written 5.8 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
˙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¨x=mg−cx−k(x+△st)
m¨x + c˙x + kx = 0 - - - - (1) (∵ mg=△st K)
Solution of the above equation:
Above equation is linear differential equation of second order whose solution is written as:
x=est - - - -(a)
Differentiating the above equation, we get
˙x=sest - - - -(b)
¨x=s2est - - - - (c)
Substituting (a), (b) and (c) in equation (1), we get
ms2est+csest+kest=0
est(ms2+cs+k)=0
Since est 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. ms2+cs+k=0 - - -(2)
As equation (2) is a quadratic expression in s, there are two values of S to be considered.
∴ s1=−b+√b2−4ac2a in general form
=−c√c2−4mk2a
=−c2m+√(c2m)2−km
∴ S1=−c2m+√(c2m)2−km
S2=−c2m−√(c2m)2−km
Hence x=es1t and x=e52t may be both the solutions of equation (1).
A general solution is formed by combination of these.
x=C1es1t+C2es2t
where c1 and c1 are arbitrary constants.
The solution is valid as long as s1 does not equal to s2