A particle of mass m and density rho sinks slowly, starting…

A particle of mass $m$ and density $\rho$ sinks slowly, starting from rest, in a viscous fluid of density $\rho_0$. The resistance of the fluid is proportional to the velocity, with proportionality constant $k$. Find the law of motion $x(t)$, where $x$ is the depth of the particle at time $t$.

Step 1: Setting up Newton's second law

Measure $x$ as depth, positive downward. Two forces act on the particle: its effective weight, reduced by buoyancy, $G = mg\left(1 - \frac{\rho_0}{\rho}\right)$.

a constant, positive quantity (since the particle sinks, its density $\rho$ exceeds the fluid's density $\rho_0$); and a resisting force proportional to the velocity, $-k\dfrac{dx}{dt}$, opposing the motion. Newton's second law gives

\[ m\ddot x = G - k\dot x. \]

Step 2: Solving for the velocity

Let $v=\dot x$. The equation becomes the first-order linear equation

\[ \dot v + \frac{k}{m}v = \frac{G}{m}. \]

Its homogeneous solution is $C_1 e^{-kt/m}$, and a constant particular solution is $v_p = G/k$ — this is precisely the terminal velocity $v_\infty$ at which the resisting force balances the effective weight. So

\[ v(t) = \frac{G}{k} + C_1 e^{-kt/m}. \]

Since the particle starts from rest, $v(0)=0$, so $C_1=-G/k$, and

\[ v(t) = \frac{G}{k}\left(1 - e^{-kt/m}\right). \]

Step 3: Integrating for the position

With $x(0)=0$, \[x(t) = \int_0^t v(\tau)\,d\tau = \frac{G}{k}\left[\tau + \frac{m}{k}e^{-k\tau/m}\right]_0^t = \frac{G}{k}\left[t - \frac{m}{k}\left(1-e^{-kt/m}\right)\right].\]

Step 4: The law of motion

\[x(t) = \frac{G}{k}\left[t - \frac{m}{k}\left(1 - e^{-kt/m}\right)\right], \qquad G = mg\left(1 - \frac{\rho_0}{\rho}\right)\]

Substitution of this $x(t)$ back into $m\ddot x - (G-k\dot x)$ reduces identically to zero, and both $x(0)=0$ and $\dot x(0)=0$ hold exactly, confirming the solution.