Thanks. This a quite overlooked concept. I was perplexed when I realised (after doing many classes on dynamics) while solving a problem that why don't I take weight of the body in the equilibrium. This video proved what I needed
@INTEGRALPHYSICS3 жыл бұрын
Great vid. This whole video is tremendously helpful once you realize that most of the l-lo-x-xo-u 's that show up are just expressing the change in length (or rate thereof) of the spring/damper.
@BetulKaplan8 жыл бұрын
omg!!! it was soo helpful dude! you teach very simple. thank you!!
@richardbloemenkamp85323 жыл бұрын
Great, I found the same solution. I coundn't find this problem with google with the forcing term on the other side of the spring and damper as the mass. It is not very hard but it is nice to see the confirmation that my calculation was correct. This solution is quite interesting for many cases where a small mass is connected through a spring and damper to a large moving mass/object that does not 'feel' the small mass. Choosing sinusoids with different frequencies for u and then computing the resultant x or (x-u) is also very interesting. In some cases u is unknown but (x-u) can be determined allowing to find u through inversion.
@monikwrite96545 жыл бұрын
Amazing explain than my classroom.....😄🙏
@Baracudus9 жыл бұрын
Nice! You missed a dt "^2" at 11:26 :)
@abhideepsingh44848 жыл бұрын
what is 'u' at 8:12?
@wolfavie8 жыл бұрын
u is u(t) which is the change in height of the ground over time
@annamauroam6 жыл бұрын
wolfavie Of the ground????? why should the ground move???
@AvinashKumar-co1hj4 жыл бұрын
@@annamauroam the ground don't move it's uneven...like giving lift to the axle is u(t)
@r.d.75753 жыл бұрын
Thanks, it's very helpful. Do you know how to calculate the energy stored in the damper ?
@richardbloemenkamp85323 жыл бұрын
Dampers don't store energy. They dissipate energy. All energy that goes into the damper turns into heat. No energy can come out of an ideal damper therefore we cannot speak of energy storage. An ideal spring is exactly the opposite. It dissipates no energy and all the energy put into it is stored and will come out when the spring returns to its non-compressed length.
@Vatea978 жыл бұрын
Hi Greg, When you sum all the forces at static equilibrium, can you neglect the damper because the speed is 0 and the damper's equation depends only on that? I'm working on a project right now and trying to model the same device except the mass is downward (as if the ground was the surface of water and the mass was sinking). The wheel would serve as a floater and both would be connected through a spring with damping properties too. Are your differential equations appliable in such case? At static equilibrium, the mass' weight would be compensated by the wheels buoyancy.
@malezacaminante95777 жыл бұрын
To the first question you ask, the answer is yes as velocity is dx/dt, in equilibrium dx/dt= 0 so all the damper term is cancelled. To the second, yes the equations are the same, the only thing you should notice is that instead the spring compressing it is expanding, but in the end the direction doesnt matter as all the forces act in opposite ways, (newton's third law)
@bilmohammed55775 жыл бұрын
what does it mean if the mass spring damper is = 0 does that mean the system is in equilibrium and is on level ground??
@jowillyntimtim82347 жыл бұрын
Why is it in your final FBD the weight of the vehicle is no more to be found.
@aamirsheikh97896 жыл бұрын
Hii sir ..can you help in finding out the dynamic parameters like stiffness, mass of spring , damping of a mass tuned liquid damper ..
@sakthimurugan38887 жыл бұрын
sir ,kindly tell me which tool or software you use for creating this video.. your video are nice and clearly expressed. thanks sir.
@malezacaminante95777 жыл бұрын
super nice
@Simagha9 жыл бұрын
Thank you so much
@ruhengluo80537 жыл бұрын
I like, a little doubt you can ask?
@disbecknell10 жыл бұрын
What program did you produce this tutorial in?
@annamauroam6 жыл бұрын
I don't get what u(t) is
@HabboBear18 жыл бұрын
I came here to see how hard engineering is. Lmfao, I didn't understand anything.
@melochin82378 жыл бұрын
this is acctualy quite easy if you know how to solve differential equations
@soumyajitsarkar23728 жыл бұрын
u need to learn basic engineering mechanics, basic mechanics on general and also basic differential equation and then natural vibration , then only u will be able to understand this clearly.This are mainly taught at the end of 2nd year or intro of 3rd year in Mech.Engg
@kashefalgita79507 жыл бұрын
HabboBear1 😂😂😂you didn't see any think
@ashaiena98477 жыл бұрын
Why is it in your final FBD the weight of the vehicle is no more to be found.