Chapter 7, Problem 7.67CP

Review. A light spring has unstressed length 15.5 cm. It is described by Hooke’s law with spring constant 4.30 N/m. One end of the horizontal spring is held on a fixed vertical axle, and the other end is attached to a puck of mass m that can mow without friction over a horizontal surface. The puck is set into motion in a circle with a period of 1.30 s. (a) Find the extension of the spring x as it depends on m. Evaluate x for (b) m = 0.070 0 kg. (c) m = 0.140 kg, (d) m = 0.180 kg, and (e) m = 0.190 kg. (f) Describe the pattern of variation of x as it depends on m.

To determine

The extension of the spring as it depends on $m$.

Answer to Problem 7.67CP

The extension of the spring is $\frac{\left(3.62m\right)\text{m}}{\left(4.30\text{kg}-23.4m\right)}$.

Explanation of Solution

Given info: The unscratched length of spring is $15.5\text{cm}$, the spring constant of the spring is $4.30\text{N}/\text{m}$ and time period of puck is $1.30\text{s}$.

The extension of spring is $x$ so the total length of spring is,

$d_{\text{total}}=d+x$

Here,

$d$ is the length of unscratched spring.

From Newton’s law, the force of an object is,

$F=ma$

Here,

$m$ is the mass of the object.

$a$ is the acceleration of the object.

From the Hooke’s law, the spring force of spring is,

$F=kx$

Here,

$k$ is the spring constant.

$x$ is the length of the spring.

Substitute $ma$ for $F$ in the above equation.

$\begin{array}{c}ma=kx\\ x=\frac{ma}{k}\end{array}$ (1)

The expression for the acceleration of the object is,

$a=\frac{v}{t}$ (2)

Here,

$v$ is the speed of the octopus.

$t$ is the time taken by the octopus.

The formula for the speed of the object is,

$v=\frac{d_{\text{total}}}{t}$

Rearrange the above equation.

$t=\frac{d_{\text{total}}}{v}$

Substitute $\frac{d_{\text{total}}}{v}$ for $t$ in equation (2).

$\begin{array}{c}a=\frac{v}{\left(\frac{d_{\text{total}}}{v}\right)}\\ =\frac{v^2}{d_{\text{total}}}\end{array}$ (3)

The speed of an object in term of oscillation is,

$v=d_{\text{total}}\omega$

Here,

$\omega$ is the oscillation frequency

Substitute $\frac{2\pi }{T}$ for $\omega$ in the above equation.

$v=d_{\text{total}}\left(\frac{2\pi }{T}\right)$

Here,

$T$ is the time period.

Substitute $d_{\text{total}}\left(\frac{2\pi }{T}\right)$ for $v$ in equation (3).

$\begin{array}{c}a=\frac{{\left(d_{\text{total}}\left(\frac{2\pi }{T}\right)\right)}^2}{d_{\text{total}}}\\ =d_{\text{total}}\frac{4{\pi }^2}{T^2}\end{array}$

Substitute $d_{\text{total}}\frac{4{\pi }^2}{T^2}$ for $a$ in equation (1).

$x=\frac{m\left(d_{\text{total}}\frac{4{\pi }^2}{T^2}\right)}{k}$

Substitute $d+x$ for $d_{\text{total}}$ in the above equation.

$\begin{array}{l}x=\frac{m\left(\left(d+x\right)\frac{4{\pi }^2}{T^2}\right)}{k}\\ kx=\frac{4{\pi }^{2}md}{T^2}+\frac{4{\pi }^{2}mx}{T^2}\\ x=\frac{\frac{4{\pi }^{2}md}{T^2}}{k-\frac{4{\pi }^{2}m}{T^2}}\end{array}$

Substitute $15.5\text{cm}$ for $d$, $4.30\text{N}/\text{m}$ for $k$ and $1.30\text{s}$ for $T$ in the above equation.

$\begin{array}{l}x=\frac{\frac{4{\pi }^{2}m\left(15.5\text{cm}\right)}{{\left(1.30\text{s}\right)}^2}}{4.30\text{N}/\text{m}-\frac{4{\pi }^{2}m}{{\left(1.30\text{s}\right)}^2}}\\ =\frac{\frac{4{\pi }^{2}m\left(15.5\times {10}^{-2}\text{m}\right)}{{\left(1.30\text{s}\right)}^2}}{\frac{\left(4.30\text{N}/\text{m}\right){\left(1.30\text{s}\right)}^2-4{\pi }^{2}m}{{\left(1.30\text{s}\right)}^2}}\\ =\frac{\left(3.62m\right)\text{kg}\cdot \text{m}}{\left(\left(4.30\text{kg}\right)-\left(23.4m\right)\text{kg}\right)}\end{array}$

Conclusion:

Therefore, the extension of the spring is $\frac{\left(3.62m\right)\text{kg}\cdot \text{m}}{\left(\left(4.30\text{kg}\right)-\left(23.4m\right)\text{kg}\right)}$.

