In a few hours I got my last exam of mechanical vibrations, I didn´t understand my school teacher so I saw all the videos until here and now I really know . I wanna thank you for sharing your knowledge in such an easy way. You're a great teacher. Greetings from Querétaro, México.
are you sure about the general Solution? Does that mean the system is unstable if c2 is not zero....I looked into other resources and I didn't see the same solution
Since the roots of the (s_1)^2 = 0. This case is called repeated roots where s_1,1 = s_1,2 = 0. Hence he used the general form that he had derieved in his earlier video of reduction of order. Hence he writes (c_1 + c_2 t)e^{-1/2*pt) where since we dont have damping p=0 and hence, c_1 + c_2 t. I recommend you see his video th-cam.com/video/f87L_lUsgI0/w-d-xo.htmlsi=esqq6lKExJYIsym6
In a few hours I got my last exam of mechanical vibrations, I didn´t understand my school teacher so I saw all the videos until here and now I really know . I wanna thank you for sharing your knowledge in such an easy way. You're a great teacher. Greetings from Querétaro, México.
Thanks for your nice compliment. I hope you passed your exam!
In 7:30 , why you did't add -k^2 to the determinant=0 ?
It is canceled by the +k^2 from multiplying the entries on the main diagonal.
are you sure about the general Solution? Does that mean the system is unstable if c2 is not zero....I looked into other resources and I didn't see the same solution
Since the roots of the (s_1)^2 = 0. This case is called repeated roots where s_1,1 = s_1,2 = 0. Hence he used the general form that he had derieved in his earlier video of reduction of order.
Hence he writes (c_1 + c_2 t)e^{-1/2*pt) where since we dont have damping p=0 and hence, c_1 + c_2 t.
I recommend you see his video th-cam.com/video/f87L_lUsgI0/w-d-xo.htmlsi=esqq6lKExJYIsym6