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FREE VIBRATIONS WITH VISCOUS DAMPING.
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Figure (a) shows a spring mass dash pot system.

enter image description here

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=mgcxk(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+b24ac2a in general form

=cc24mk2a

=c2m+(c2m)2km

S1=c2m+(c2m)2km

S2=c2m(c2m)2km

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

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