Chapter 12, Problem 46P

(a)

To determine

The position of $x$ at a moment $84.45\text{s}$ later.

Answer to Problem 46P

The position of $x$ at a moment $84.45\text{s}$ later is $\underset{\_}{15.8\text{cm}}$

Explanation of Solution

The general expression for the position of the object is given by,

    $x=A\cos \omega t$        (I)

Here, $x$ is the position of the object, $A$ is the amplitude, $\omega$ is the angular frequency, $t$ is the time.

Write the expression for the force due to a spring.

    $F=kx$        (II)

Here, $F$ is the force acting due to spring, $k$ is the spring constant, $x$ is the compression distance.

Write the expression for the force acting on an object.

    $F=mg$        (III)

Here, $g$ is the acceleration due to gravity, $m$ is the mass of the object.

Equate equation (II) and (III) and solve for $k$.

    $\begin{array}{c}kx=mg\\ k=\frac{mg}{k}\end{array}$        (IV)

Take the $x$ axis as pointing downward,

Write the expression for the angular frequency.

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

Conclusion:

Substitute $450\text{g}$ for $m$ and $35.0\text{cm}$ for $x$ in equation (IV) to find $k$.

    $\begin{array}{c}k=\frac{\left(450\text{g}\times \frac{1\text{kg}}{{10}^3\text{g}}\right)9.8\text{m}/{\text{s}}^2}{35.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}}\\ =12.6\text{N}/\text{m}\end{array}$

Substitute $12.6\text{N}/\text{m}$ for $k$, $450\text{g}$ for $m$ in equation (V) to find $\omega$.

    $\begin{array}{c}\omega =\sqrt{\frac{12.6\text{N}/\text{m}}{450\text{g}\times \frac{1\text{kg}}{{10}^3\text{g}}}}\\ =5.29\text{rad}/\text{s}\end{array}$

Substitute $18.0\text{cm}$ for $A$, $5.29$ for $\omega$ and $84.4\text{s}$ for $t$ in equation (I) to find $x$.

    $\begin{array}{c}x=\left(18.0\text{cm}\right)\cos \left(5.29\text{rad}/\text{s}\right)\left(84.4\text{s}\right)\\ =15.8\text{cm}\end{array}$

Therefore, The position of $x$ at a moment $84.45\text{s}$ later is $\underset{\_}{15.8\text{cm}}$

(b)

To determine

The distance traveled by the vibrating object in part (a).

Answer to Problem 46P

The distance traveled by the vibrating object in part (a) is $\underset{\_}{51.1\text{m}}$.

Explanation of Solution

The distance traveled is equal to the number of oscillations multiplied by the distance traveled in a single time period.

The time period is given by,

    $T=\frac{2\pi }{\omega }$        (VI)

Here, $T$ is the time period.

The number of oscillations made by the spring is given by,

    $n=\frac{t}{T}$        (VI)

Here, $t$ is the total time of the oscillations.

The distance travelled by the mass in one complete oscillation is $4A$.

The total distance traveled in $84.4\text{s}$ is,

    $d=4An$        (VII)

Conclusion:

Substitute $5.29\text{rad}$ for $\omega$ in equation (VI) to find $T$.

    $\begin{array}{c}T=\frac{2\pi }{5.29}\\ =1.1874\text{s}\end{array}$

Substitute $1.1874\text{s}$ for $T$, $84.4\text{s}$ for $t$ in equation (VI) to find $n$.

    $\begin{array}{c}n=\frac{84.4\text{s}}{1.1874\text{s}}\\ =71.079\end{array}$

Substitute e $0.18\text{m}$ for $A$ and $71.079$ for $n$ in equation (

  $\begin{array}{c}d=4\left(0.18\text{m}\right)\left(71.079\right)\\ =51.1769\text{m}\\ \text{=51}\text{.2}\text{m}\end{array}$

Therefore, The distance traveled by the vibrating object in part (a) is $\underset{\_}{51.2\text{m}}$.

