So this is a simple problem but I'm just getting stumped. The question is:

A particle not connected to a spring, moving in a straight line, is subject to a retardation force of magnitude $\beta(\frac{dx}{dt})^n$, with $\beta > 0$.
a) Show that if 0 < n < 1, the particle will come to rest in a finite time. How far will the particle travel, and when will it stop?

So I think this would be the starting equation: $m\frac{d^2x}{dt^2}+\beta (\frac{dx}{dt})^n=0$ and the particle will stop when $\frac{dx}{dt}=0$ but that's all I got..I don't really know what to do from here. The question asks for n in a range so that's kind of throwing me off. Any ideas? Thanks


2 Answers 2


Hint: Split the problem in two:

  1. First solve the 1st order ODE $m\frac{dv}{dt}=-\beta v^n$ for the velocity $v(t)$ as a function of time $t$. (The exercise formulation doesn't say so, but there are physical reasons to believe that the velocity $v\geq 0$ should be assumed to be non-negative).

  2. Next integrate the velocity $v(t)$ wrt. $t$ to find the position $x(t)$ as a function of time $t$.


I too think

$m\dfrac{d^2x}{dt^2} + \beta(\dfrac{dx}{dt})^n = 0, \tag{1}$

is the right equation with which to start. Then observe that (1) implies

$m(\dfrac{dx}{dt})^{-n}\dfrac{d^2x}{dt^2} + \beta = 0, \tag{2}$

and that

$\dfrac{d}{dt}(\dfrac{dx}{dt})^{1 - n} = (1 - n)(\dfrac{dx}{dt})^{- n}\dfrac{d^2x}{dt^2}, \tag{3}$

whence (2) may be written

$\dfrac{m}{1 - n}\dfrac{d}{dt}(\dfrac{dx}{dt})^{1 - n} + \beta = 0 \tag{4}$


$\dfrac{m}{1 - n}\dfrac{d}{dt}(\dfrac{dx}{dt})^{1 - n} = -\beta. \tag{5}$

Note that all these calculations are valid for $n$ in the range $0 < n < 1$. It should also be observed that the assumption $dx / dt \ge 0$ is implicit in (1) and what follows, since $(dx/dt)^n$ only makes sense for the given range of $n$ if this is the case. These things being said, if we integrate (5) 'twixt $t_0$ and $t$, and divide through by $m / (1 - n)$ we find that

$(\dfrac{dx}{dt})^{1 - n}(t) - (\dfrac{dx}{dt})^{1 - n}(t_0) = -\dfrac{1 - n}{m}\beta(t - t_0), \tag{6}$


$(\dfrac{dx}{dt})^{1 - n}(t) = (\dfrac{dx}{dt})^{1 - n}(t_0) -\dfrac{1 - n}{m}\beta(t - t_0). \tag{7}$

Suppose the particle has velocity $v_0 = (dx/dt)(t_0)$ at $t = t_0$. Then (7) becomes

$(\dfrac{dx}{dt})^{1 - n}(t) = v_0^{1 - n} -\dfrac{1 - n}{m}\beta(t - t_0), \tag{8}$

and if we set the left-hand side equal to $0$, then we can solve for the value $t_r$ of $t$ at which the particle has zero velocity, i.e, when it "comes to rest":

$v_0^{1 - n} -\dfrac{1 - n}{m}\beta(t_r - t_0) = 0 \Rightarrow t_r = t_0 + \dfrac{ m}{\beta(1 - n)}v_0^{1 - n}. \tag{9}$

The distance $d$ the particle will travel is given by integrating $dx / dt$ between $t_0$ and $t_r$; indeed, from equation (8) we have

$\dfrac{dx}{dt} = (v_0^{1 - n} - \dfrac{1 - n}{m}\beta (t -t_0))^{1 / (1 - n)}, \tag{10}$

and thus

$x_r - x_0 = \int_{t_0}^{t_r} (\dfrac{dx}{dt})dt = \int _{t_0}^{t_r} (v_0^{1- n} - \dfrac{1-n}{m}\beta(t - t_0))^{1 / (1 - n)}dt; \tag{11}$

the right-hand integral occurring in (11) looks at first glance formidable, but becomes quite tractable under the substitution

$u = v_0^{1 - n} - \dfrac{1 -n}{m}\beta(t - t_0), \tag{12}$

for then

$du = -\dfrac{1-n}{m}\beta dt, \tag{13}$

so that

$dt = -\dfrac{m}{\beta(1 - n)}du, \tag{14}$

and the integral on the right-hand side of (11) becomes

$-\dfrac{m}{\beta(1 - n)}\int _{v_0^{1 - n}}^0 u^{1 / (1 - n)}du = -\dfrac{m}{\beta(1 - n)}(\dfrac{1 - n}{2 - n}u^{\frac{2 - n}{1 - n}}\mid_{v_0^{1 - n}}^0 = \dfrac{m}{\beta(2 - n)}v_0^{2 - n}, \tag{15}$

since $u(t_0) = v_0^{1 - n}$ and $u(t_r) = 0$. The distance the particle travels is thus

$d = \dfrac{m}{\beta(2 - n)}v_0^{2 - n}. \tag{16}$

Well, I think I got all these calculations right. Gotta run to work right now, will check it all again later!

Hope this helps! Cheerio,

and as always,

Fiat Lux!!!


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