Chapter 6, Problem 65P

(a)

To determine

The extension of the spring for a mass of $m$.

Answer to Problem 65P

The extension of the spring for a mass of $m$ is $\underset{\_}{x=\frac{4{\pi }^{2}m{L}_0/T^2}{k-4{\pi }^{2}m{L}_0/T^2}}$.

Explanation of Solution

Write the expression for centripetal force of the mass $m$ attached at the end of the spring.

    $F=\frac{m{v}^2}{r}$        (I)

Here, $F$ is the centripetal force, $m$ is the mass, $v$ is the velocity, $r$ is the radius of the orbit.

Write the expression for velocity in terms of time period.

    $v=\frac{2\pi r}{T}$        (II)

Here, $v$ is the velocity, $r$ is the radius of the orbit, and $T$ is the time period.

Write the expression for force from hooks law.

    $F=kx$        (III)

Here, $F$ is the force, $k$ is the spring constant, and $x$ is the distance.

Use equation (II) and (III) in equation (I) and rearrange.

    $\begin{array}{c}kx=\frac{m{\left(\frac{2\pi r}{T}\right)}^2}{r}\\ k{T}^{2}x=4{\pi }^{2}mr\end{array}$        (IV)

Write the expression for radius of the pluck’s motion.

    $r=L_0+x$        (V)

Use equation (V) in equation (IV), to find $x$.

    $\begin{array}{c}k{T}^{2}x=4{\pi }^{2}m\left(L_0+x\right)\\ kx=\frac{\left(4{\pi }^{2}m{L}_0\right)}{T^2}+\frac{x\left(4{\pi }^{2}m\right)}{T^2}\\ x\left(k-\frac{4{\pi }^{2}m}{T^2}\right)=\frac{4{\pi }^{2}m{L}_0}{T^2}\\ x=\frac{4{\pi }^{2}m{L}_0/T^2}{k-4{\pi }^{2}m{L}_0/T^2}\end{array}$        (VI)

Conclusion:

Therefore, the extension of the spring for a mass of $m$ is $\underset{\_}{x=\frac{4{\pi }^{2}m{L}_0/T^2}{k-4{\pi }^{2}m{L}_0/T^2}}$.

(b)

To determine

The extension of the spring for the mass $0.0700\text{kg}$.

Answer to Problem 65P

The extension of the spring for the mass $0.0700\text{kg}$ is $\underset{\_}{0.0951\text{m}}$.

Explanation of Solution

Substitute $4.30\text{N}/\text{m}$ for $k$, $0.155\text{m}$ for $L_0$, $1.30\text{s}$ for $T$, in equation (VI), to find $x$.

    $\begin{array}{c}x=\frac{4{\left(3.14\right)}^2\left(0.155\text{m}\right)m/{\left(1.30\text{s}\right)}^2}{4.30\text{N}/\text{m}-4{\left(3.14\right)}^{2}m\left(0.155\text{m}\right)/{\left(1.30\text{s}\right)}^2}\\ =\frac{\left(3.62\text{m}/{\text{s}}^2\right)m}{4.30\text{kg}/{\text{s}}^2-\left(23.361/{\text{s}}^2\right)m}\\ =\frac{\left(3.62\text{m}\right)m}{\left[4.30\text{kg}-\left(23.36\right)\right]m}\frac{1/{\text{s}}^2}{1/{\text{s}}^2}\\ =\frac{\left(3.62\text{m}\right)m}{4.30\text{kg}-\left(23.4\right)m}\end{array}$        (VII)

Conclusion:

Substitute $0.070\text{kg}$ for $m$ in equation (VII), to find $x$.

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

Therefore, the extension of the spring for the mass $0.0700\text{kg}$ is $\underset{\_}{0.0951\text{m}}$.

(c)

To determine

The extension of the spring for the mass $0.140\text{kg}$.

Answer to Problem 65P

The extension of the spring for the mass $0.140\text{kg}$ is $\underset{\_}{0.492\text{m}}$.

Explanation of Solution

From equation (VII).

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

Conclusion:

Substitute $0.140\text{kg}$ for $m$ in equation (VII), to find $x$.

    $\begin{array}{c}x=\frac{\left(3.62\text{m}\right)\left(0.140\text{kg}\right)}{4.30\text{kg}-\left(23.36\right)\left(0.140\text{kg}\right)}\\ =0.492\text{m}\end{array}$

Therefore, the extension of the spring for the mass $0.140\text{kg}$ is $\underset{\_}{0.492\text{m}}$.

(d)

To determine

The extension of the spring for the mass $0.180\text{kg}$.

Answer to Problem 65P

The extension of the spring for the mass $0.180\text{kg}$ is $\underset{\_}{6.85\text{m}}$.

Explanation of Solution

From equation (VII).

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

Conclusion:

Substitute $0.180\text{kg}$ for $m$ in equation (VII), to find $x$.

    $\begin{array}{c}x=\frac{\left(3.62\text{m}\right)\left(0.180\text{kg}\right)}{4.30\text{kg}-\left(23.36\right)\left(0.180\text{kg}\right)}\\ =\frac{0.652}{0.0952}\text{m}\\ =\text{6}\text{.85}\text{m}\end{array}$

Therefore, the extension of the spring for the mass $0.180\text{kg}$ is $\underset{\_}{6.85\text{m}}$.

(e)

To determine

The extension of the spring for the mass $0.190\text{kg}$.

Answer to Problem 65P

For the mass $0.190\text{kg}$ the spring cannot constrain the motion, this situation is impossible.

Explanation of Solution

From equation (VII) the spring extension is given by

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

Conclusion:

Substitute $0.190\text{kg}$ for $m$ in equation (VII), to find $x$.

    $\begin{array}{c}x=\frac{\left(3.62\text{m}\right)\left(0.190\text{kg}\right)}{4.30\text{kg}-\left(23.36\right)\left(0.190\text{kg}\right)}\\ =\frac{0.652}{0}\text{m}\\ =\infty \end{array}$

Therefore, For the mass $0.190\text{kg}$ the extension of the spring goes to infinity. The spring cannot constrain the motion, this situation is impossible

(f)

To determine

To explain the pattern of variation of $x$ as it depends on $m$.

Answer to Problem 65P

The extension of the spring is directly proportional to the mass $m$ for the mass in few grams. For the mass $0.184\text{kg}$ and more the extension starts to diverge to infinity.

Explanation of Solution

The extension of the spring is directly proportional to the mass $m$, for the lighter mass in few grams. When the mass is further more increased from $0.184\text{kg}$ the extension starts to diverge to infinity.

Conclusion:

Therefore, the extension of the spring is directly proportional to the mass $m$ for the mass in few grams. For the mass $0.184\text{kg}$ and more the extension starts to diverge to infinity.