Question

A block of mass 1 kg is attached to a spring of force constant k = 25/4 N/m. It is pulled x = 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 #75316 – Math – Differential Equations

Question

A block of mass $1\,\mathrm{kg}$ is attached to a spring of force constant $k = 25/4\,\mathrm{N/m}$. It is pulled $x = 0.3\,\mathrm{m}$ 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 $-4\,\mathrm{v}$ (where $v$ is the instantaneous velocity of the block). Determine the position $x(t)$ of the block at time $t$.

Solution:

On a block of mass $1\,\mathrm{kg}$, there are three forces:

The force due to gravity: $F_g = m \cdot g = 1 \cdot x''$

$g = 9.8\,\mathrm{m/s^2}$ – gravitational acceleration

The force of elasticity of a spring: $F_s = -k \cdot x$

$$k = 25/4\,\mathrm{m/s^2}$$ $$m$$ $$\frac{N}{m}$$

- spring modulus

The damping force: $F_d = -d \cdot v = -4 \cdot v = -4 \cdot x'$

$$d = 4$$

- damping coefficient

According to Newton’s second law, taking into account the signs of the projections of forces, we obtain the following differential equation with initial conditions:

$$x'' + 4x' + \frac{25}{4}x = 0 \quad \text{IVP: } x(0) = 0.3, \quad x'(0) = 0$$

Assuming that the solution has the form $a \cdot e^{r \cdot t}$ we get $(m \cdot r^2 + d \cdot r + k) \cdot a \cdot e^{r \cdot t} = 0$

Upon solving for the roots of the characteristic equation we get the following:

$$r_{1,2} = \frac{-d \pm \sqrt{d^2 - 4mk}}{2m} = \frac{-4 \pm \sqrt{16 - 25}}{2} = -2 \pm \frac{3}{2}i$$

so the displacement is:

$$x(t) = C_1 e^{-2 \cdot t} \cos(3t) + C_2 e^{-2 \cdot t} \sin(3t)$$ $$x'(t) = C_1 (-2 \cdot e^{-2 \cdot t} \cos(3t) - 3 \cdot e^{-2 \cdot t} \sin(3t)) + C_2 (3 \cdot e^{-2 \cdot t} \cos(3t) - 2 \cdot e^{-2 \cdot t} \sin(3t))$$

We use the initial conditions for finding constants $C_1$ and $C_2$:

$$\left\{ \begin{array}{c} C_1 e^{-2 \cdot 0} \cos(0) + C_2 e^{-2 \cdot 0} \sin(0) = 0.3 \\ C_1 (-2 \cdot e^{-2 \cdot 0} \cos(0) - 3 \cdot e^{-2 \cdot 0} \sin(0)) + C_2 (3 \cdot e^{-2 \cdot 0} \cos(0) - 2 \cdot e^{-2 \cdot 0} \sin(0)) = 0 \end{array} \right.$$ $$\left\{ \begin{array}{c} C_1 = 0.3 \\ -2C_1 + 3C_2 = 0 \end{array} \right. \quad \rightarrow \quad C_1 = 0.3, \quad C_2 = 0.2$$

Then the position $x(t)$ of the block at time $t$:

$$x(t) = 0.3e^{-2 \cdot t}\cos(3t) + 0.2e^{-2 \cdot t}\sin(3t)$$

Answer: $x(t) = 0.3e^{-2 \cdot t}\cos(3t) + 0.2e^{-2 \cdot t}\sin(3t)$

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