Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 14, Problem 34P

(a)

To determine

The distance travelled by the particle during the time t=0sto2s .

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The time period of the particle is 8.0s .

The amplitude of the oscillation of the particle is 12cm .

Formula used:

The position of the particle is given as:

  x=Acos(ωt+δ)   ............. (1)

Here, x is the position of the particle, A is the amplitude, ω is the angular frequency, t is the time and δ is the phase constant.

Write the expression for the angular frequency of oscillation.

  ω=2πT   ............. (2)

Substitute 8.0s for T in equation (2)

  ω=2π8.0sω=π4s-1

Write the expression for the initial position of the particle with amplitude and phase constant.

  x0=Acosδ

Simplify the above equation we get.

  δ=cos1(x0A)   ............. (3)

Substitute 0 for x0 in equation (3).

  δ=cos1(0A)δ=π2

Substitute π2 for δ , 12cm for A and π4s-1 for ω in equation (1)

  x=Acos(ωt+δ)x=(12cm)cos(π4s -1t+π2)

Now the distance particle travels in initial time tinitial and final time tfinal is:

  x=(12cm)cos[(π4s1)tfinal+π2]((12cm)cos[( π 4 s 1)tinitial+π2])   ............. (4)

Calculation:

Substitute 2.0s for tfinal and 0s for tinitial in equation (4).

  x=(12cm)cos[( π 4 s 1)2.0s+π2](( 12cm)cos[( π 4 s 1 )0s+ π 2])x=(12cm)cos[( π 4 s 1)2.0s+π2](( 12cm)cosπ2)x=12cm

Conclusion:

The distance the particle travels at t=0sto2s is x=12cm .

(b)

To determine

The distance travelled by the particle at time t=0.2sto0.4s.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The time period of the particle is 8.0s .

The amplitude of the oscillation of the particle is 12cm .

Formula used:

The position of the particle is given as:

  x=Acos(ωt+δ)   ............. (1)

Here, x is the position of the particle, A is the amplitude, ω is the angular frequency, t is the time and τ is the phase constant.

Calculation:

Substitute 4.0s for tfinal and 2.0s for tinitial in equation (4).

  x=(12cm)cos[( π 4 s 1)4.0s+π2](( 12cm)cos[( π 4 s 1 )2s+ π 2])x=(12cm)cos[3π2](12cm)cos(π)x=12cm

Conclusion:

The position of the particle is 12cm .

(c)

To determine

The distance travelled by the particle t=0sto1s .

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The time period of the particle is 0.8s .

The amplitude of the oscillation of the particle is 12cm .

Formula used:

The position of the particle is given as:

  x=Acos(ωt+δ)

Calculation:

Substitute 1s for tfinal and 0s for tinitial in equation (4).

  x=(12cm)cos[( π 4 s 1)1.0s+π2](( 12cm)cos[( π 4 s 1 )0s+ π 2])x=(12cm)cos( 3π4)x=8.48cm

Conclusion:

The position of the particle is 8.48cm .

(d)

To determine

The distance travelled by the particle t=1.0sto2.0s .

(d)

Expert Solution
Check Mark

Explanation of Solution

Given:

The time period of the particle is 0.8s .

The amplitude of the oscillation of the particle is 12cm .

Formula used:

The position of the particle is given as:

  x=Acos(ωt+δ)

Calculation:

Substitute 2.0s for tfinal and 1.0s for tinitial in equation (4).

  x=(12cm)cos[( π 4 s 1)2.0s+π2](( 12cm)cos[( π 4 s 1 )1.0s+ π 2])x=(12cm)cos[π](( 12cm)cos[ 3π 4])x=3.52cm

Conclusion:

The position of the particle is 3.52cm .

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Chapter 14 Solutions

Physics for Scientists and Engineers

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