(c)

To determine

The position of the object $84.4\text{s}$ later when an object of mass $440\text{g}$ is bung on it at rest.

Answer to Problem 46P

The position of the object $84.4\text{s}$ later when an object of mass $440\text{g}$ is bung on it at rest is $\underset{\_}{-15.9\text{cm}}$

Explanation of Solution

Substitute $0.440\text{g}$ for $m$, $9.80\text{m}/{\text{s}}^2$ for $g$ and $35.5\text{cm}$ for $x$ in equation (IV) to find $k$.

        $\begin{array}{c}k=\frac{\left(440\text{g}\times \frac{1\text{kg}}{{10}^3\text{g}}\right)9.8\text{m}/{\text{s}}^2}{35.5\text{cm}\times \frac{1\text{m}}{100\text{cm}}}\\ =12.1\text{N}/\text{m}\end{array}$

Substitute $12.1\text{N}/\text{m}$ for $k$, $440\text{g}$ for $m$ in equation (V) to find $\omega$.

    $\begin{array}{c}\omega =\sqrt{\frac{12.1\text{N}/\text{m}}{440\text{g}\times \frac{1\text{kg}}{{10}^3\text{g}}}}\\ =5.244\text{rad}/\text{s}\end{array}$

Substitute $18.0\text{cm}$ for $A$, $5.244$ for $\omega$ and $84.4\text{s}$ for $t$ in equation (I) to find $x$.

    $\begin{array}{c}x=\left(18.0\text{cm}\right)\cos \left(5.24\text{rad}/\text{s}\right)\left(84.4\text{s}\right)\\ =-15.9\text{cm}\end{array}$

Therefore, the position of $x$ at a moment $84.45\text{s}$ later is $\underset{\_}{-15.9\text{cm}}$

Conclusion:

Therefore, the position of the object $84.4\text{s}$ later when an object of mass $440\text{g}$ is bung on it at rest is $\underset{\_}{-15.9\text{cm}}$

(d)

To determine

The distance traveled by the object in part (c)

Answer to Problem 46P

The distance traveled by the object in part (c) is $\underset{\_}{50.8\text{cm}}$.

Explanation of Solution

The distance traveled is equal to the number of oscillations multiplied by the distance traveled in a single time period.

The time period is given by,

    $T=\frac{2\pi }{\omega }$

The number of oscillations made by the spring is given by,

    $n=\frac{t}{T}$

The distance travelled by the mass in one complete oscillation is $4A$.

The total distance traveled in $84.4\text{s}$ is,

    $d=4An$

Conclusion:

Substitute $5.244\text{rad}$ for $\omega$ in equation (VI) to find $T$.

    $\begin{array}{c}T=\frac{2\pi }{5.24}\\ =1.199\text{s}\end{array}$

Substitute $1.199\text{s}$ for $T$, $84.4\text{s}$ for $t$ in equation (VI) to find $n$.

    $\begin{array}{c}n=\frac{84.4\text{s}}{1.199\text{s}}\\ =70.39\end{array}$

Substitute e $0.18\text{m}$ for $A$ and $71.079$ for $n$ in equation (

  $\begin{array}{c}d=4\left(0.18\text{m}\right)\left(70.39\right)\\ =50.8\text{m}\end{array}$

Therefore, the distance traveled by the vibrating object in part (c) is $\underset{\_}{50.8\text{m}}$.

(e)

To determine

The reason for the different answers to part (a) and (c) when the initial data in parts (a) and (c) are so similar and the answers to parts (b) and (d) are relatively close.

Answer to Problem 46P

The answers in parts a and b are different because of the difference in angular velocities.

Explanation of Solution

Diverging patterns of oscillations which starts out in phase but becoming completely out of phase changes the answers in part (a) and (b). Since the angular velocity is changing it will lead to the change in the position of the phase. But this phase difference is higher when compared to the angular velocity difference as it is multiplied by time also. Thus it yields larger change in the position.

For parts c and d, the distance traveled depends only on the angular velocity such that their difference is not so large.