Chapter 3, Problem 3.1P

(a)

To determine

The natural frequency $v_0$ and the period ${\tau }_0$.

Answer to Problem 3.1P

The natural frequency $v_0$ is $1.6\text{ Hz}$ and the period ${\tau }_0$ is $0.63\text{ sec}$.

Explanation of Solution

Write the formula to find the natural frequency

    $v_0=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$        (I)

Here, $v_0$ is the natural frequency, $k$ is the spring constant and $m$ is the mass.

Write the formula to find the period

    ${\tau }_0=\frac{1}{v_0}$        (II)

Here, ${\tau }_0$ is the period.

Conclusion:

Substitute ${10}^4\text{dyne}/\text{cm}$ for $k$ and ${10}^2\text{ gram}$ for $m$ in equation (I) to find the value of $v_0$

$\begin{array}{c}{v}_0=\frac{1}{2\pi }\sqrt{\frac{{10}^4\text{dyne}/\text{cm}}{{10}^2\text{ gram}}}\\ =\frac{1}{2\pi }\sqrt{\frac{{10}^4\text{gram}\cdot \text{cm}/{\text{sec}}^2\cdot \text{cm}}{{10}^2\text{ gram}}}\\ =\frac{1}{2\pi }\sqrt{{10}^2{\text{ sec}}^{-2}}\\ =\frac{10}{2\pi }{\text{ sec}}^{-1}\\ =1.6\text{ Hz}\end{array}$

Thus, the natural frequency is $1.6\text{ Hz}$.

Substitute $1.6\text{ Hz}$ for $v_0$ in equation (II) to find the value of ${\tau }_0$

$\begin{array}{c}{\tau }_0=\frac{1}{1.6\text{ Hz}}\\ =0.63\text{ sec}\end{array}$

Thus, the period is $0.63\text{ sec}$.

(b)

To determine

The total energy.

Answer to Problem 3.1P

The total energy is $4.5\times {10}^4\text{ erg}$.

Explanation of Solution

Write the formula to find the energy

    $E=\frac{1}{2}k{A}^2$        (III)

Here, $E$ is the kinetic energy of the system, $A$ is the distance the mass moved.

Conclusion:

Substitute ${10}^4\text{dyne}/\text{cm}$ for $k$ and $3\text{cm}$ for $m$ in equation (III) to find the value of $E$

$\begin{array}{c}E=\frac{1}{2}\times {10}^4\text{dyne}/\text{cm}\times {\left(3\text{cm}\right)}^2\\ =4.5\times {10}^4\text{ erg}\end{array}$

Thus, the total energy is $4.5\times {10}^4\text{ erg}$.

(c)

To determine

The maximum speed.

Answer to Problem 3.1P

The maximum speed is $30\text{cm}/\text{sec}$.

Explanation of Solution

The maximum velocity of the system is attained when the total energy of the system is equal to the kinetic energy of the system.

Write the formula to find the kinetic energy

    $E=\frac{1}{2}m{v}^2$        (IV)

Here, $v$ is the maximum velocity.

Conclusion:

Substitute $4.5\times {10}^4\text{ erg}$ for $E$ and $100\text{ gram}$ for $m$ in equation (IV) to find the value of $v_{\max }$

$\begin{array}{c}\frac{1}{2}m{v}_{\max }^2=E\\ v_{\max }=\sqrt{\frac{2E}{m}}\\ =\sqrt{\frac{2\times 4.5\times {10}^4\text{ erg}}{100\text{ gram}}}\\ =30\text{cm}/\text{sec}\end{array}$

Thus, the maximum velocity is $30\text{cm}/\text{sec}$.