Chapter 13, Problem 62P

(a)

To determine

To show: The time interval for a transverse pulse to travel the length of the rope is $\Delta t=2\sqrt{\frac{L}{mg}}\left(\sqrt{M+m}\right)-\sqrt{M}$.

Answer to Problem 62P

The time interval for a transverse pulse to travel the length of the rope is $\Delta t=2\sqrt{\frac{L}{mg}}\left(\sqrt{M+m}\right)-\sqrt{M}$.

Explanation of Solution

The free body diagram of the suspended object with rope is shown below.

Figure (1)

From free body diagram, the tension is equal to the weight of the body.

$T=Mg+\frac{mg}{L}x$

Here,

$T$ is the tension in the rope.

$m$ is the unit mass of the rope.

$L$ is the length of the rope.

$M$ is the mass of the suspended object.

$g$ is the acceleration due to gravity.

Formula to calculate the linear mass density is,

$\mu =\frac{m}{L}$

Here,

$\mu$ is the linear mass density.

Formula to calculate the speed of the wave is,

$v=\sqrt{\frac{T}{\mu }}$ (1)

Here, $v$ is the speed of the wave.

Substitute $\frac{m}{L}$ for $\mu$ and $Mg+\frac{mg}{L}x$ for $T$ in equation (1).

$\begin{array}{c}v=\sqrt{\frac{Mg+\frac{mg}{L}x}{\frac{m}{L}}}\\ =\sqrt{g}\left(\frac{ML}{m}+x\right)\end{array}$

Formula to calculate the time taken to cover the smallest distance is,

$\begin{array}{c}v=\frac{dx}{dt}\\ dt=\frac{dx}{v}\end{array}$ (2)

Here,

$dx$ is the smallest distance.

$dt$ is the time taken to cover the smallest.

Substitute $\sqrt{g}\left(\frac{ML}{m}+x\right)$ for $v$ in equation (2).

$\begin{array}{c}dt=\frac{dx}{\sqrt{g}\left(\frac{ML}{m}+x\right)}\\ =\frac{1}{\sqrt{g}}{\left(\frac{ML}{m}+x\right)}^{-\frac{1}{2}}dx\end{array}$ (3)

Integrate the right hand side of equation (3) from $0$ to $L$ and left hand side from $0$ to $t$ for total time.

$\begin{array}{c}{\displaystyle \underset{0}{\overset{t}{\int }}dt}={\displaystyle \underset{0}{\overset{L}{\int }}\frac{1}{\sqrt{g}}{\left(\frac{ML}{m}+x\right)}^{-\frac{1}{2}}dx}\\ \left(t-0\right)=\frac{2}{\sqrt{g}}\times \sqrt{\frac{L}{m}}\left(\left(M+m\right)-\sqrt{M}\right)\\ \Delta t=2\sqrt{\frac{L}{mg}}\left(\sqrt{M+m}-\sqrt{M}\right)\end{array}$

Conclusion:

Therefore, the time interval for a transverse pulse to travel the length of the rope is $\Delta t=2\sqrt{\frac{L}{mg}}\left(\sqrt{M+m}\right)-\sqrt{M}$ is verified.

(b)

To determine

To show:The expression in part (a) reduces to the result of Problem 48 when $M=0$.

Answer to Problem 62P

the expression in part (a) reduces to the result of Problem 48 when $M=0$ is $\frac{1}{2}\sqrt{\frac{L}{g}}$.

Explanation of Solution

The expression for the problem 48 is $\Delta t=\frac{1}{2}\sqrt{\frac{L}{g}}$.

The given expression in part (a) for total time is,

$\Delta t=2\sqrt{\frac{L}{mg}}\left(\sqrt{M+m}\right)-\sqrt{M}$ (4)

Substitute $0$ for $M$ in equation (4).

$\begin{array}{c}\Delta t=2\sqrt{\frac{L}{mg}}\left(\sqrt{0+m}\right)-\sqrt{0}\\ =\frac{1}{2}\sqrt{\frac{L}{g}}\end{array}$

Conclusion:

Therefore, the expression in part (a) reduces to the result of Problem 48 when $M=0$ is verified above.

(c)

To determine

To show:The expression in part (a) reduces to $\Delta t=2\sqrt{\frac{mL}{Mg}}$ for $m<<M$.

Answer to Problem 62P

The expression in part (a) reduces to $\Delta t=2\sqrt{\frac{mL}{Mg}}$ for $m<<M$.

Explanation of Solution

The given expression in part (a) for total time is,

$\begin{array}{c}\Delta t=2\sqrt{\frac{L}{mg}}\left(\sqrt{M+m}\right)-\sqrt{M}\\ =2\sqrt{\frac{L}{mg}}\left[M^{\frac{1}{2}}{\left(1+\frac{m}{M}\right)}^{\frac{1}{2}}-M^{\frac{1}{2}}\right]\\ =2\sqrt{\frac{L}{mg}}\left[M^{\frac{1}{2}}{\left(1+\frac{m}{M}\right)}^{\frac{1}{2}}-1\right]\end{array}$ (5)

Apply the binomial theorem in the equation (5).

$\begin{array}{c}\Delta t=2\sqrt{\frac{L}{mg}}\left[M^{\frac{1}{2}}{\left(1+\frac{m}{M}\right)}^{\frac{1}{2}}-1\right]\\ =2\sqrt{\frac{L}{mg}}\left[M^{\frac{1}{2}}\left(1+\frac{m}{2M}-1\right)\right]\\ =2\sqrt{\frac{L}{mg}}\sqrt{M}\left(\frac{m}{2M}\right)\\ =\sqrt{\frac{mL}{Mg}}\end{array}$

Conclusion:

Therefore, the expression in part (a) reduces to $\Delta t=2\sqrt{\frac{mL}{Mg}}$ for $m<<M$.