Chapter 15, Problem 47P

(a)

To determine

The frequency of the damped oscillation.

Answer to Problem 47P

The frequency of the damped oscillation is $\underset{\_}{7.00\text{Hz}}$.

Explanation of Solution

Given that the mass of the oscillating object is $10.65\text{kg}$, the spring constant is $2.05\times {10}^4\text{N/m}$, the damping coefficient is $3.00\text{N}\cdot \text{s/m}$.

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

  ${\omega }_0=\sqrt{\frac{k}{m}}$                                                                                                                  (I)

Here, ${\omega }_0$ 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.

  $\omega =\sqrt{{\omega }_0^2-{\left(\frac{b}{2m}\right)}^2}$                                                                                                  (II)

Here, $\omega$ is the angular frequency of the damped oscillator, $b$ is the damping coefficient.

Use equation (I) in (II).

  $\omega =\sqrt{\frac{k}{m}-{\left(\frac{b}{2m}\right)}^2}$                                                                                                  (III)

Write the expression for the frequency of the damped oscillator.

  $f=\frac{\omega }{2\pi }$                                                                                                                   (IV)

Here, $f$ is the frequency.

Use equation (III) in (IV).

  $f=\frac{1}{2\pi }\sqrt{\frac{k}{m}-{\left(\frac{b}{2m}\right)}^2}$                                                                                              (V)

Conclusion:

Substitute $10.65\text{kg}$ for $m$, $2.05\times {10}^4\text{N/m}$ for $k$, and $3.00\text{N}\cdot \text{s/m}$ for $b$ in equation (V) to find $f$.

  $\begin{array}{c}f=\frac{1}{2\pi }\sqrt{\frac{2.05\times {10}^4\text{N/m}}{10.65\text{kg}}-{\left(\frac{3.00\text{N}\cdot \text{s/m}}{2\left(10.65\text{kg}\right)}\right)}^2}\\ =6.98\text{Hz}\\ \approx 7.00\text{Hz}\end{array}$

Therefore, the frequency of the damped oscillation is $\underset{\_}{7.00\text{Hz}}$.

(b)

To determine

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

Answer to Problem 47P

The amplitude of oscillation decrease in each cycle by $\underset{\_}{2.00\%}$.

Explanation of Solution

Write the general expression for the damped oscillation.

  $x=A_0{e}^{-bt/2m}\cos \left(\omega t+\varphi \right)$                                                                                         (VI)

Here, $x$ is the displacement, $A_0$ is the initial amplitude, and $\varphi$ is the phase angle.

In one complete cycle (period $T$), the amplitude changes from $A_0$ to $A_0{e}^{-bT/2m}$. Thus, the fractional decrease in amplitude can be computed as,

  $\begin{array}{c}\frac{\Delta A}{A_0}=\frac{A_0-A_0{e}^{-bT/2m}}{A_0}\\ =1-e^{-bT/2m}\end{array}$                                                                                             (VII)

Write the expression for the period of the oscillator.

  $T=\frac{1}{f}$                                                                                                                  (VIII)

Conclusion:

Substitute $6.98\text{Hz}$ for $f$ in equation (VIII) to find $T$.

  $\begin{array}{c}T=\frac{1}{6.98\text{Hz}}\\ =0.143\text{s}\end{array}$

Substitute $0.143\text{s}$ for $T$, $3.00\text{N}\cdot \text{s/m}$ for $b$, and $10.65\text{kg}$ for $m$ in equation (VII) to find the fractional decrease in amplitude.

  $\begin{array}{c}\frac{\Delta A}{A_0}=1-e^{-\left(3.00\text{N}\cdot \text{s/m}\right)\left(0.143\text{s}\right)/2\left(10.65\text{kg}\right)}\\ =0.0200\\ =2.00\%\end{array}$

Therefore, the amplitude of oscillation decrease in each cycle by $\underset{\_}{2.00\%}$.

(c)

To determine

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

Answer to Problem 47P

The time interval that elapses while the energy of the system drops to $5.00\%$ of its initial value is $\underset{\_}{10.6\text{s}}$.

Explanation of Solution

Given that the mass of the oscillating object is $10.65\text{kg}$, the spring constant is $2.05\times {10}^4\text{N/m}$, the damping coefficient is $3.00\text{N}\cdot \text{s/m}$. 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.

  $E=E_0{e}^{-2bt/2m}$                                                                                                          (IX)

Here, $E$ is the energy at time $t$, $E_0$ is the initial energy.

Since the energy of the system drops to $5.00\%$ of its initial value, $E=0.05E_0$. Thus, equation (IX) can be modified as,

  $0.05E_0=E_0{e}^{-2bt/2m}$                                                                                                  (X)

Solve equation (X) for $t$.

  $\begin{array}{c}\ln \frac{1}{0.05}=\frac{bt}{m}\\ t=\frac{m}{b}\ln \left(\frac{1}{0.05}\right)\end{array}$                                                                                            (XI)

Conclusion:

Substitute $3.00\text{N}\cdot \text{s/m}$ for $b$, and $10.65\text{kg}$ 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.

  $\begin{array}{c}t=\left(\frac{10.65\text{kg}}{3.00\text{N}\cdot \text{s/m}}\right)\ln \left(\frac{1}{0.05}\right)\\ =10.6\text{s}\end{array}$

Therefore, the time interval that elapses while the energy of the system drops to $5.00\%$ of its initial value is $\underset{\_}{10.6\text{s}}$.