Question

A block of mass 1 kg is attached to a spring of force constant Nm .
4

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 #76698, Physics / Other

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

Solution:

According to Newton's second law,

$$ma = -kx + F_{frict}$$ $$a = \ddot{x}, \quad F_{frict} = -4v = -4\dot{x}$$

Thus,

$$m\ddot{x} + 4\dot{x} + kx = 0$$

We obtain the following differential equation:

$$\ddot{x} + 4\dot{x} + \frac{25}{4}x = 0$$

Initial conditions:

$$\left\{ \begin{array}{l} x(0) = 0.3\,m \\ \dot{x}(0) = 0 \end{array} \right.$$

Assume a solution of the form

$$x = e^{\lambda t}$$

we get

$$\lambda^2 + 4\lambda + \frac{25}{4} = 0$$ $$\lambda_{1,2} = \frac{-4 \pm \sqrt{16 - 25}}{2} = -2 \pm 1.5i$$

so the displacement is:

$$x(t) = C_1 e^{-2t} \cos(1.5t) + C_2 e^{-2t} \sin(1.5t)$$ $$\dot{x}(t) = C_1 e^{-2t} (-2 \cos(1.5t) - 1.5 \sin(1.5t)) + C_2 e^{-2t} (-2 \sin(1.5t) + 1.5 \cos(1.5t))$$

We use initial conditions to find $C_1$ and $C_2$:

$$C_1 \cos(0) + C_2 \sin(0) = 0.3$$ $$C_1 (-2 \cos(0) - 1.5 \sin(0)) + C_2 (-2 \sin(0) + 1.5 \cos(0)) = 0$$ $$C_1 \cdot 1 + C_2 \cdot 0 = 0.3$$ $$C_1 (-2) + C_2 (1.5) = 0$$

So,

$$C_1 = 0.3$$ $$C_2 = \frac{2 \cdot 0.3}{1.5} = 0.4$$

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

$$x(t) = 0.3 e^{-2t} \cos(1.5t) + 0.4 e^{-2t} \sin(1.5t)$$

Answer: $x(t) = 0.3 e^{-2t} \cos(1.5t) + 0.4 e^{-2t} \sin(1.5t)$.

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