Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
9th Edition
ISBN: 9781305932302
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 15, Problem 47P

(a)

To determine

The frequency of the damped oscillation.

(a)

Expert Solution
Check Mark

Answer to Problem 47P

The frequency of the damped oscillation is 7.00Hz_.

Explanation of Solution

Given that the mass of the oscillating object is 10.65kg, the spring constant is 2.05×104N/m, the damping coefficient is 3.00Ns/m.

Write the expression for the angular frequency of the undamped oscillation.

  ω0=km                                                                                                                  (I)

Here, ω0 is the angular frequency of the undamped oscillator, k is the spring constant, and m is the mass of the oscillator.

Write the expression for the angular frequency of the damped oscillator.

  ω=ω02(b2m)2                                                                                                  (II)

Here, ω is the angular frequency of the damped oscillator, b is the damping coefficient.

Use equation (I) in (II).

  ω=km(b2m)2                                                                                                  (III)

Write the expression for the frequency of the damped oscillator.

  f=ω2π                                                                                                                   (IV)

Here, f is the frequency.

Use equation (III) in (IV).

  f=12πkm(b2m)2                                                                                              (V)

Conclusion:

Substitute 10.65kg for m, 2.05×104N/m for k, and 3.00Ns/m for b in equation (V) to find f.

  f=12π2.05×104N/m10.65kg(3.00Ns/m2(10.65kg))2=6.98Hz7.00Hz

Therefore, the frequency of the damped oscillation is 7.00Hz_.

(b)

To determine

The percentage by which the amplitude of oscillation decrease in each cycle.

(b)

Expert Solution
Check Mark

Answer to Problem 47P

The amplitude of oscillation decrease in each cycle by 2.00%_.

Explanation of Solution

Write the general expression for the damped oscillation.

  x=A0ebt/2mcos(ωt+ϕ)                                                                                         (VI)

Here, x is the displacement, A0 is the initial amplitude, and ϕ is the phase angle.

In one complete cycle (period T), the amplitude changes from A0 to A0ebT/2m. Thus, the fractional decrease in amplitude can be computed as,

  ΔAA0=A0A0ebT/2mA0=1ebT/2m                                                                                             (VII)

Write the expression for the period of the oscillator.

  T=1f                                                                                                                  (VIII)

Conclusion:

Substitute 6.98Hz for f in equation (VIII) to find T.

  T=16.98Hz=0.143s

Substitute 0.143s for T, 3.00Ns/m for b, and 10.65kg for m in equation (VII) to find the fractional decrease in amplitude.

  ΔAA0=1e(3.00Ns/m)(0.143s)/2(10.65kg)=0.0200=2.00%

Therefore, the amplitude of oscillation decrease in each cycle by 2.00%_.

(c)

To determine

The time interval that elapses while the energy of the system drops to 5.00% of its initial value.

(c)

Expert Solution
Check Mark

Answer to Problem 47P

The time interval that elapses while the energy of the system drops to 5.00% of its initial value is 10.6s_.

Explanation of Solution

Given that the mass of the oscillating object is 10.65kg, the spring constant is 2.05×104N/m, the damping coefficient is 3.00Ns/m. Given that the energy of the system drops to 5.00% of its initial value.

Since the energy is proportional to the square of the amplitude, the fractional rate of decrease of energy is twice as fast as the amplitude.

Write the expression for the energy of the damped oscillator.

  E=E0e2bt/2m                                                                                                          (IX)

Here, E is the energy at time t, E0 is the initial energy.

Since the energy of the system drops to 5.00% of its initial value, E=0.05E0. Thus, equation (IX) can be modified as,

  0.05E0=E0e2bt/2m                                                                                                  (X)

Solve equation (X) for t.

  ln10.05=btmt=mbln(10.05)                                                                                            (XI)

Conclusion:

Substitute 3.00Ns/m for b, and 10.65kg for m in equation (XI) to find the time interval t that elapses while the energy of the system drops to 5.00% of its initial value.

  t=(10.65kg3.00Ns/m)ln(10.05)=10.6s

Therefore, the time interval that elapses while the energy of the system drops to 5.00% of its initial value is 10.6s_.

