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Obliv
Obliv
01:15
I'm blanking out, if I'm graphing points with a best fit line in the form of f vs N
and the equation is $f = \frac{\sqrt{F}}{2L\sqrt{\mu}}N$
I know F and L as constants, i'm trying to find $\mu$ experimentally by analyzing the graph and the slope is equal to $\frac{1}{\sqrt{\mu}}$ right
so all I need to do is take the reciprocal of the slope and square it to get the experimental value?
 
6 hours later…
ACuriousMind
ACuriousMind
06:51
@Obliv yes
or, well, that is if you graphed $2Lf$ vs. $\sqrt{F}N$. If you graphed $f$ vs. $N$ then the slope is of course $\frac{\sqrt{F}}{2L\sqrt{\mu}}$ and you need to first use your knowledge of $F$ and $L$ to arrive at $1/\sqrt{\mu}$ in isolation
@Feynman_00 nah, I don't think that's in QoGS.
Feynman_00
Feynman_00
07:28
I see
 
3 hours later…
PM 2Ring
PM 2Ring
10:58
@Jim He didn't take the hint. He's written another answer promoting his "theory": physics.stackexchange.com/a/728942/123208
 
7 hours later…
Obliv
Obliv
17:38
what causes the damping of a wave? Is it entropy?
Jim
Jim
@PM2Ring sigh
Obliv
Obliv
18:15
imgur.com/a/B9xZZBO guys i substituted the sum-to-product identity like so
and I got a $-\omega t$ in the cos since it's a - b / 2 so it's two $-\omega$
is this just a matter of $-\omega t = \omega t$
for a cosine function
Feynman_00
Feynman_00
19:13
@Obliv yes.

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