Chapter 15, Problem 15.12P

(a) A hanging spring stretches by 35.0 cm when an object of mass 450 g is hung on it at rest. In this situation, we define its position as x = 0. The object is pulled down an additional 18.0 cm and released from rest to oscillate without friction. What is its position x at a moment 84.4 s later? (b) Find the distance traveled by the vibrating object in part (a), (c) What If? Another hanging spring stretches by 35.5 cm when an object of mass 440 g is hung on it at rest. We define this new position as x = 0. This object is also pulled down an additional 18.0 cm and released from rest to oscillate without friction. Find its position 84.4 s later, (d) Find the distance traveled by the object in part (c). (e) Why are the answers to parts (a) and (c) so different when the initial data in parts (a) and (c) are so similar and the answers to parts (b) and (d) are relatively close? Does this circumstance reveal a fundamental difficulty in calculating the future?

To determine

The position $x$ of the object after the time of $84.4\text{s}$.

Answer to Problem 15.12P

The position $x$ of the object is $15.8\text{cm}$.

Explanation of Solution

Given info: The distance through which spring stretches is $35.0\text{cm}$, the mass of an object attached to the spring is $450\text{g}$, the position of the mass initially $x$ is $0$, the additional distance through which object is pulls the spring is $18.0\text{cm}$, the time is $84.4\text{s}$.

Write the equation of position of an object attached to a spring.

$x=A\cos \left(\omega t\right)$ (1)

Here,

$x$ is the position of an object.

$A$ is the amplitude which is the maximum distance at which the object reaches.

$\omega$ is the angular frequency.

$t$ is the time.

Write the equation for angular frequency.

$\omega =\sqrt{\frac{k}{m}}$ (2)

Here,

$k$ is the force constant of the spring.

$m$ is the mass of an object.

Write the equation of force generated due to stretch in spring.

$F=kx\text{'}$

Here,

$x\text{'}$ is the distance through which the spring stretches.

Write the equation of force due to the weight of the object.

$F=mg$

Here,

$g$ is the acceleration due to gravity.

Substitute $kx\text{'}$ for $F$ in above equation to find

$\begin{array}{c}kx\text{'}=mg\\ k=\frac{mg}{x\text{'}}\end{array}$

Substitute $\frac{mg}{x\text{'}}$ for $k$ in equation (2) to find $\omega$.

$\begin{array}{c}\omega =\sqrt{\frac{\left(\frac{mg}{x\text{'}}\right)}{m}}\\ \omega =\sqrt{\frac{g}{x\text{'}}}\end{array}$

Substitute $\sqrt{\frac{g}{x\text{'}}}$ for $\omega$ in equation (1) to find $x$.

$x=A\cos \left(\sqrt{\frac{g}{x\text{'}}}t\right)$ (3)

Substitute $18.0\text{cm}$ for $A$, $9.81\text{m}/{\text{s}}^2$ for $g$, $35.0\text{cm}$ for $x\text{'}$ and $84.4\text{s}$ for $t$ in above equation to find $x$.

$\begin{array}{c}x=\left(18.0\text{cm}\right)\cos \left(\sqrt{\frac{\left(9.81\text{m}/{\text{s}}^2\right)}{\left\{35.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right\}}}\left(84.4\text{s}\right)\right)\\ =\left(18.0\text{cm}\right)\cos \left(446.6\text{rad}\right)\\ =15.8\text{cm}\end{array}$

Conclusion:

Therefore, the position $x$ of the object is $15.8\text{cm}$.

To determine

The distance travelled by the object in part (a).

Answer to Problem 15.12P

The distance travelled by the object is $51.14\text{m}$.

Explanation of Solution

Given info: The distance through which spring stretches is $35.0\text{cm}$, the mass of an object attached to the spring is $450\text{g}$, the position of the mass initially $x$ is $0$, the additional distance through which object is pulls the spring is $18.0\text{cm}$, the time is $84.4\text{s}$.

Write the equation of distance travelled by the object for complete oscillation.

$d=Dn+\left(A-x\right)$

Here,

$d$ is the distance travelled by the object for complete oscillation.

$D$ is the total distance travelled by the object in one oscillation.

$n$ is the number of complete oscillation.

The total distance $D$ travelled by the object in one oscillation is four times the amplitude.

Substitute $4A$ for $D$ in above equation to find $d$.

$d=4An+\left(A-x\right)$ (4)

The number of complete oscillation for angular displacement of $446.6\text{rad}$ is,

$\begin{array}{c}n=\frac{446.6\text{rad}}{2\pi \text{rad}}\\ =71\end{array}$

Substitute   $18.0\text{cm}$ for $A$, $71$ for $n$ and $15.8\text{cm}$ for $x$ in equation (4) to find $d$.

