Great job. No BS, clear, and perfect pace. It is ironic that students pay Universities the big $$$, for the privilege of learning online by themselves and for free.
Hi, here are the full steps: (-w^2m1 + k1 + k2)*A1 - k2*A2 = 0 Bring the k2 A2 term to the other side: (-w^2m1 + k1 + k2)*A1 = k2*A2 Divide by A2 on both sides: (A1/A2) * (-w^2m1 + k1 + k2) = k2 Divide by the (-w^2m1 + k1 + k2) on both sides: A1/A2 = k2 / (-w^2m1 + k1 + k2) Does that help? :)
That's a really good question! The answer is a bit complicated and involves some linear algebra. The problem can be expressed as: Solve the following equation to find X [B]X = 0 Where B is the big matrix and X = [A1; A2]. Clearly one solution is X = 0 (since [B]0 = 0), but this is trivial since it corresponds to the solution that A1 = 0 and A2 = 0 (no motion). We want to solve for nontrivial solutions to this. IF [B] was invertible then: we could invert [B] by both sides to get [B]^-1 [B] X= [B]^-1 0 X = [B]^-1 0 X = 0 but this is trivial and we don't want to solve for trivial solutions So the only way for nontrivial solutions to exist is when [B] is NOT invertible. A matrix is only not invertible if the det([B]) = 0 Hope that makes sense :)
Thank you for this video, one of the best videos I've seen about vibrations! Do you also have examples of torsional systems? Like for example two discs with a connecting shaft in order to determine their natural frequencies/mode shapes?
Great video, Does this translate to Multi degree of freedoms other that 2? Why do we have other methods it this works so well? Like dunkerly rayliegh matrix iteration etc?
Matthew James 1 second ago Hey Randy! Sorry for the late reply. Just a quick clarification, at 3:33 i'm not saying that w2 = A1/A2. What I was saying is that the amplitude ratio evaluated at w = w2 yields the following result. We know that it is actually A1/A2 because (-w^2m1+k1+k2)A1 - k2A2 = 0 (-w^2m1+k1+k2)A1 = k2A2 A1 = A2 k2/(-w^2m1+k1+k2) A1/A2 = k2/(-w^2m1+k1+k2) Hope that helps :)
Hi there, sadly I do not have a video that includes damping and external forces for 2DoF yet. But the process is very similar. If you're interested in solving for the steady state solution to an harmonic force acting F=F0 cos(wt) on the blocks then the equations of motion will yield a mass matrix [M], a stiffness matrix [K], a damping matrix [C] and a force vector {F}. [M]x.. + [C]x. + [K]x = {F}. Then, the way you solve this is by assuming a solution in the form x1 = Re(X1 e^-iwt) and x2 = Re(X2 e^iwt) and modeling the harmonic force as F = F0 e^iwt. Then substituting this in you will get [M]X + [C]X + [K]X = {F0} (-w^2[M] + iw[C] + [K])X = {F0} You can then invert the matrix, and solve for X. Notice X will be complex. Then find the steady state displacements using x1 = Re(X1 e^-iwt) and x2 = Re(X2 e^iwt). I've skipped a few of the mathematical steps and justifications, but that's the big picture. Hope that helps. This summer I plan to make a lot more vibration videos :)
Oh and I almost forgot to say that this is just for the steady state response! To find the total response, you need to add it to the transient solution (x = x_transient + x_steady state). x_transient can be found by setting the external force to 0 such that [M]x.. + [C]x. + [K]x = 0. Then guess a solution in the form x1 = A1e^lambda t and x2 = A2e^lambda t. Substituting this in will yield (lambda^2 [M] + lambda [C] + [K])A= 0 Since the RHS is 0, the determinate of the matrix must be 0. Solving for lambda will yield 4 complex solutions. then using the superposition theorem, the total transient solution is: x = A e^lambda1 t + Be^lambda2 t + Ce^lambda3 t + De^lambda4 t where A B C and D are constants to be determined from the initial positions and velocities of both blocks. The whole term can also be simplified accordingly using eulers formula cos(y)+isin(y) = e^iy. Once you have the transient solution and the steady state solution, then you can literally just add them up to get your total response. :D Just a minor side-note though: usually Engineers are only interested in the steady state response, so you'll usually only be asked to find that in an exam :)
