@@DdoubleB03 I am explaining u.In case of 1st derivative of acceleration,we get dv/dt.After then, when we differenciate the velocity again we gets lastly d^2x/dt^2. this is how the "second derivative" comes from.
4:11 Isn't there a +C when you are integrating the velocity and if there is, why is there a +C on the other side of the equation? Don't the +C's cancel each other out?
No, there are different constants on each side, they don’t cancel each other out, you can call them for example c1 and c2, you subtract c1 on both sides and ur left with c2-c1 in the right hand side, and a constant minus another constant is a new constant, which you denote just c
He integrated w but with respect to t so w becomes a constant and gets multiplied by it and if you take its derivative it becomes w hence lhs equal to rhs
It doesn't matter. Sin and Cos can be used interchangeably if you change the phase constant. So if phi is 90 degrees, then it becomes cos and phi becomes 0
If the particle starts from its mean position we take sine as position is 0 but when the particle in shm starts from its maximum displacement A then we say it as cos
To find acceleration of a particle at maximum position from its equilibrium state from Newtown law we use a=F÷m F=-kx we take k÷m as beta during derivation we get root beta and we denote it as omega . So omega square is k÷m
X is a variable which we are analysing its behaviour as the time goes. So it is not always the same thing as the Amplitude. If they're same at a point, it means that the object moved as far as possible from the starting point which have now became as big as the position vector of the amplitude. So yeah it would be 0 in this case but you're forgetting that this term is inside of the cos function, so cos(0) would be 1 and x = A (Amplitude) as intented to be.
At last, somone who explains it... Thank u!!!
it's an ok explanation. He doesn't explain where the "second derivative" comes from.
@@DdoubleB03 I am explaining u.In case of 1st derivative of acceleration,we get dv/dt.After then, when we differenciate the velocity again we gets lastly d^2x/dt^2. this is how the "second derivative" comes from.
Unbelievable.! You taught the topic very easily within a few minutes.
I saw other videos but this one mad most sense to me
4:11 Isn't there a +C when you are integrating the velocity and if there is, why is there a +C on the other side of the equation? Don't the +C's cancel each other out?
No, there are different constants on each side, they don’t cancel each other out, you can call them for example c1 and c2, you subtract c1 on both sides and ur left with c2-c1 in the right hand side, and a constant minus another constant is a new constant, which you denote just c
Very good at explaining ,,,
How is this easier than adding? You teach well. Thank you.
At 03:20, he mentioned about the chain rule. Which part is it? Why dv/dt = dv/dx*dx/dt? Can someone explain?
Part of chain rule the dx and dx cancel out and we get dv/dt
It is best on you tube 🎉
0:40 Isnt x away from the equilibrium position
Please in the course of finding the integral why didn't you integrate (w)?,
I think because the integration only depends on x and t, then ω becomes constant that can be placed outside of integral
He integrated w but with respect to t so w becomes a constant and gets multiplied by it and if you take its derivative it becomes w hence lhs equal to rhs
ω here is a constant watch video again he takes ω=sqrt(k/m)
Please sir which software did you used to record your videos🙏
At which case shm equation will be in cosine and when it will be in Sine ?? Please clarify
@Luxuryhitler not necessary...sin and cosine functions can be converted to each other interchangeably..
Hi
It doesn't matter. Sin and Cos can be used interchangeably if you change the phase constant. So if phi is 90 degrees, then it becomes cos and phi becomes 0
isnt it suppose to be Acos(wt+Ö)
If the particle starts from its mean position we take sine as position is 0 but when the particle in shm starts from its maximum displacement A then we say it as cos
@@suchitawasnik3633thanks
Depends of the initial point. They're actually the same thing
How ω² = k/m , can anyone explain
the w itself is root of k/m then k/m equals to the ω²
@@eymendediler5357but how do you derive w = k/m? is this purely definitional?
@@mailingbox F = mw^2x = kx
@@mailingboxIt is purely definitional.
To find acceleration of a particle at maximum position from its equilibrium state from Newtown law we use a=F÷m F=-kx we take k÷m as beta during derivation we get root beta and we denote it as omega . So omega square is k÷m
Nice explanation bro
Good learning from continu
omg this was so helpful thank you.
😮Thank you so much
Im amazed by this form, thanks a lot bro !
Like it
How c= 1/2 A^2w^2
If you choose the moment where the velocity is 0, you can find the exact value of c by eliminating the v variable
Thanks you sir in this concept is completed 😀 simple harmonic motion
Thanks dude
TOP G
this video doesn't even make sense.
@@DdoubleB03 it helped me thu, i do not know about u
A=amplitude
X=Amplitude
X=A
Sqrt(A^2 - X^2)
Sqrt(0)?????
X is a variable which we are analysing its behaviour as the time goes. So it is not always the same thing as the Amplitude. If they're same at a point, it means that the object moved as far as possible from the starting point which have now became as big as the position vector of the amplitude. So yeah it would be 0 in this case but you're forgetting that this term is inside of the cos function, so cos(0) would be 1 and x = A (Amplitude) as intented to be.
Nice
This tutorial need another tutorial
What’s c mean
That's an integration constant
Isn't it cos
you can use cos or use sine. Both works.
your channel need to make SEO
sin(sin^-1(x/A)) =/= x/A
??
It is
Sin*(Sin^-1) is Sin*(1/Sin) by considering basic maths , you will get the rest X/A .