Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 16, Problem 5PQ

(a)

To determine

The position, velocity and acceleration of each of the simple harmonic oscillator’s at time t=0s .

(a)

Expert Solution
Check Mark

Answer to Problem 5PQ

At t=0s, simple harmonic oscillator’s position is 3.83×102m , its velocity is 0.751m/s and acceleration is 4.14m/s2 .

Explanation of Solution

Write the expression for the velocity of the simple harmonic oscillator.

  vy(t)=(0.850m/s2)sin(10.4t5.20)                                                                    (I)

Here, vy(t) is the velocity of the simple harmonic oscillator.

Write the general equation of acceleration of a simple harmonic oscillator.

  ay(t)=amaxcos(ωt+φ)                                                                                       (II)

Here, ay(t) is the acceleration of the simple harmonic oscillator, amax is the maximum acceleration, ω is the angular frequency and φ is the initial phase.

Write the relation between maximum acceleration and maximum displacement.

  amax=ymaxω2                                                                                                        (III)

Here, ymax is the maximum displacement.

Write the general expression for the velocity of simple harmonic oscillator.

  vy(t)=vmaxsin(ωt+φ)                                                                                        (IV)

Here, vy(t) is the velocity of the simple harmonic oscillator and vmax is the maximum velocity.

Write the expression for the maximum velocity.

  vmax=ymaxω                                                                                                            (V)

Substitute (V) in (IV) to get relation of vy(t).

  vy(t)=ymaxωsin(ωt+φ)                                                                                    (VI)

Write the expression for the displacement of the simple harmonic oscillator.

  y(t)=ymaxcos(ωt+φ)                                                                                        (VII)

Conclusion:

Compare equation (I) and (IV) to get ω .

  ω=10.4rad/s

Compare equation (I) and (IV) to get vmax .

  vmax=0.850m/s2

Compare equation (I) and (IV) to get φ .

  φ=5.20

Substitute 10.4rad/s for ω and 0.850m/s2 for vmax in equation (V) to get ymax .

  0.850m/s2=ymax(10.4rad/s)ymax=0.850m/s2(10.4rad/s)

Substitute (0.850m/s2)(10.4rad/s) for ymax and 10.4rad/s for ω in equation (III) to get amax.

  amax=(0.850m/s2)(10.4rad/s)(10.4rad/s)2

Substitute (0.850m/s2)(10.4rad/s)(10.4rad/s)2 for amax, 5.20 for φ and 10.4rad/s for ω in equation (II) to get ay(t).

  ay(t)=((0.850m/s2)(10.4rad/s))(10.4rad/s)2cos((10.4rad/s)t5.20)=(8.84m/s2)cos((10.4rad/s)t5.20)                     (VIII)

Substitute (0.850m/s2)(10.4rad/s) for ymax, 5.20 for φ and 10.4rad/s for ω in equation (V) to get y(t).

  y(t)=(0.850m/s2)(10.4rad/s)cos((10.4rad/s)t5.20)                                                  (IX)

Consider t=0 .

Substitute 0 for t in equation (IX) to get y(t) .

  y(t)=(0.850m/s2)(10.4rad/s)cos((10.4rad/s)(0s)5.20)=3.83m×102m

Substitute 0 for t in equation (VIII) to get ay(t) .

  ay(t)=(8.84m/s2)cos((10.4rad/s)(0)5.20)=4.14m/s2

Substitute 0 for t in equation (I) to get vy(t) .

  vy(t)=(0.850m/s2)sin(10.4(0s)5.20)=0.751m/s

Therefore, at t=0s, simple harmonic oscillators position is 3.83×102m , its velocity is 0.751m/s and acceleration is 4.14m/s2 .

(b)

To determine

The position, velocity and acceleration of simple harmonic oscillator at t=0.500s .

(b)

Expert Solution
Check Mark

Answer to Problem 5PQ

At t=0.500s, simple harmonic oscillators position is 8.17×102m , its velocity is 0m/s and its acceleration is 8.84m/s2 .

Explanation of Solution

Use equation (I) to calculate velocity, equation (VIII) to calculate acceleration and equation (IX) to calculate position of simple harmonic oscillator.

Conclusion:

Consider t=0.500s .

Substitute 0.500s for t in equation (IX) to get y(t) .

  y(t)=(0.850m/s2)(10.4rad/s)cos((10.4rad/s)(0.500s)5.20)=8.17m×102m

Substitute 0.500s for t in equation (VIII) to get ay(t) .

  ay(t)=(8.84m/s2)cos((10.4rad/s)(0.500s)5.20)=8.84m/s2

Substitute 0.500s for t in equation (I) to get vy(t) .

  vy(t)=(0.850m/s2)sin(10.4(0.500s)5.20)=0m/s

Therefore, at t=0.500s, simple harmonic oscillators position is 8.17×102m , its velocity is 0m/s and its acceleration is 8.84m/s2 .

(c)

To determine

The position, velocity and acceleration of simple harmonic oscillator at t=2.00s .

(c)

Expert Solution
Check Mark

Answer to Problem 5PQ

At t=2.00s, simple harmonic oscillators position is 8.13m×102m , its velocity is 9.16×102m/s and its acceleration is 8.79m/s2 .

Explanation of Solution

Use equation (I) to calculate velocity, equation (VIII) to calculate acceleration and equation (IX) to calculate position of simple harmonic oscillator.

Conclusion:

Consider t=2.00s .

Substitute 2.00s for t in equation (IX) to get y(t) .

  y(t)=(0.850m/s2)(10.4rad/s)cos((10.4rad/s)(2.00s)5.20)=8.13m×102m

Substitute 2.00s for t in equation (VIII) to get ay(t) .

  ay(t)=(8.84m/s2)cos((10.4rad/s)(2.00s)5.20)=8.79m/s2

Substitute 2.00s for t in equation (I) to get vy(t) .

  vy(t)=(0.850m/s2)sin(10.4(2.00s)5.20)=9.16×102m/s

Therefore, at t=2.00s, simple harmonic oscillators position is 8.13×102m , its velocity is 9.16×102m/s and its acceleration is 8.79m/s2 .

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Chapter 16 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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