Pht100 Overdamped And Critically Damped - Module 1 Lecture 3


In this class, we are going to discuss different cases in damned harmonic oscillations over damped, critically damped and under damped cases, and I move the case square less than omega0 in this k. Square is k. Square is greater than omega 0 square, otherwise equations, minus k, plus root of k, square, minus omega, zero square, t, plus b into e, raised to minus k square, minus omega, zero square, t, right quadratic equation programming. Okay, starting stretching is starting from the maximum position x equal to. In a maximum position velocity videos. This is condition. One first, conditional equal to zero x equal to n versus our first condition. Okay, number condition, t, equal to zero x equal to a. And then the second condition in the v is equal to zero.

Either d, x by d, t, zero l and done DX by DT at every one x by DT equal to 01 x equal to 0. Okay. There is zero. One x step and either minus k, plus root of k, square, minus omega, zero square into a plus minus k, minus root of k, square, minus omega square into b. And that. Will change the differential you can do minus k, a plus an into root of k, square, minus omega, 0 square, plus minus k, b, minus b into root of k, square, minus omega 0 square in Italy, a plus b, a minus k, a plus root of k, square, minus omega, 0 square into a minus b in the k. N, a minus k n, an inside like the remaining k, an equal to a minus b into root of k, square minus omega, 0 square negative, a minus b equal to k. A divided by root of k, square minus omega, 0 square and SLI, a minus b, a plus b and a minus b, a by. Root of k, square, minus omega, 0 square unknown, a plus b, plus a minus b, equal to a plus b and a minus b in the AK plus root of k, square, minus omega zero square, I know solving submit I'm going to do a b and a plus b, minus a minus okay and a term, a plus b, minus of a minus b number minus two, minus one plus.

I with a minus one okay, a plus b, minus a'm going to eat the negative. One of the plus b, minus of a minus b, a plus b, minus a minus plus a plus b, equal to a minus k, an into root of k square. Minus omega zero square.

And there is a b is equal to an into one minus root of k by root of k. Square minus omega. Zero square is x of t is equal to a by 2 into 1 plus k by root k, square minus omega 0 square into e, raised to minus k, plus root of k, square, minus omega, 0 square into t, plus a by 2 into 1 minus k by root k, square minus omega 0 square into e, raised to minus k, minus root of k, square, minus omega 0 square into t, any solution x erased minus. Kt, other exponentially, decreasing on in the time x. Coordinate x often the exponentially, decreasing on indication over damped. In the case okay power down to the name is root of number alpha 1, minus k, plus root of k, square, minus omega, 0 square lay and then alpha 2, t and then alpha 1 minus k. Substitute a raised to minus k, t, plus b, e, raised to minus k, t, a plus b into e, raised to minus. Thank you.

Dated : 25-Apr-2022

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