I have question, The initial conditions of any vibrating system are to be applied to a: 1) Transient solution 2) Steady-State solution 3) Total solution 4) None of them Thanks I am waiting ur answers guys
Hey mate sorry for the late reply! Finding the modeshape for the third natural frequency is a bit tough to do by hand so I've written a MATLAB script for you. Notice in my code I didn't use det(-w^2M +K) = 0. I could have done! But it would have gotten messy. Instead I wrote -w^2 MX + KX = 0 as w^2X = inv(M)KX, which shows that the eigenvalues of the matrix inv(M)K are the natural frequency squared, and the mode shapes are the eigenvectors. Here is my code: clc %Equations of Motion %m1x1.. +(k1+k2)x1 - k2x2 =0; %m2x2.. -k2x1 +(k2+k3)x2 -k3x3 =0; %m3x3.. -k3x2 +k3x3 =0; syms k1 k2 k3 m1 m2 m3 x1 x2 x3 w1 w2 w3 M = [m1 0 0; 0 m2 0; 0 0 m3]; K = [k1+k2 -k2 0; -k2 k2+k3 -k3; 0 -k3 k3]; %solves eigenvector (X) and natural frequency squared (w^2) [X, natfreq_sqrt]= eig(inv(M)*K); w = sqrt(diag(natfreq_sqrt)); %natural freq w1 = w(1) w2 = w(2) w3 = w(3) %modeshapes X1 = X(:,1) X2 = X(:,2) X3 = X(:,3)
That is the best video I have seen explaining different modes and different natural frequencies. Thank you so much
Glad you liked it :)
This just summed up my 1 week lecture in 5 mins. Thank you !!
absolute beast. took me hours a few weeks ago looking at my prof's lecture slides to somewhat learn it. you just cleared everything up in 5 mins!
3 hours to my exam then I see this.. thank you🙏🏾
Great job. No BS, clear, and perfect pace. It is ironic that students pay Universities the big $$$, for the privilege of learning online by themselves and for free.
This video taught me more than my whole vibration course at uni
This is actually useful and concise
way better than my lecturer
what a legend. Great Explanation!!
Amazing video! Very helpful. You're very good at what you do, keep it up!
Thanks. Very explanatory
You are welcome!
also at 3:22 would it be -k2 on top of the amplitude ratio? in your above equation its a negative k
Hi, here are the full steps:
(-w^2m1 + k1 + k2)*A1 - k2*A2 = 0
Bring the k2 A2 term to the other side:
(-w^2m1 + k1 + k2)*A1 = k2*A2
Divide by A2 on both sides:
(A1/A2) * (-w^2m1 + k1 + k2) = k2
Divide by the (-w^2m1 + k1 + k2) on both sides:
A1/A2 = k2 / (-w^2m1 + k1 + k2)
Does that help? :)
Jesus Christ. Didn't think it was that simple. Thanks!
Phenomenal! Thank you so much!!
Why is it that taking the determinant of that final amplitude matrix gives us our natural frequencies?
That's a really good question! The answer is a bit complicated and involves some linear algebra. The problem can be expressed as: Solve the following equation to find X
[B]X = 0
Where B is the big matrix and X = [A1; A2]. Clearly one solution is X = 0 (since [B]0 = 0), but this is trivial since it corresponds to the solution that A1 = 0 and A2 = 0 (no motion). We want to solve for nontrivial solutions to this.
IF [B] was invertible then:
we could invert [B] by both sides to get
[B]^-1 [B] X= [B]^-1 0
X = [B]^-1 0
X = 0 but this is trivial and we don't want to solve for trivial solutions
So the only way for nontrivial solutions to exist is when [B] is NOT invertible.
A matrix is only not invertible if the det([B]) = 0
Hope that makes sense :)
Thank you for this video, one of the best videos I've seen about vibrations! Do you also have examples of torsional systems? Like for example two discs with a connecting shaft in order to determine their natural frequencies/mode shapes?
what an explanation thanks so much!
Awesome work ! needed for my vibs course, thanks for the help
no worries! Glad you found it helpful!
HOW WE WILL SOLVE IT BY MATRIX METHOD WITHOUT USING SUPPORT OF DIFFERENTIAL EQUATION
Great video, Does this translate to Multi degree of freedoms other that 2? Why do we have other methods it this works so well? Like dunkerly rayliegh matrix iteration etc?
I think omega(2) is written as A2/A1 actually. A1/A2 would have the k2 term in the denominator.
