Question

A block of mass 1 kg is attached to a spring of force constant k=25/4N/m.It is
pulled 0.3 m from its equilibrium position and released from rest. This spring‐block
apparatus is submerged in a viscous fluid medium which exerts a damping force of
– 4 v (where v is the instantaneous velocity of the block). Determine of the position x(t)
of the block at time t.

Answer 1

Answer on Question #73083- Physics / Mechanics | Relativity

A block of mass $m = 1$ kg is attached to a spring of force constant $k = 25 / 4$ N/m. It is pulled $x_0 = 0.3$ m from its equilibrium position and released from rest. This spring-block apparatus is submerged in a viscous fluid medium which exerts a damping force of $F_{\mathrm{frict}} = -4v$ (where $v$ is the instantaneous velocity of the block). Determine of the position $x(t)$ of the block at time $t$.

Solution:

The Newton's second law

$$m a = - k x + F _ {\mathrm {f r i c t}}$$ $$m \ddot {x} (t) + 4 \dot {x} (v) + k x = 0$$ $$\ddot {x} + 4 \dot {x} + \frac {2 5}{4} x = 0$$

Solution

$$x \sim e ^ {\lambda t}$$ $$\lambda^ {2} + 4 \lambda + \frac {2 5}{4} = 0$$ $$D = 1 6 - 2 5 = - 9$$ $$\lambda = \frac {- 4 \pm 3 i}{2} = - 2 \pm 1. 5 i$$ $$x (t) = e ^ {- 2 t} A \cos (1. 5 t)$$ $$x (0) = A = 0. 3$$

Finally

$$x (t) = 0. 3 e ^ {- 2 t} \cos (1. 5 t)$$

Answer $x(t) = 0.3e^{-2t}\cos (1.5t)$

Answer provided by https://www.AssignmentExpert.com