Question

A mass of 2 kg is fixed to an end of a spring with spring constant k = 128 Nm−1
and the
system is placed inside a fluid. It is set into vibration from its equilibrium position with
an initial speed of 0.6 ms−1
. If the damping force due to the fluid is 40v(t) N where v is
the instantaneous speed of the mass, determine the position of the mass as a function of
time t.

Answer 1

Answer on Question #62065 – Math – Differential Equations

Question

A mass of $2\,\mathrm{kg}$ is fixed to an end of a spring with spring constant $k = 128\,\mathrm{Nm}^{-1}$ and the system is placed inside a fluid. It is set into vibration from its equilibrium position with an initial speed of 0.6 ms$^{-1}$. If the damping force due to the fluid is $40\,\mathrm{v(t)}$ N, where $\mathbf{v}$ is the instantaneous speed of the mass, determine the position of the mass as a function of time $t$.

Solution

Using Newton’s second law we have:

$$m \ddot{x} = - \lambda \dot{x} - k x;$$ $$2 \ddot{x} = - 40 \dot{x} - 128 x;$$ $$\ddot{x} + 20 \dot{x} + 64 x = 0;$$

Now we need to solve characteristic equation:

$$\lambda^2 + 20 \lambda + 64 = 0 \Rightarrow \lambda_1 = -16, \lambda_2 = -4.$$

The general solution of the differential equation (1) is

$$x(t) = C_1 e^{-16t} + C_2 e^{-4t}.$$

Now we can use the initial conditions for position and speed:

$$x(t = 0) = 0 = C_1 + C_2;$$ $$\dot{x}(t = 0) = 0.6 = -16 C_1 - 4 C_2;$$ \left\{ \begin{array}{l} C_1 + C_2 = 0 \\ -16 C_1 - 4 C_2 = 0.6 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} C_2 = - C_1 \\ -16 C_1 - 4 C_2 = 0.6 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} C_2 = - C_1 \\ 16 C_2 - 4 C_2 = 0.6 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} C_2 = - C_1 \\ 12 C_2 = 0.6 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} C_2 = - C_1 \\ 2 C_2 = 0.6 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} C_2 = - C_1 \\ 2 C_2 = 0.6 \end{array} \right. \end{array} \right. $$\left\{ \begin{array}{l} C_2 = - C_1 \\ C_2 = 0.6 / 12 \end{array} \right. \Rightarrow C_1 = -0.05, C_2 = 0.05;$$

Position of the mass as a function of time $t$ is

$$x(t) = -0.05 e^{-16t} + 0.05 e^{-4t}.$$

Answer: $x(t) = -0.05 e^{-16t} + 0.05 e^{-4t}$.

www.AssignmentExpert.com