Added: 3 years ago
From: khanacademy
Views: 55,098
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  • This is just AMAZING..not the physics part but the way you explain the physics part. My professor will try to "explain" this to us tomorrow but i have hunch it won't be this good.

  • Awesome but I wish this was HD

  • Wow, finally I know where that final equation comes from. Professors are just like "if you solve the differential equation you get yadda yadda yadda." I'm only just taking differential equations this semester (first class tomorrow) so I didn't really know what they were talking about. THANK YOU!

  • People like you deserve heaven!!

  • don't move the pointer so much I'm dizzy :D

  • Can someone explain clearly why x(t) intuitively is equal to Acos(wt) ? Am I just lacking intuition here?

  • AT 8:07 WE HAVE HE ELIMINATE FACTORS ON EACH SIDE, BUT WHY WHY NOT ELIMINATE COSWT ASWELL?

  • @drmo92 You have to account for when coswt is equal to zero. The minus sign can be eliminated because it is constantly -1 and the A can be eliminated because amplitude is never zero. But coswt can possibly be zero. And we all know you can't divide by zero... Or the world will end..

  • @drmo92 he does :)

  • @lprk94 kinda lol.

  • YouTube should give you a bigger blackboard!!

  • What a nice guess you had!!...

  • I find one of these 10 minute videos more helpful than an hour lesson from my teacher. Thanks!

  • Excellent teaching

  • Is cosine an arbitrary function you just chose or is it THE function that always applies to any example?

  • @indrazor1 hey sorry for a late reply, but I think Cosine is the function that is to be used, if the spring were to have a maximum displacement at t = 0. Not sure though.

  • when u do the derivative of Acos(wt) how do you know which variables are constants? I'm guessing A is constant cause the function never changes amplitude, and time is always the same, so those would be the constants.

  • @Morelloo1 time is the independent variable of the function, its not a constant. If it were it's derivative would have been zero and not 1 like we found. However small omega and A are constants, you can see why A is a constant and omega was assumed so that we could find the (K/m)^1/2 for the function.

  • why is x(t)=Acos("omega" t) but not just Acos(t) ?

    please help!

  • @vietpride4life1990 If it is just A cos(t), the period must be 2Pi. If x(t) = Acos("omega"t), the period is 2Pi/omega. Since we know the period is not necessarily 2Pi, we make the function A cos("omega"t)

  • omfg this is ingenius!!!

  • Note that 8000 people went on to part 3. You seem to be very wrong.

  • @DarkKnightBob1o1 I DO :D

  • שיחקת אותה גבר

  • Why did you use a cosine function  while my book uses sine functions?

  • @me9787

    Might still be helpful to reply. That's might be because in Sal's example, the spring is not at rest at t=0. it is streched to A. Hence, at t=0, x=A. In your book the mobile might be at rest at t=0.

    If the mobile is not streched at t=0, then x at t=0 will be zero and x as a function of time must be ressemble a sine function because sine of 0=0 whereas cosine 0 is never zero. Cosine of zero is one or A*1.

    See Sal's videos on trigonometry and especially amplitude.

  • Is there also a video about a pendulum, where a mass is hung at a wire (don't know the english expression, german is "Fadenpendel") by Sal?

    I would be interested in the reason, why the differential equation there cannot be solved by calculus or with elementary functions.

  • finally i know something about this.. can i ask how' the (omega X T ) came out of this equation?

  • At what time in the video?

  • w(omega) = 2pi/T This is the ratio between 2pi (the period of a general cos graph and T (the period of a certain cos graph). The general equation for a cos graph is A+Bcos(Cx+D) where C=2pi/T where A, B, C and D are all constants. In this graph omega = C so omega is a constant. A is where the graph is vertically translated and this graph is in the centre position. B the amplitude of the graph and D is how much the graph moves along the x-axis. This graph has not moved on the x-axis at all so D=0

  • this a great vid.

    20,000 views, wh0ot!

  • Gracias por la explicación.

  • Good work, but you've forgotten the phase parameter (which can say where's the body at some time when we detect its motion starting NOT from equilibrium point). I.e. the final solution looks: x(t) = Asin(wt + phi). if we know that at t0 the body is for example A/2 from its equilibrium and moves for example to left. Then we have phase shift of 5PI/6.

    Also I've liked much the finding of w = sqrt(k / m) which is done strictly by substituting.

  • My professor was terrible at explaining this, but everything you said made perfect sense-- thanks.

  • wwooow you are soo smart!!!!! i wish i had your brains,  but you sound older, im only 13 :]

  • you are the MAT GOD!!!!!!

    thank you so much... i teach me so much

    u helpd me get on A+ on my last test by one of your other videos

    THANKS MAN!!!

  • this is a great service keep up the great work khan

  • you are god of mathematics!!!

    thankyou for your video and time ....

    you will help so much of us/

  • Thank God for his slow-ness so we slow people can understand. Since you're Einstein, what are you doing here? Write a Nobel Prize Theory and leave us alone.

  • Mr. Kahn Great channel! could you please arrange your posts so that one may find the appropriate Parts sequentially

  • Nice work,

    You should say Omega for w and Kappa for K.

    Where's the Φ

  • T = teletubbies

  • HAHAHAHAHAHAHAHAHAHAHAHA

  • ah, remind of me simpler days before the complex analysis and partial differential equations...

  • if only i had found these videos before my physics exams 2 days ago XD oh well...

  • nerds rule the world

  • this is trippy looking

  • if X is a function perspective of the second phase if motion is only a parabola of gravity and time the sequernce for time travel :)

  • where is part 1??

  • Thank you Khan, i enjoy your videos

  • understanding what he's saying is completely different than being able to do it on your own.

    Maybe other teachers aren't as good as khan.

  • the hard thing about physics usually is to know what to do in a question...understanding wtf the question is asking is one thing but knowing how to manipulate the available data to fit the equations you have....you have to see that this is equal to this and this has the same force as this

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