To determine

The extension in spring for $m=0.070\text{kg}$.

Answer to Problem 7.67CP

The extension in spring is $0.0951\text{m}$.

Explanation of Solution

Given info: The mass of puck is $0.070\text{kg}$.

From part (a), the extension in spring is,

$\frac{\left(3.62m\right)\text{kg}\cdot \text{m}}{\left(\left(4.30\text{kg}\right)-\left(23.4m\right)\text{kg}\right)}$

Substitute $0.070\text{kg}$ for $m$ in the above equation.

$\begin{array}{c}x=\frac{\left(3.62\left(0.070\text{kg}\right)\right)\text{m}}{\left(4.30\text{kg}-23.4\left(0.070\text{kg}\right)\right)}\\ \approx 0.0951\text{m}\end{array}$

Conclusion:

Therefore, the extension in spring is $0.0951\text{m}$.

To determine

The extension in spring for $m=0.140\text{kg}$.

Answer to Problem 7.67CP

The extension in spring is $0.492\text{m}$.

Explanation of Solution

Given info: The mass of puck is $0.140\text{kg}$.

From part (a), the extension in spring is,

$\frac{\left(3.62m\right)\text{kg}\cdot \text{m}}{\left(\left(4.30\text{kg}\right)-\left(23.4m\right)\text{kg}\right)}$

Substitute $0.140\text{kg}$ for $m$ in the above equation.

$\begin{array}{c}x=\frac{\left(3.62\left(0.140\text{kg}\right)\right)\text{m}}{\left(4.30\text{kg}-23.4\left(0.140\text{kg}\right)\right)}\\ \approx 0.492\text{m}\end{array}$

Conclusion:

Therefore, the extension in spring is $0.492\text{m}$.

To determine

The extension in spring for $m=0.180\text{kg}$.

Answer to Problem 7.67CP

The extension in spring is $6.85\text{m}$.

Explanation of Solution

Given info: The mass of puck is $0.180\text{kg}$.

From part (a), the extension in spring is,

$\frac{\left(3.62m\right)\text{kg}\cdot \text{m}}{\left(\left(4.30\text{kg}\right)-\left(23.4m\right)\text{kg}\right)}$

Substitute $0.180\text{kg}$ for $m$ in the above equation.

$\begin{array}{c}x=\frac{\left(3.62\left(0.180\text{kg}\right)\right)\text{m}}{\left(4.30\text{kg}-23.4\left(0.180\text{kg}\right)\right)}\\ \approx 6.85\text{m}\end{array}$

Conclusion:

Therefore, the extension in spring is $6.85\text{m}$.

To determine

The extension in spring for $m=0.190\text{kg}$.

Answer to Problem 7.67CP

The situation for extension in spring for $m=0.190\text{kg}$ is impossible.

Explanation of Solution

Given info: The mass of puck is $0.190\text{kg}$.

From part (a), the extension in spring is,

$\frac{\left(3.62m\right)\text{kg}\cdot \text{m}}{\left(\left(4.30\text{kg}\right)-\left(23.4m\right)\text{kg}\right)}$

Substitute $0.190\text{kg}$ for $m$ in the above equation.

$\begin{array}{c}x=\frac{\left(3.62\left(0.190\text{kg}\right)\right)\text{m}}{\left(4.30\text{kg}-23.4\left(0.190\text{kg}\right)\right)}\\ \approx -4.96\text{m}\end{array}$

The extension in spring is the distance of spring after extension and distance cannot measure in left side $x$-axis that means distance cannot be negative.

Conclusion:

Therefore, the situation for extension in spring for $m=0.190\text{kg}$ is impossible.

To determine

The variation of $x$ as it depends on $m$.

Answer to Problem 7.67CP

The value of $x$ is diverge to infinity for $m=0.184\text{kg}$.

Explanation of Solution

From part (a), the extension in spring is,

$\frac{\left(3.62m\right)\text{kg}\cdot \text{m}}{\left(\left(4.30\text{kg}\right)-\left(23.4m\right)\text{kg}\right)}$

From the above expression, it is shown that the extension in spring is directly proportional to the mass of puck so the extension in spring increases as $m$ increases.

After a certain point the extension of spring diverge to infinity.

The mass of puck when the extension of spring diverges to infinity,

$\begin{array}{c}\frac{1}{0}=\frac{3.62m}{\left(4.30\text{kg}-23.4m\right)}\\ 4.30\text{kg}-23.4m=0\\ m=\frac{4.30\text{kg}}{23.4}\\ m\approx 0.184\text{kg}\end{array}$

Conclusion:

Therefore, the value of $x$ is diverge to infinity for $m=0.184\text{kg}$.