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Determine a differential equation governing a falling body in resisting medium

Modify the equations (Example 2 on page 314 of Apostol) for the velocity of a falling body in a resisting medium if the resistance of the medium is proportional to v^2 instead of to v. Prove that the resulting differential equation can be written in each of the following forms:

    \[ \frac{ds}{dv} = \frac{m}{k} \frac{v}{c^2 - v^2}, \qquad \frac{dt}{dv} = \frac{m}{k} \frac{1}{c^2-v^2}, \]

where c = \sqrt{\frac{mg}{k}}. By integrating these find the following formulas for v:

    \begin{align*}  v^2 &= \frac{mg}{k} \left( 1 - e^{\frac{-2ks}{m}} \right) \\[10pt]  v &= c \frac{e^{bt} - e^{-bt}}{e^{bt} + e^{-bt}} = c \tanh (bt), \end{align*}

where b = \sqrt{\frac{kg}{m}}. Determine the value of v as t \to +\infty.

Starting with the equation

    \[ ma =mg - kv \]

in example 2 and modifying it so that the resistance is proportional to v^2 we have

    \begin{align*}  ma = mg - kv^2 && \implies && mv' &= mg - kv^2 \\  && \implies && v' = g - \frac{k}{m} v^2. \end{align*}

Using the chain rule as we did in the previous exercise we know

    \[ \frac{dv}{dt} = v \cdot \frac{dv}{ds}. \]


    \begin{align*}  &&v \cdot \frac{dv}{ds} &= g - \frac{k}{m} v^2 \\[9pt]  \implies && \frac{dv}{ds} &= \frac{g}{v} - \frac{k}{m} v \\[9pt]  \implies && \frac{dv}{ds} &= \frac{gm-kv^2}{mv} \\[9pt]  \implies && \frac{ds}{dv} &= \frac{mv}{gm-kv^2}. \end{align*}

Letting c = \sqrt{\frac{mg}{k}} we then have

    \[ \frac{ds}{dv} = \frac{m}{k} \left( \frac{v}{c^2-v^2} \right). \]

This is the first requested equation.


    \begin{align*}  v' = g - \frac{k}{m} v^2 && \implies && \frac{dv}{dt} &= \frac{mg-kv^2}{m} \\[9pt]  && \implies && \frac{dt}{dv} &= \frac{m}{mg-kv^2} \\[9pt]  && \implies && \frac{dt}{dv} &= \frac{m}{k} \left( \frac{1}{c^2-v^2} \right). \end{align*}

Integrating the first equation we find,

    \begin{align*}  && \frac{ds}{dv} &= \frac{m}{k} \frac{v}{c^2-v^2} \\[9pt]  \implies && s &= \frac{m}{k} \int \frac{v}{c^2-v^2} \, dv \\[9pt]  \implies && s &= -\frac{m}{2k} \log (c^2 - v^2) \\[9pt]  \implies && e^{-\frac{2ks}{m}} &= c^2 - v^2 \\[9pt]  \implies && v^2 &= \frac{mg}{k} - e^{-2ks}{m}. \end{align*}

Integrating the second equation,

    \begin{align*}  && \frac{dt}{dv} &= \frac{m}{k} \cdot \frac{1}{c^2-v^2} \\[9pt]  \implies && t &= \frac{m}{k} \int \frac{1}{c^2-v^2} \, dv \\[9pt]  \implies && t &= \frac{1}{2} \frac{m}{k} \frac{1}{c} log \left(\frac{c+v}{c-v}\right) \\[9pt]  \implies && 2bt &= \log \left( \frac{c+v}{c-v} \right) &(b = \sqrt{\frac{kg}{m}} )\\[9pt]  \implies && e^{2bt} &= \frac{c+v}{c-v} \\[9pt]  \implies && v &= c \cdot \frac{e^{2bt} - 1}{e^{2bt} + 1} \\[9pt]  \implies && v &= c \cdot \tanh (bt). \end{align*}

This implies

    \[ \lim_{t \to \infty} v = 1. \qquad \blacksquare\]


  1. S says:

    The starting condition is v(0)=0, so there is no typo in the book, but there is in the proposed solution for the first integral here.

    The constant of the first integral is \frac{m}{k} \log c. The constant of the second integral is 0.

  2. Evangelos says:

    RoRi, I noticed that you approached the first integration (that of \frac{ds}{dv}) the same way that I did. Namely, working with the assumption that Tom made a typo in the formula for v^2

    The way that you and I solved the problem, the formula is as such

        \begin{equation*} \begin{split} \(v^2\)  &= \(c^2\) - \(e^\frac{-2ks}{m} \)  \\          &= \frac{gm}{k} - \(e^\frac{-2ks}{m} \)  \end{split} \end{equation*}

    But the book has it written as such, so that the factor \frac{gm}{k} distributes.

        \begin{equation*} \begin{split} \(v^2\)  &= \(c^2\) - \(e^\frac{-2ks}{m} \)  \\          &= \frac{gm}{k}(1 - \(e^\frac{-2ks}{m} \) ) \end{split} \end{equation*}

    It threw me for a loop when I solved the equation. I thought it might have been a clever way to get rid of the constant of integration, but I couldn’t figure out a way to get it to distribute the way it was written in the book

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