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

Question

Want to see more full solutions like this?

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.

Want to see more full solutions like this?

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

Students have asked these similar questions

ՍՈՈՒ XVirginia Western Community Coll x P Course Home X + astering.pearson.com/?courseld=13289599#/ Figure y (mm) x=0x = 0.0900 m All ✓ Correct For either the time for one full cycle is 0.040 s; this is the period. Part C - ON You are told that the two points x = 0 and x = 0.0900 m are within one wavelength of each other. If the wave is moving in the +x-direction, determine the wavelength. Express your answer to two significant figures and include the appropriate units. 0 t(s) λ = Value m 0.01 0.03 0.05 0.07 Copyright © 2025 Pearson Education Inc. All rights reserved. 日 F3 F4 F5 1775 % F6 F7 B F8 Submit Previous Answers Request Answer ? × Incorrect; Try Again; 3 attempts remaining | Terms of Use | Privacy Policy | Permissions | Contact Us | Cookie Settings 28°F Clear 4 9:23 PM 1/20/2025 F9 prt sc F10 home F11 end F12 insert delete 6 7 29 & * ( 8 9 0 t = back Ο

Part C Find the height yi from which the rock was launched. Express your answer in meters to three significant figures. Learning Goal: To practice Problem-Solving Strategy 4.1 for projectile motion problems. A rock thrown with speed 12.0 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 19.0 m before hitting the ground. From what height was the rock thrown? Use the value g = 9.800 m/s2 for the free-fall acceleration. PROBLEM-SOLVING STRATEGY 4.1 Projectile motion problems MODEL: Is it reasonable to ignore air resistance? If so, use the projectile motion model. VISUALIZE: Establish a coordinate system with the x-axis horizontal and the y-axis vertical. Define symbols and identify what the problem is trying to find. For a launch at angle θ, the initial velocity components are vix=v0cosθ and viy=v0sinθ. SOLVE: The acceleration is known: ax=0 and ay=−g. Thus, the problem becomes one of…

Phys 25

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.

Want to see more full solutions like this?

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

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.

Want to see more full solutions like this?

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

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.

Want to see more full solutions like this?

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.