 the particle is at +y0.

Thus, the sketch for the condition y→=(y0cos2πtT)j^ is given in figure 1.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 2, Problem 18PQ

Answer 7

Chapter 2, Problem 18PQ

(a)

To determine

Sketch for the condition y→=(y0cos2πtT)j^.

Answer 8

(a)

Expert Solution

Answer to Problem 18PQ

The sketch for the condition y→=(y0cos2πtT)j^ is given in figure 1.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write an expression for the condition.

y→=(y0cos2πtT)j^

Here, y→ is the instantaneous position, y0 is the initial position, t is the time and T is the time period.

Here, y axis points upwards. The time starts at t=0. there will not be any negative value for time. At time t=0 the particle is at +y0.

Thus, the sketch for the condition y→=(y0cos2πtT)j^ is given in figure 1.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 2, Problem 18PQ

Answer 13

(b)

To determine

The position of the particle at t=0,12T,T,32T,2T and 52T.

(b)

Expert Solution

Answer to Problem 18PQ

The position of the particle at t=0, 12T, T,32T, 2T and 52T are y0j^, −y0j^, y0j^, −y0j^, y0j^ and −y0j^ respectively and the positions are marked in figure 1.

Explanation of Solution

Write an expression for the condition.

y→=(y0cos2πtT)j^ (I)

Conclusion:

For t=0:

Substitute 0 for t in equation (I).to find y→.

y→=(y0cos2π(0)T)j^=(y0cos(0))j^=y0j^

For t=T2:

Substitute T/2 for t in equation (I).to find y→.

y→=(y0cos2π(T2)T)j^=(y0cos(π))j^=−y0j^

For t=T:

Substitute T for t in equation (I).to find y→.

y→=(y0cos2π(T)T)j^=(y0cos(2π))j^=y0j^

For t=3T2:

Substitute 3T/2 for t in equation (I).to find y→.

y→=(y0cos2π(3T2)T)j^=(y0cos(3π))j^=−y0j^

For t=2T:

Substitute 2T for t in equation (I).to find y→.

y→=(y0cos2π(2T)T)j^=(y0cos(4π))j^=y0j^

For t=5T2:

Substitute 5T/2 for t in equation (I).to find y→.

y→=(y0cos2π(5T2)T)j^=(y0cos(5π))j^=−y0j^

Thus, the particle can be seen at y0j^, −y0j^, y0j^, −y0j^, y0j^ and −y0j^ when the time is t=0, 12T, T,32T, 2T and 52T respectively and the positions are marked in figure 1.

Answer 14

(b)

To determine

The position of the particle at t=0,12T,T,32T,2T and 52T.

Answer 15

(b)

Expert Solution

Answer to Problem 18PQ

The position of the particle at t=0, 12T, T,32T, 2T and 52T are y0j^, −y0j^, y0j^, −y0j^, y0j^ and −y0j^ respectively and the positions are marked in figure 1.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write an expression for the condition.

y→=(y0cos2πtT)j^ (I)

Conclusion:

For t=0:

Substitute 0 for t in equation (I).to find y→.

y→=(y0cos2π(0)T)j^=(y0cos(0))j^=y0j^

For t=T2:

Substitute T/2 for t in equation (I).to find y→.

y→=(y0cos2π(T2)T)j^=(y0cos(π))j^=−y0j^

For t=T:

Substitute T for t in equation (I).to find y→.

y→=(y0cos2π(T)T)j^=(y0cos(2π))j^=y0j^

For t=3T2:

Substitute 3T/2 for t in equation (I).to find y→.

y→=(y0cos2π(3T2)T)j^=(y0cos(3π))j^=−y0j^

For t=2T:

Substitute 2T for t in equation (I).to find y→.

y→=(y0cos2π(2T)T)j^=(y0cos(4π))j^=y0j^

For t=5T2:

Substitute 5T/2 for t in equation (I).to find y→.

y→=(y0cos2π(5T2)T)j^=(y0cos(5π))j^=−y0j^

Thus, the particle can be seen at y0j^, −y0j^, y0j^, −y0j^, y0j^ and −y0j^ when the time is t=0, 12T, T,32T, 2T and 52T respectively and the positions are marked in figure 1.