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Next to consider the free damped motion.

Â Free damped motion means there is some damping but no external force,

Â so that our differential equation,

Â becomes x double prime plus 2 lambda x prime plus omega square x is equal to 0.

Â And its characteristic equation is,

Â as you can read it from this,

Â r squared plus 2 lambda times r plus omega squared and that is equal to 0.

Â The root to sub this quadratic equation is given by,

Â R is equal to negative lambda plus or

Â minus square root of lambda squared minus omega squared.

Â So, depending on the sign of this discriminant lambda squared minus omega squared.

Â When this is a positive and when this quantity is 0,

Â and when this is a negative,

Â we have a three distinct cases as the following.

Â The general solution, depending on the sign of this,

Â the discriminant, lambda squared minus omega squared,

Â we have are three distinct cases like, first,

Â when the lambda squared minus omega squared is a positive,

Â which we call as overdamped.

Â Then, we have a two distinct real roots for the characters to

Â equation given by this quantity so that the general solution,

Â the motion of the masses x of t is equals to e to the negative lambda t times c

Â one exponential square root of lambda squared minus omega squared t and plus c two times

Â e to the negative square root of lambda squared minus omega squared t. The second case,

Â when the lambda squared minus omega squared is equal to 0,

Â then if this is equal to 0 then we have only one double real root,

Â r is equal to negative lambda.

Â In that case we quote it as a critically damped,

Â and the general solution we know is given by the x of t is equal to e to

Â the minus lambda t times c one plus c two t. Finally,

Â if this quantity is a negative,

Â if this lambda squared minus omega squared is a negative,

Â which we call as the underdamped case,

Â the solution will be x of t is equal to e to the minus of

Â lambda t times c one cosine square root of omega squared

Â minus lambda squared t plus c

Â two times sine square root of omega squared minus lambda squared

Â t. So we have those three distinct cases

Â depending on this discriminant lambda squared minus omega squared.

Â Graphically, you can see that the first,

Â the overdamped case, which is the Case 1,

Â Case 1 down there,

Â in this case the typical graph is given by,

Â by this first graph,

Â which is the overdamped cases.

Â Second, in the second case,

Â in which the lambda squared minus omega squared is equal to zero.

Â So, that solution is given by this one.

Â Its typical graph is given by this second graph,

Â and in the third case when lambda squared minus omega squared is negative,

Â so that underdamped cases in which the general solution

Â involves not only the exponential term but the cosine and the sine term two,

Â then this case the graph looks like the last one.

Â This is the graph for the underdamped cases.

Â What we have to observe from this three cases,

Â all these three cases are the following.

Â 4:49

Those motion x of t is non-oscillatory or three in the Cases 1 and 2.

Â But there is some oscillation in the Case 3 as you can see clearly from the picture.

Â Second, in all three cases,

Â in all three cases as you can see,

Â as you can see from the solution,

Â from the solution this in the first the case,

Â in front of we have exponential minus lambda t. In the second case,

Â we also have exponential minus lambda t. In the third case,

Â we also have an exponential minus lambda t,

Â which goes to 0 exponentially when tan goes to infinity.

Â So, in all three cases we expect,

Â the limit of x of t,

Â when t turns to infinity is equal to 0.

Â And the Case 2, the second graph,

Â this one is called the critically damped.

Â It's called a critically dumped because any small decrease in the damping force,

Â results in the oscillatory motion.

Â If you decrease the damping force a little bit,

Â then the sign of the discriminant becomes the

Â negative so that your general solution will involve the sine or cosine.

Â In the last Case 3, of which the solution is given by this one.

Â For this second term in the parenthesis,

Â you can combine this two into one using the trigonometry identity again in the following.

Â The last case, the solution can be written as,

Â A to e times exponential negative lambda t, times

Â the cosine square root of omega squared minus lambda squared t minus phi,

Â where as the user the capital A is equal to square root of c one squared plus

Â c two squared and A times the cosine phi is the c one and the A times the sine phi,

Â that is equals to c two.

Â Look at this equation (7).

Â Because of this exponential term,

Â the function x of t is not periodic at all.

Â But we have, by looking at this a second term, cosine term,

Â this is a periodic function which has a period two

Â pi over square root of omega squared minus lambda squared.

Â So this is not genuine,

Â the periodic function but,

Â because of this second term,

Â we call this quantity two pi over

Â square root of omega squared minus lambda squared, the quasi period,

Â and it's a reciprocal,

Â that is square root of omega squared minus lambda squared

Â over two pi the quasi frequency.

Â In Fact, the quasi period is

Â the time interval between any two successive maxima of x of t or minima.

Â As you can see from the picture,

Â from this point to another minimum point,

Â so from this to that,

Â the time needed, that is a quasi period.

Â And finally, let's consider the forced damped motion.

Â So we now consider the non homogeneous differential equation.

Â To say this is the equal to,

Â you can remind x double prime t plus two lambda x the prime of t and

Â plus omega square x of t and that is equal to g. And here we are we are shown that,

Â there is some external force, f,

Â which is not necessarily zero.

Â So that g is not equal to 0.

Â So that's a non-homogeneous second order constant co-efficients differential equation.

Â And we know that,

Â the general solution of this non-homogeneous differential equation,

Â x of t can be decomposed as the sum of x of sub c of t plus x of sub p of t,

Â Where x of sub c of t is a general solution of

Â the corresponding homogeneous differential equation when g is equal to 0.

Â So, in other words, this is a general solution of the free undamped motion.

Â On the other hand, the x of p of t is

Â any particular solution of ordinary differential equation.

Â What I mean is,

Â x of c of t is a general solution of

Â corresponding homogeneous equation that is equal to 0.

Â From this you are going to get x of c of t and x of p of t is

Â the particular or solution of this given non-homogeneous differential equation.

Â In particular, if this external force is the periodic force,

Â given by a cos gamma t plus b sin gamma t,

Â which is a periodic,

Â then there is a particular solution of the form.

Â x of p of t equals to a cos alpha gamma t plus B times the sine gamma of t.

Â We know the form of this particular solution by

Â the method of the undetermined coefficients.

Â And also we know that the corresponding homogeneous solutions say,

Â x of c of t, the complementary solution,

Â it goes to 0, when t turns to infinity.

Â So, that means what?

Â For the solution given by x is equal to x of c plus x of p,

Â let t turns to infinity.

Â And we know that this turns to zero so that for large enough the time,

Â x behave like the x of t of p,

Â for t big enough.

Â For t large enough, the our solution x of t,

Â it just to behaves like x of p of t. In this sense,

Â we call the complementary solution x of c of t, a transient solution,

Â solution value for the finite time and the particular solution x p of t,

Â a steady state solution of differential equation.

Â