$\begin{array}{c}d=4\left(18.0\text{cm}\right)\left(71\right)+\left(18.0\text{cm}-15.8\text{cm}\right)\\ =5114.2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ \approx 51.14\text{m}\end{array}$

Conclusion:

Therefore, the distance travelled by the object is $51.14\text{m}$.

To determine

The position of the object.

Answer to Problem 15.12P

The position $x$ of the object is $-15.9\text{cm}$.

Explanation of Solution

Given info: The distance through which spring stretches is $35.0\text{cm}$, the mass of an object attached to the spring is $450\text{g}$, the position of the mass initially $x$ is $0$, the additional distance through which object is pulls the spring is $18.0\text{cm}$, the time is $84.4\text{s}$, the distance through which another hanging spring stretches is $35.5\text{cm}$, the another mass attached to this spring at rest is $440\text{g}$, the new position $x$ is $0$, the additional stretch is $18.0\text{cm}$, the time after this condition is $84.4\text{s}$.

Substitute $18.0\text{cm}$ for $A$, $9.81\text{m}/{\text{s}}^2$ for $g$, $35.5\text{cm}$ for $x\text{'}$ and $84.4\text{s}$ for $t$ in equation (3) to find $x$.

$\begin{array}{c}x=\left(18.0\text{cm}\right)\cos \left(\sqrt{\frac{\left(9.81\text{m}/{\text{s}}^2\right)}{\left\{35.5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right\}}}\left(84.4\text{s}\right)\right)\\ =\left(18.0\text{cm}\right)\cos \left(443.4\text{rad}\right)\\ =-15.9\text{cm}\end{array}$

Conclusion:

Therefore, the position $x$ of the object is $-15.9\text{cm}$.

To determine

The distance travelled by the object in part (c).

Answer to Problem 15.12P

The distance travelled by the object is $50.8\text{cm}$.

Explanation of Solution

Given info: The distance through which spring stretches is $35.0\text{cm}$, the mass of an object attached to the spring is $450\text{g}$, the position of the mass initially $x$ is $0$, the additional distance through which object is pulls the spring is $18.0\text{cm}$, the time is $84.4\text{s}$, the distance through which another hanging spring stretches is $35.5\text{cm}$, the another mass attached to this spring at rest is $440\text{g}$, the new position $x$ is $0$, the additional stretch is $18.0\text{cm}$, the time after this condition is $84.4\text{s}$.

Write the equation of distance travelled by the object for complete oscillation.

$d=4An$ (5)

The number of complete oscillation for angular displacement of $443.67\text{rad}$ is,

$\begin{array}{c}n=\frac{443.67\text{rad}}{2\pi \text{rad}}\\ =70.61\end{array}$

Substitute  $18.0\text{cm}$ for $A$ and $70.61$ for in equation (5) to find $d$.

$\begin{array}{c}d=4\left(18.0\text{cm}\right)\left(70.61\right)\\ =5083.92\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ \approx 50.8\text{m}\end{array}$

Conclusion:

Therefore, the distance travelled by the object is $50.8\text{cm}$.

To determine

The reason that the answer of parts (a) and (c) is different when the initial data in parts (a) and (c) are similar and the answers of parts (b) and (d) are relatively close and whether this circumstance give the fundamental difficulty.

Answer to Problem 15.12P

The oscillation patterns diverge from each other, starts out in phase and becomes out of phase completely. It is impossible to make future predictions with the known data.

Explanation of Solution

Given info: The distance through which spring stretches is $35.0\text{cm}$, the mass of an object attached to the spring is $450\text{g}$, the position of the mass initially $x$ is $0$, the additional distance through which object is pulls the spring is $18.0\text{cm}$, the time is $84.4\text{s}$, the distance through which another hanging spring stretches is $35.5\text{cm}$, the another mass attached to this spring at rest is $440\text{g}$, the new position $x$ is $0$, the additional stretch is $18.0\text{cm}$, the time after this condition is $84.4\text{s}$.

The pattern of oscillation diverges completely from each other even if the initial data are same. Starts out in phase initially, but becomes completely out of phase.

To calculate the future predictions, exact data of the present is required which is impossible. It is difficult to make prediction with the known data.

Thus, the oscillation patterns diverge from each other, starts out in phase and becomes out of phase completely. It is impossible to make future predictions with the known data.

Conclusion:

Therefore, the oscillation patterns diverge from each other, starts out in phase and becomes out of phase completely. It is impossible to make future predictions with the known data.