Second order differential equation for spring-mass systems

[@IntegrandoADNdelasMatematicas]

Your classes is helping me to clarify some concepts

[@bevinmaultsby]

I'm glad to hear that!

[@maxrybold1531]

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]

You're very welcome!

[@JulianKimbleUW]

What a life saver. Best video I've seen on the topic!

[@bevinmaultsby]

Great to hear! Thanks for watching

[@samblake9953]

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

[@bevinmaultsby]

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

[@thomashowe1920]

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

[@bevinmaultsby]

Thank you! I'm glad you enjoyed it.

[@NamelessProducts]

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]

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

[@Ivan-mp6ff]

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

[@bevinmaultsby]

You’re welcome! Glad you enjoyed it

[@rachelb7500]

This was so clear and helpful! Thank you so much for sharing

[@bevinmaultsby]

You're welcome!

[@ShakilAhmedBhuiyan]

Very precise lecture. Very easy to understand.

[@bevinmaultsby]

Thank you!

[@tusharnath2408]

Excellent Video. Thank you for it.

[@bevinmaultsby]

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

[@nowardchaselenkana8596]

clear and straight forward... cheers Doc

[@bevinmaultsby]

Glad you enjoyed it!

[@Ivan-mp6ff]

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]

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]

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]

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

[@Jacoblikesyoutube]

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]

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]

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

[@bevinmaultsby]

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

[@Jacoblikesyoutube]

@@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.

[@navanithnavanith773]

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

[@bevinmaultsby]

Glad it helped! Springs are fun :)

[@Durr-E-Shehwar-lf3ln]

Thanks for such explanation ❤️

[@bevinmaultsby]

You are very welcome!

[@padraiggluck2980]

Nice presentation. I don’t understand where the t comes from in ex.3 in (c1 + c2*t).

[@bevinmaultsby]

Does this help? kzbin.info/www/bejne/ZpqZgKlsoa96b7c

[@padraiggluck2980]

@ Yes, thank you.

[@reaganjustice]

why cant you be my teacher for diff equ? thank you so much

[@bevinmaultsby]

You're very welcome!

[@kaartikvij9328]

thanks a lot mam, u really did grt

[@bevinmaultsby]

Thank you! Glad you liked it

[@enigmath0630]

Excellent!

[@bevinmaultsby]

Glad you liked it!

[@neeb]

Thank you

[@bevinmaultsby]

You’re welcome!

[@mattbabik8417]

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

[@bevinmaultsby]

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]

@@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]

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

[@xghostable8850]

Noice video

[@bevinmaultsby]

Thank you!