written 4.9 years ago by |

Figure (a) shows a spring mass dash pot system.

The spring and dash pot are in parallel.

Let, k = stiffness of spring, N/m

m = mass of the body, kg

c = damping coefficient, Ns/m

x = displacement of body from mean position

$\dot{x}$ = velocity of the body.

The damped resistance at any instant is equal to cx, where x is the velocity and c is the damping co efficient.

From Newton's second law of motion, the equation of motion is written as

$m \ddot{x} = mg - cx - k(x + \triangle_{st})$

m$\ddot{x}$ + c$\dot{x}$ + kx = 0 - - - - (1) ($\because$ $mg = \triangle_{st} \ K$)

**Solution of the above equation:**

Above equation is linear differential equation of second order whose solution is written as:

$x = e^{st}$ - - - -(a)

Differentiating the above equation, we get

$\dot{x} = se^{st}$ - - - -(b)

$\ddot{x} = s^2 e^{st}$ - - - - (c)

Substituting (a), (b) and (c) in equation (1), we get

$ms^2 e^{st} + cse^{st} + ke^{st} = 0$

$e^{st} (ms^2 + cs + k) = 0$

Since $e^{st}$ is always a positive number, equation (a) will be a solution of the differential equation (1) only if the expression in the parentheses is equal to zero.

i.e. $ms^2 + cs + k = 0$ - - -(2)

As equation (2) is a quadratic expression in s, there are two values of S to be considered.

$\therefore$ $s_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$ in general form

$= \frac{-c \sqrt{c^2 - 4mk}}{2a}$

$= \frac{-c}{2m} + \sqrt{ (\frac{c}{2m})^2 - \frac{k}{m}}$

$\therefore$ $S_1 = \frac{-c}{2m} + \sqrt{ (\frac{c}{2m})^2 - \frac{k}{m}}$

$S_2 = \frac{-c}{2m} - \sqrt{ (\frac{c}{2m})^2 - \frac{k}{m}}$

Hence $x = e^{s1t}$ and $x = e^{52t}$ may be both the solutions of equation (1).

A general solution is formed by combination of these.

$x = C_1e^{s1t} + C_2e^{s2t}$

where $c_1$ and $c_1$ are arbitrary constants.

The solution is valid as long as $s_1$ does not equal to $s_2$