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And you posted equation 7.5.In the equation 7.4, the author is taking v0=√(C/M)*x
Sorry, I made a typo, I posted the correct one.And you posted equation 7.5.
Yes.What is the total energy of a simple spring oscillator made up of?
Yes.This is for a spring oscillator, right?
It doesn't say much, cause it's trying to explain spring oscillators in a general way.Could you please define the terms in the equation you posted?
v_{0}=√(C/M)*x
If you do the integral you got:Maybe I missed it, but I don't see this in any of the stuff you posted.
If you are going to post typeset equations, it would be worthwhile to learn a little LaTeX. Info => HowTo => https://www.physicsforums.com/help/latexhelp/If you do the integral you got: View attachment 231123. Then -M*v_{0}^{2}=C*x^{2}. So v0=√(-C/M)*x.
Edit: i just realized i forgot to put the minus.
If you are going to post typeset equations, it would be worthwhile to learn a little LaTeX. Info => HowTo => https://www.physicsforums.com/help/latexhelp/
For example, $$\frac 1 2 Mv^2 - \frac 1 2 Mv_0^2$$
If you do the integral you got:. Then -M*v02=C*x2. So v0=√(-C/M)*x.
If you do the integral you got ##\frac{1}{2} Mv^2 - \frac{1}{2} Mv_0^2 =##.
Then ##-M v^2_0 = C x^2##
But we got that ## x = A \sin(\phi)##
(5.12) integral. And yes, i just realized that I cropted it in a wrong way, but basically: $$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2.$$Do what integral? Also, this only half of an equation!
Cause we have thatI don't see how you got this.
I'm just taking that from the text book. I already posted that part but i will do it again if that helps u:Actually, ##x = A \sin \left( \omega_0 t + \phi \right)##.
Cause we have that
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2 = \frac 1 2 Mv^2-\frac 1 2 Cx^2,$$
I'm just taking that from the text book.
But we got that ## x = A \sin(\phi)##
Sorry, sorry, I'm really sorry, i just made a lot of typos, I meantI agree with
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2 ,$$
but I do not understand the second equals sign, I do not understand why
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2$$
is also equal to
$$\frac 1 2 Mv^2-\frac 1 2 Cx^2 .$$
Sorry, i just didn't read "at t=0", which is a huge mistake, but even like that i think is kind of weird cause he have thatNo, you have not taken it from the textbook. You wrote (and used)
That's because
$$ \Delta (T+V) =\Delta E = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2,$$
cause here ##\Delta V=0.##
Thank you so much! I finally get it, all of my questions were between pages 149 and 150. We will see conservation of energy in three classes, that's why I was so confused about that.No,
$$\Delta V = \frac{1}{2} C x^2 - \frac{1}{2} C x_0^2 .$$
See equation (5.20).