Matthew James
1 second ago
Hey Randy! Sorry for the late reply. Just a quick clarification, at 3:33 i'm not saying that w2 = A1/A2. What I was saying is that the amplitude ratio evaluated at w = w2 yields the following result.
We know that it is actually A1/A2 because
(-w^2m1+k1+k2)A1 - k2A2 = 0
(-w^2m1+k1+k2)A1 = k2A2
A1 = A2 k2/(-w^2m1+k1+k2)
A1/A2 = k2/(-w^2m1+k1+k2)
Hope that helps :)
Just an amazing explanation.. Thanks
Glad it was helpful!
the simulation software please ?? is that python pygame ???
Great video. Do you, by chance, have one that includes damping?
Hi there, sadly I do not have a video that includes damping and external forces for 2DoF yet. But the process is very similar. If you're interested in solving for the steady state solution to an harmonic force acting F=F0 cos(wt) on the blocks then the equations of motion will yield a mass matrix [M], a stiffness matrix [K], a damping matrix [C] and a force vector {F}. [M]x.. + [C]x. + [K]x = {F}. Then, the way you solve this is by assuming a solution in the form x1 = Re(X1 e^-iwt) and x2 = Re(X2 e^iwt) and modeling the harmonic force as F = F0 e^iwt. Then substituting this in you will get
[M]X + [C]X + [K]X = {F0}
(-w^2[M] + iw[C] + [K])X = {F0}
You can then invert the matrix, and solve for X. Notice X will be complex.
Then find the steady state displacements using x1 = Re(X1 e^-iwt) and x2 = Re(X2 e^iwt).
I've skipped a few of the mathematical steps and justifications, but that's the big picture. Hope that helps. This summer I plan to make a lot more vibration videos :)
Oh and I almost forgot to say that this is just for the steady state response! To find the total response, you need to add it to the transient solution (x = x_transient + x_steady state). x_transient can be found by setting the external force to 0 such that [M]x.. + [C]x. + [K]x = 0. Then guess a solution in the form x1 = A1e^lambda t and x2 = A2e^lambda t. Substituting this in will yield
(lambda^2 [M] + lambda [C] + [K])A= 0
Since the RHS is 0, the determinate of the matrix must be 0.
Solving for lambda will yield 4 complex solutions. then using the superposition theorem, the total transient solution is:
x = A e^lambda1 t + Be^lambda2 t + Ce^lambda3 t + De^lambda4 t
where A B C and D are constants to be determined from the initial positions and velocities of both blocks. The whole term can also be simplified accordingly using eulers formula cos(y)+isin(y) = e^iy.
Once you have the transient solution and the steady state solution, then you can literally just add them up to get your total response. :D Just a minor side-note though: usually Engineers are only interested in the steady state response, so you'll usually only be asked to find that in an exam :)
I have question,
The initial conditions of any vibrating system are to be applied to a:
1) Transient solution
2) Steady-State solution
3) Total solution
4) None of them
Thanks I am waiting ur answers guys
3) total solution :D
i love u
What will be the mode shape for 3rd block
Hey mate sorry for the late reply! Finding the modeshape for the third natural frequency is a bit tough to do by hand so I've written a MATLAB script for you. Notice in my code I didn't use det(-w^2M +K) = 0. I could have done! But it would have gotten messy. Instead I wrote -w^2 MX + KX = 0 as w^2X = inv(M)KX, which shows that the eigenvalues of the matrix inv(M)K are the natural frequency squared, and the mode shapes are the eigenvectors. Here is my code:
clc
%Equations of Motion
%m1x1.. +(k1+k2)x1 - k2x2 =0;
%m2x2.. -k2x1 +(k2+k3)x2 -k3x3 =0;
%m3x3.. -k3x2 +k3x3 =0;
syms k1 k2 k3 m1 m2 m3 x1 x2 x3 w1 w2 w3
M = [m1 0 0; 0 m2 0; 0 0 m3];
K = [k1+k2 -k2 0; -k2 k2+k3 -k3; 0 -k3 k3];
%solves eigenvector (X) and natural frequency squared (w^2)
[X, natfreq_sqrt]= eig(inv(M)*K);
w = sqrt(diag(natfreq_sqrt));
%natural freq
w1 = w(1)
w2 = w(2)
w3 = w(3)
%modeshapes
X1 = X(:,1)
X2 = X(:,2)
X3 = X(:,3)
Great video, thanks
Glad you liked it :)
Great video
Thanks :D
great!!