It is an homogeneous second ODE. Using x(t)= exp(st) then differentiate twice gives x(t)' = s*exp(st) and x(t)"= s^2*exp(st). Substituting this in x" + d/m*x' + k/m*x=0 gives --> s^2*exp(st) + d/m*(s*exp(st)) + k/m*exp(st) = 0 . Now we take exp(st) common as its a common factor in all terms. This results in (s^2 + d/m*s + k/m)*exp(st). The natural frequency is given as wn=sqrt(k/m) --> wn^2=k/m. Damping factor is given as J=d/2mwn. rewrite this in d/m gives d/m=2Jwn. Substituting wn^2 and d/m in (s^2 + d/m*s + k/m)*exp(st) gives --> (s^2 +2JwnS + wn^2)exp(st)=0. As exp(st) will never reach zero, the terms in the brackets must equal 0 in order for the equation to be true. I hope my explanations helps and makes sense:)
I didn't get how the solution part works at 10:26
How did the quadratic eqn with s^2 + ... gets multiplied to the exponent and equated?
It is an homogeneous second ODE. Using x(t)= exp(st) then differentiate twice gives x(t)' = s*exp(st) and x(t)"= s^2*exp(st). Substituting this in x" + d/m*x' + k/m*x=0 gives --> s^2*exp(st) + d/m*(s*exp(st)) + k/m*exp(st) = 0 .
Now we take exp(st) common as its a common factor in all terms. This results in (s^2 + d/m*s + k/m)*exp(st). The natural frequency is given as wn=sqrt(k/m) --> wn^2=k/m. Damping factor is given as J=d/2mwn. rewrite this in d/m gives d/m=2Jwn. Substituting wn^2 and d/m in (s^2 + d/m*s + k/m)*exp(st) gives --> (s^2 +2JwnS + wn^2)exp(st)=0. As exp(st) will never reach zero, the terms in the brackets must equal 0 in order for the equation to be true.
I hope my explanations helps and makes sense:)
@@robbybos1999 wow, got it now! thanks so much buddy!
That was a very clear explanation!
Natural frequency isnt it 1 over time? , 1/t=f
Thank you!!