Question

a particle of mass m falls freely under gravity in a liquid that offers a resistive force proportional to its velocity : fres= -gama dx/dt solve it.

Answer 1

Answer on Question #78062 – Math – Differential Equations

Question

A particle of mass $m$ falls freely under gravity in a liquid that offers a resistive force proportional to its velocity:

$$f_{res} = -\gamma \frac{dx}{dt}$$

Solve it.

Solution

The net force is

$$F_{net} = mg - \gamma \frac{dx}{dt}$$

The differential equation from Newton’s Law

$$F_{net} = ma$$ $$a = \frac{d^2 x}{dt^2} = g - \frac{\gamma}{m} \frac{dx}{dt}$$

Let $v(t) = \frac{dx}{dt}$. Then $a = \frac{dv}{dt}$

$$\frac{dv}{dt} = g - \frac{\gamma}{m} v$$ $$\frac{dv}{g - \frac{\gamma}{m} v} = dt$$ $$\int \frac{1}{v - \frac{mg}{\gamma}} dv = -\frac{\gamma}{m} \int dt$$ $$\ln \left| v - \frac{mg}{\gamma} \right| = -\frac{\gamma}{m} t + \ln C$$ $$v - \frac{mg}{\gamma} = C e^{-\frac{\gamma}{m} t}$$ $$v = \frac{mg}{\gamma} + C e^{-\frac{\gamma}{m} t}$$

When $t = 0, v(0) = v_0$

$$v_0 = \frac{mg}{\gamma} + C = C = v_0 - \frac{mg}{\gamma}$$ $$v = \frac{mg}{\gamma} + \left(v_0 - \frac{mg}{\gamma}\right) e^{-\frac{\gamma}{m} t}$$ $$\frac{dx}{dt} = v = \frac{mg}{\gamma} + \left(v_0 - \frac{mg}{\gamma}\right) e^{-\frac{\gamma}{m} t}$$ $$dx = \left(\frac{mg}{\gamma} + \left(v_0 - \frac{mg}{\gamma}\right) e^{-\frac{\gamma}{m} t}\right) dt$$ $$x = \frac{mg}{\gamma} t - \frac{m}{\gamma} \left(v_0 - \frac{mg}{\gamma}\right) e^{-\frac{\gamma}{m} t} + C_1$$

When $t = 0, x(0) = x_0$

$$x_0 = -\frac{m}{\gamma} \left(v_0 - \frac{mg}{\gamma}\right) e^{-\frac{\gamma}{m} t} + C_1 = C_1 = x_0 + \frac{m}{\gamma} \left(v_0 - \frac{mg}{\gamma}\right)$$ $$x = \frac{mg}{\gamma} t - \frac{m}{\gamma} \left(v_0 - \frac{mg}{\gamma}\right) e^{-\frac{\gamma}{m} t} + x_0 + \frac{m}{\gamma} \left(v_0 - \frac{mg}{\gamma}\right)$$ $$v = v_0 e^{-\frac{\gamma}{m} t} + \frac{mg}{\gamma} \left(1 - e^{-\frac{\gamma}{m} t}\right)$$ $$x = x_0 + v_0 \frac{m}{\gamma} \left(1 - e^{-\frac{\gamma}{m} t}\right) - \frac{m^2 g}{\gamma^2} \left(1 - e^{-\frac{\gamma}{m} t}\right) + \frac{mg}{\gamma} t$$

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