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 87P

(a)

To determine

The amplitude of oscillation.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Mass of object is 2.00kg .

Linear damping constant is 2.00kg/s .

Value of force constant is 400N/m .

Value of maximum force is 10.0N .

Angular frequency is 10.0rad/s .

Formula used:

Write expression for amplitude of damped oscillation.

  A=Fmaxm2( ω 0 2 ω 2 )+b2ω2

Here, Fmax is the maximum force, m is the mass of object, b is the damping constant and ω is the angular frequency.

Substitute km for ω0 in above expression.

  A=Fmaxm2( k m ω 2 )+b2ω2   …… (1)

Calculation:

Substitute 10N for Fmax , 2kg for m , 400N/m for k , 10.0rad/s for ω and 2.00kg/s for b in equation (1).

  A=10N ( 2kg ) 2 ( 400N/m 2kg ( 10.0 rad/s ) 2 )+ ( 2.00 kg/s ) 2 ( 10.0 rad/s ) 2 =0.0498m

Conclusion:

The amplitude of oscillation is 0.0498m .

(b)

To determine

The frequency at which resonance occurs.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

Mass of object is 2.00kg .

Linear damping constant is 2.00kg/s .

Value of force constant is 400N/m .

Value of maximum force is 10.0N .

Angular frequency is 10.0rad/s .

Formula used:

Write expression for condition of resonance.

  ω=ω0

Substitute km for ω0 in above expression.

  ω=km   …… (2)

Calculation:

Substitute 2kg for m and 400N/m for k in equation (2).

  ω= 400N/m 2kg=14.14rad/s

Conclusion:

Thus, the frequency at which resonance occurs is 14.14rad/s .

(c)

To determine

The amplitude of oscillation at resonance.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

Mass of object is 2.00kg .

Linear damping constant is 2.00kg/s .

Value of force constant is 400N/m .

Value of maximum force is 10.0N .

Angular frequency is 10.0rad/s .

Formula used:

Write expression for amplitude of damped oscillation.

  A=Fmaxm2( ω 0 2 ω 2 )+b2ω2   …… (3)

Here, Fmax is the maximum force, m is the mass of object, b is the damping constant and ω is the angular frequency.

Write expression for condition of resonance.

  ω=ω0

Substitute ω for ω0 in equation (3)

  A=Fmaxm2( ω 2 ω 2 )+b2ω2

Solve above expression.

  A=Fmaxb2ω2   …… (4)

Calculation:

Substitute 10N for Fmax , 10.0rad/s for ω and 2.00kg/s for b in equation (4).

  A=10N ( 2.00 kg/s ) 2 ( 10.0 rad/s ) 2 =0.0354m

Conclusion:

Thus, the amplitude of oscillation at resonance is 0.0354m .

(d)

To determine

The width of resonance curve.

(d)

Expert Solution
Check Mark

Explanation of Solution

Given:

Mass of object is 2.00kg .

Linear damping constant is 2.00kg/s .

Value of force constant is 400N/m .

Value of maximum force is 10.0N .

Angular frequency is 10.0rad/s .

Formula used:

Write expression for width of resonance curve.

  Δω=bm   …… (5)

Here, Δω is the width of resonance curve.

Calculation:

Substitute 2kg for m and 2.00kg/s for b in equation (5).

  Δω=2.00kg/s2kg=1rad/s

Conclusion:

Thus, the width of the resonance curve is 1rad/s .

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Chapter 14 Solutions

Physics for Scientists and Engineers

Ch. 14 - Prob. 11PCh. 14 - Prob. 12PCh. 14 - Prob. 13PCh. 14 - Prob. 14PCh. 14 - Prob. 15PCh. 14 - Prob. 16PCh. 14 - Prob. 17PCh. 14 - Prob. 18PCh. 14 - Prob. 19PCh. 14 - Prob. 20PCh. 14 - Prob. 21PCh. 14 - Prob. 22PCh. 14 - Prob. 23PCh. 14 - Prob. 24PCh. 14 - Prob. 25PCh. 14 - Prob. 26PCh. 14 - Prob. 27PCh. 14 - Prob. 28PCh. 14 - Prob. 29PCh. 14 - Prob. 30PCh. 14 - Prob. 31PCh. 14 - Prob. 32PCh. 14 - Prob. 33PCh. 14 - Prob. 34PCh. 14 - Prob. 35PCh. 14 - Prob. 36PCh. 14 - Prob. 37PCh. 14 - Prob. 38PCh. 14 - Prob. 39PCh. 14 - Prob. 40PCh. 14 - Prob. 41PCh. 14 - Prob. 42PCh. 14 - Prob. 43PCh. 14 - Prob. 44PCh. 14 - Prob. 45PCh. 14 - Prob. 46PCh. 14 - Prob. 47PCh. 14 - Prob. 48PCh. 14 - Prob. 49PCh. 14 - Prob. 50PCh. 14 - Prob. 51PCh. 14 - Prob. 52PCh. 14 - Prob. 53PCh. 14 - Prob. 54PCh. 14 - Prob. 55PCh. 14 - Prob. 56PCh. 14 - Prob. 57PCh. 14 - Prob. 58PCh. 14 - Prob. 59PCh. 14 - Prob. 60PCh. 14 - Prob. 61PCh. 14 - Prob. 62PCh. 14 - Prob. 63PCh. 14 - Prob. 64PCh. 14 - Prob. 65PCh. 14 - Prob. 66PCh. 14 - Prob. 67PCh. 14 - Prob. 68PCh. 14 - Prob. 69PCh. 14 - Prob. 70PCh. 14 - Prob. 71PCh. 14 - Prob. 72PCh. 14 - Prob. 73PCh. 14 - Prob. 74PCh. 14 - Prob. 75PCh. 14 - Prob. 76PCh. 14 - Prob. 77PCh. 14 - Prob. 78PCh. 14 - Prob. 79PCh. 14 - Prob. 80PCh. 14 - Prob. 81PCh. 14 - Prob. 82PCh. 14 - Prob. 83PCh. 14 - Prob. 84PCh. 14 - Prob. 85PCh. 14 - Prob. 86PCh. 14 - Prob. 87PCh. 14 - Prob. 88PCh. 14 - Prob. 89PCh. 14 - Prob. 90PCh. 14 - Prob. 91PCh. 14 - Prob. 92PCh. 14 - Prob. 93PCh. 14 - Prob. 94PCh. 14 - Prob. 95PCh. 14 - Prob. 96PCh. 14 - Prob. 97PCh. 14 - Prob. 98PCh. 14 - Prob. 99PCh. 14 - Prob. 100PCh. 14 - Prob. 101PCh. 14 - Prob. 103PCh. 14 - Prob. 104PCh. 14 - Prob. 105PCh. 14 - Prob. 106P
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