Second order differential equation for spring-mass systems

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Let's look at modeling the motion of a spring-mass system (a harmonic oscillator) using a second-order differential equation. From Newton's Second Law, we arrive at mx'' + cx' + kx = 0 (or a forcing function), where x(t) is the position of the spring-mass over time, m is the mass, c is the damping coefficient, and k is the constant from Hooke's Law. We focus on the effects of damping and how to detect what kind of damping a spring-mass system has based on the roots of the characteristic equation.
Four types of spring motion are discussed based on the roots: undamped (no damping force), underdamped (small damping), critically damped (damping force just prevents oscillation), and overdamped (large damping). Different damping leads to different behaviors, which we can illustrate with MATLAB simulations.
#mathematics #math #differentialequations #ordinarydifferentialequations #stemeducation #harmonicoscillator #hookeslaw #physics #matlab #matlabsimulation #iitjammathematics

Опубликовано:

29 сен 2024

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Комментарии : 40

@thomashowe1920 10 месяцев назад

Clear and concise. I wish I had access to this when I took this class.

@bevinmaultsby 10 месяцев назад

Thank you! I'm glad you enjoyed it.

@kaartikvij9328 4 месяца назад

thanks a lot mam, u really did grt

@bevinmaultsby 4 месяца назад

Thank you! Glad you liked it

@Ivan-mp6ff 3 месяца назад

so pleasurable to watch, informative and detailed. Pretty in all aspects. Thank you

@bevinmaultsby 3 месяца назад

You’re welcome! Glad you enjoyed it

@xghostable8850 2 месяца назад

Noice video

@bevinmaultsby 2 месяца назад

Thank you!

@NamelessProducts 5 месяцев назад

Im trying to learn qualitative analysis of nonlinear 2nd order differential equations and all the examples so far have been in springs. This helped a lot. Thank you.

@bevinmaultsby 5 месяцев назад

Excellent! The reason is that spring-mass systems are nice harmonic oscillators (when a system experiences a restoring force proportional to its displacement).

@maxrybold1531 5 месяцев назад

I used this tutorial to brush up my understanding of characteristic equations that describe the behavior of a spring mass damper system to confirm simulation results, via Desmos, of essentially the same system outlined in a SolidWorks tutorial textbook. Great explanations, thanks!

@bevinmaultsby 5 месяцев назад

You're very welcome!

@navanithnavanith773 4 месяца назад

Very clear and precise explanation , helped me understand the concept very quickly. You saved my semester marks 😀😀😀

@bevinmaultsby 4 месяца назад

Glad it helped! Springs are fun :)

@Ivan-mp6ff 3 месяца назад

At about 23:23, critical damping, what would be the corresponding units of C1 and C2 in order to be consistent with the dimension of LHS of the equation, i.e.distance? I am trying to do a dimensional analysis on it. Thank you.

@bevinmaultsby 3 месяца назад

Great question! Assuming we are working standard units, C1 would be meters, and C2 would be meters/second. Exponential functions are dimensionless, so we don't associated any units to the first term. Then it would need to be m + (m/s)s. Hope that helps!

@Ivan-mp6ff 3 месяца назад

Thank you for the prompt reply. I was interested in the units because it may shed light on where could they have come from. I am a medical doctor interested in linking engineering science to medical science and your quality uploads help tremendously. Now that I have confirmed by your helped that they are of different units, when I model vibration and natural frequency to living tissues, I know these seemingly arbitrary constants actually come from different sources. Thank you once again for being my virtual tutor Hope your generosity will continue to grace me with more knowledge that will benefit my patients in the near future.

@bevinmaultsby 3 месяца назад

@@Ivan-mp6ff What interesting concepts you must be studying. I'm glad my videos are helpful!

@nowardchaselenkana8596 5 месяцев назад

clear and straight forward... cheers Doc

@bevinmaultsby 5 месяцев назад

Glad you enjoyed it!

@tusharnath2408 6 месяцев назад

Excellent Video. Thank you for it.

@bevinmaultsby 6 месяцев назад

You're very welcome, I'm glad you enjoyed it!

@Jacoblikesyoutube 7 месяцев назад

I'm a little lost on the step at 16:45, the last step of the first example. x(t) = cos(2t) because it's the only value at the initial condition that equals 0? So in another situation if both trig functions provided a non-zero output, we might end up with x(t) = c_1 * cos(2t) + c_2 * sin(2t)? Is it effectively always x(t) = c_1 * cos(2t) + c_2 * sin(2t) but the result in the first example simplifies to x(t) = cos(2t)?

@bevinmaultsby 7 месяцев назад

Yes--you're understanding this correctly. The general form of the solution is x(t) = c_1* cos(2t) + c_2*sin(2t), where the coefficients c_1 and c_2 are determined by initial conditions. In this particular scenario, with x(0)=1 and x'(0)=0, it turns out that c_1=1 and c_2=0. Here's a different scenario you can work through: if x(0)=2 and x'(0)=1, then c_1 = 2 and c_2 = 1/2. Then the solution would be x(t) = 2*cos(2t) + 1/2 * sin(2t). Does that help?

@Jacoblikesyoutube 7 месяцев назад

@@bevinmaultsby Yeah that makes sense! In this scenario you would need to also handling it like the second example that was underdamped?

@bevinmaultsby 7 месяцев назад

@@Jacoblikesyoutube Maybe, what do you mean by handling? I want to make sure you're making the right connection between the examples.

@Jacoblikesyoutube 7 месяцев назад

@@bevinmaultsby My understanding is that the key difference between the 4 examples is the damping coefficient. In the scenario of your earlier reply where c_1 = 2 and c_2 = 1/2 then the damping effect would be underdamped and thus we would have to find the complex roots values and proceed in a method similar to the 2nd example.

@samblake9953 7 месяцев назад

Awesome stuff! Super clear and I love the fade outs and ins!!

@bevinmaultsby 7 месяцев назад

Thanks so much! I'm glad you enjoyed it.

@ShakilAhmedBhuiyan 9 месяцев назад

Very precise lecture. Very easy to understand.

@bevinmaultsby 9 месяцев назад

Thank you!

@enigmath0630 5 месяцев назад

Excellent!

@bevinmaultsby 5 месяцев назад

Glad you liked it!

@neeb 6 месяцев назад

Thank you

@bevinmaultsby 6 месяцев назад

You’re welcome!

@mattbabik8417 8 месяцев назад

I get 12/35 and 2/35 for the last problem when I put it into wolfram alpha to solve

@bevinmaultsby 8 месяцев назад

Hmm, I just checked f[t_] := (12/37) Cos[t] + (2/37) Sin[t] f''[t] + .5 f'[t] + 4 f[t] // FullSimplify and got cos(t). How did you evaluate it?

@mattbabik8417 8 месяцев назад

@@bevinmaultsby oops i missed a minus sign. Sorry for doubting 🙏. Amazing video though. I am trying to understand the math behind MR elastography calculations and this helped a lot on the differential side.

@bevinmaultsby 8 месяцев назад

No worries... I'm glad this was helpful, what an interesting subject to study! Good luck.