Single Degree of Freedom (SDOF) System with Gravitational Effects
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- Опубликовано: 12 сен 2024
- A mathematical description of the vibrations of a mass-spring system oscillating within a gravitational field. How to account for gravitational effects.
Well done! and I hate to say it but equation 4 should be PLUS ky not MINUS. otherwise the solution would have no cosine terms. you had it right in the line right above it but you didn't change the sign of ky when you moved it to the left side of the equation.
Yep. You're correct. Thank for catching that.
Thanks for sharing though I was getting so confused about where my frame of reference is and where the gravity went.
Great! Thanks so much, I've been struggling a bit with vibrations and I started to realize that I didn't get my basics completely right, these videos are helping me out pretty greatly, thanks again!
Not all heroes wear capes.
keep it up man.
I have simply assumed here that the mass has been displaced by an amount , "y" and that "y" can further be decomposed into two components, ONE COMPONENT due to the external/gravitational load and ANOTHER COMPONENT due to the internal/dynamic loads of the system. Based upon the assumed displacement (and acceleration and velocity) of "m", we will then find the resulting loads using Newton's 2nd Law. In this example I am trying to show why we can ignore the gravitational effects on the mass IF we shift the equilibrium point (the zero point) from "0" to x0.
This is almost identical to what I showed in a previous video which ignores displacements due to external loads. (ruclips.net/video/5Q_A15h7YgQ/видео.html).
Thank you. however can you do one for Multiple Degree of Freedom (MDOF) System with Gravitational Effects
I have found what I need for my modeling subject's test, Gotta suscribe
Awesome !
Awesome explanation.
thanks!
Perfect
How can you explain that the spring elongated an extra distance "y" if no external force was applied? This I still have trouble understanding, can you explain?
This is straight from vibration theory. We assume that the mass has displaced by a small amount and examine the force(s) that this would produce within the system. The reason that it displaces by the extra amount "y" AND at a velocity of ''y_dot" is due to the fact that this is a dynamic problem - in effect, this extra displacement results from the mass's momentum. The reason we have ignored the velocity (y_dot) is that there is no damping present.
good question
18.31 why X0 = mg/k Can u determine this ? Cause i think X0 qual 0
You can find the equilibrium position by setting the spring force equal to the weight. So, k x0 = mg. X0 cannot be equal to zero because of the explanation I gave at the start of the video. If there is no gravity present, then the equilibrium position is at x=0 (based on my choice of coordinate system). When gravity is added, the equilibrium position must be nonzero.
I get lost in plugging in equation 9 to equation 7!
This is how we go about finding the particular solution using the method of undetermined coefficients...this goes back to your calculus classes on differential equations.
The right-hand side of eqn 7 has a term that is just a constant. In this case, we assume a particular solution that is a constant (technically it is a zeroth order polynomial). This is done in equation 9. To find the value of d0, we substitute eqn 9 into eqn 7 - remember that the derivatives of Xp are zero (since it is a constant). The substitution yields k.d0 = mg which can be solved for d0.
I hope this covers your question. Also, I explain this in a little more detail in this video: ruclips.net/video/GLd1MgiTne4/видео.html