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Chapter 15 Solutions

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term

Ch. 15 - Prob. 5OQCh. 15 - Prob. 6OQCh. 15 - Prob. 7OQCh. 15 - Prob. 8OQCh. 15 - Prob. 9OQCh. 15 - Prob. 10OQCh. 15 - Prob. 11OQCh. 15 - Prob. 12OQCh. 15 - Prob. 13OQCh. 15 - Prob. 14OQCh. 15 - Prob. 15OQCh. 15 - Prob. 16OQCh. 15 - Prob. 17OQCh. 15 - Prob. 1CQCh. 15 - Prob. 2CQCh. 15 - Prob. 3CQCh. 15 - Prob. 4CQCh. 15 - Prob. 5CQCh. 15 - Prob. 6CQCh. 15 - Prob. 7CQCh. 15 - Prob. 8CQCh. 15 - Prob. 9CQCh. 15 - Prob. 10CQCh. 15 - Prob. 11CQCh. 15 - Prob. 12CQCh. 15 - Prob. 13CQCh. 15 - A 0.60-kg block attached to a spring with force...Ch. 15 - Prob. 2PCh. 15 - Prob. 3PCh. 15 - Prob. 4PCh. 15 - The position of a particle is given by the...Ch. 15 - A piston in a gasoline engine is in simple...Ch. 15 - Prob. 7PCh. 15 - Prob. 8PCh. 15 - Prob. 9PCh. 15 - Prob. 10PCh. 15 - Prob. 11PCh. 15 - Prob. 12PCh. 15 - Review. A particle moves along the x axis. It is...Ch. 15 - Prob. 14PCh. 15 - A particle moving along the x axis in simple...Ch. 15 - The initial position, velocity, and acceleration...Ch. 15 - Prob. 17PCh. 15 - Prob. 18PCh. 15 - Prob. 19PCh. 15 - You attach an object to the bottom end of a...Ch. 15 - Prob. 21PCh. 15 - Prob. 22PCh. 15 - Prob. 23PCh. 15 - Prob. 24PCh. 15 - Prob. 25PCh. 15 - Prob. 26PCh. 15 - Prob. 27PCh. 15 - Prob. 28PCh. 15 - A simple harmonic oscillator of amplitude A has a...Ch. 15 - Review. A 65.0-kg bungee jumper steps off a bridge...Ch. 15 - Review. A 0.250-kg block resting on a...Ch. 15 - Prob. 32PCh. 15 - Prob. 33PCh. 15 - A seconds pendulum is one that moves through its...Ch. 15 - A simple pendulum makes 120 complete oscillations...Ch. 15 - A particle of mass m slides without friction...Ch. 15 - A physical pendulum in the form of a planar object...Ch. 15 - Prob. 38PCh. 15 - Prob. 39PCh. 15 - Consider the physical pendulum of Figure 15.16....Ch. 15 - Prob. 41PCh. 15 - Prob. 42PCh. 15 - Prob. 43PCh. 15 - Prob. 44PCh. 15 - A watch balance wheel (Fig. P15.25) has a period...Ch. 15 - Prob. 46PCh. 15 - Prob. 47PCh. 15 - Show that the time rate of change of mechanical...Ch. 15 - Show that Equation 15.32 is a solution of Equation...Ch. 15 - Prob. 50PCh. 15 - Prob. 51PCh. 15 - Prob. 52PCh. 15 - Prob. 53PCh. 15 - Considering an undamped, forced oscillator (b =...Ch. 15 - Prob. 55PCh. 15 - Prob. 56APCh. 15 - An object of mass m moves in simple harmonic...Ch. 15 - Prob. 58APCh. 15 - Prob. 59APCh. 15 - Prob. 60APCh. 15 - Four people, each with a mass of 72.4 kg, are in a...Ch. 15 - Prob. 62APCh. 15 - Prob. 63APCh. 15 - An object attached to a spring vibrates with...Ch. 15 - Prob. 65APCh. 15 - Prob. 66APCh. 15 - A pendulum of length L and mass M has a spring of...Ch. 15 - A block of mass m is connected to two springs of...Ch. 15 - Prob. 69APCh. 15 - Prob. 70APCh. 15 - Review. A particle of mass 4.00 kg is attached to...Ch. 15 - Prob. 72APCh. 15 - Prob. 73APCh. 15 - Prob. 74APCh. 15 - Prob. 75APCh. 15 - Review. A light balloon filled with helium of...Ch. 15 - Prob. 78APCh. 15 - A particle with a mass of 0.500 kg is attached to...Ch. 15 - Prob. 80APCh. 15 - Review. A lobstermans buoy is a solid wooden...Ch. 15 - Prob. 82APCh. 15 - Prob. 83APCh. 15 - A smaller disk of radius r and mass m is attached...Ch. 15 - Prob. 85CPCh. 15 - Prob. 86CPCh. 15 - Prob. 87CPCh. 15 - Prob. 88CPCh. 15 - A light, cubical container of volume a3 is...
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