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 6PQ

(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 6PQ

At t=0s, simple harmonic oscillator’s position is 3.68×103m, its velocity is 7.22×102m/s and acceleration is 0.398m/s2 .

Explanation of Solution

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

  ay(t)=(0.850m/s2)cos(10.4t5.20)                                                                    (I)

Here, ay(t) is the acceleration of the simple harmonic oscillator.

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

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

Here, amax is the maximum acceleration, ω is the angular frequency and φ is the initial phase.

Write the relation between maximum acceleration and amplitude.

  amax=ymaxω2                                                                                                        (III)

Here, ymax is the amplitude.

Write the expression for the velocity of simple harmonic oscillator.

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

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

Write the expression for the maximum velocity.

  vmax=ymaxω                                                                                                            (V)

Put equation (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)

Here, y(t) is the position of the simple harmonic oscillator.

Conclusion:

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

  ω=10.4rad/s

Compare equation (I) and (II) to get amax .

  amax=0.850m/s2

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

  φ=5.20

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

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

Substitute (0.850m/s2)(10.4rad/s)2 for ymax, 5.20 for φ and 10.4rad/s for ω in equation (VI) to get vy(t).

  vy(t)=((0.850m/s2)(10.4rad/s)2)(10.4rad/s)sin((10.4rad/s)t5.20)=((0.850m/s2)(10.4rad/s))sin((10.4rad/s)t5.20)                       (VIII)

Substitute (0.850m/s2)(10.4rad/s)2 for ymax, 5.20 for φ and 10.4rad/s for ω in equation (VII) to get y(t).

  y(t)=(0.850m/s2)(10.4rad/s)2cos((10.4rad/s)t5.20)                                                  (IX)

Consider t=0 .

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

  y(0)=(0.850m/s2)(10.4rad/s)2cos((10.4rad/s)(0s)5.20)=3.68m×103m

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

  vy(0)=((0.850m/s2)(10.4rad/s))sin((10.4rad/s)(0)5.20)=7.22×102m/s

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

  ay(0)=(0.850m/s2)cos(10.4(0s)5.20)=0.398m/s2

Therefore, at t=0s, simple harmonic oscillators position is 3.68×103m, its velocity is 7.22×102m/s and acceleration is 0.398m/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 6PQ

At t=0.500s, simple harmonic oscillator’s position is 7.86×103m, its velocity is 0m/s and its acceleration is 0.850m/s2 .

Explanation of Solution

Use equation (I) to calculate acceleration, equation (VIII) to calculate velocity and equation (IX) to calculate position of simple harmonic oscillator at t=0.500s .

Conclusion:

Consider t=0.500s .

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

  y(0.500s)=(0.850m/s2)(10.4rad/s)2cos((10.4rad/s)(0.500s)5.20)=7.86m×103m

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

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

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

  ay(0.500s)=(0.850m/s2)cos(10.4(0.500s)5.20)=0.850m/s2

Therefore, at t=0.500s, simple harmonic oscillator’s position is 7.86×103m, its velocity is 0m/s and its acceleration is 0.850m/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 6PQ

At t=2.00s, simple harmonic oscillator’s position is 7.81×103m, its velocity is 8.81×103m/s and its acceleration is 0.845m/s2 .

Explanation of Solution

Use equation (I) to calculate acceleration, equation (VIII) to calculate velocity 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(2.00s)=(0.850m/s2)(10.4rad/s)2cos((10.4rad/s)(2.00s)5.20)=7.81×103m

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

  vy(2.00s)=((0.850m/s2)(10.4rad/s))sin((10.4rad/s)(2.00s)5.20)=8.81×103m/s

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

  ay(2.00s)=(0.850m/s2)cos(10.4(2.00s)5.20)=0.845m/s2

Therefore, at t=2.00s, simple harmonic oscillator’s position is 7.81×103m, its velocity is 8.81×103m/s and its acceleration is 0.845m/s2 .

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Chapter 16 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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