VID00129

[@omarbenlamhidi6591]

love your videos and they were amazing in understanding the concepts that are skimmed over in college classes, the E&M videos were better than most of professors explanations and the way you relate and derive is crystal clear. I do however have a question about this approximation: why is in necessary to say that it approximately equals tan(theta)? the sin(theta) already equals = opposite / hypotenuse which should be x/L , no? Whereas the tangent would be opposite / adjacent = the x / perpendicular part of L on the y axis. I do see that they would be approximately the same when theta is tiny, but what's the point when sin(theta) is ready already. Maybe I misunderstood and not even sure if you still look at these, a response would be appreciated. If not, God Bless you and your family sir! I wish you had more videos!!!

[@Phobos221B]

Correct, i was wondering the same thing

[@ermitz90]

Hmm...feel like sine theta = opp/hyp right?? ^^ btw, nice vid.

[@NamNguyen-gd2yf]

I think you made a mistake here. tan(theta) = opp/adj , *not* opp/hyp (that's sin(theta) ). So we don't need to use tan(theta) at all.

[@_mallol_5828]

10/10

[@surendrakverma555]

Thanks Sir 👍

[@branum1983]

Why does the sine of theta need to be compared to tangent of theta. When sine is already opp/hyp=x/L? Also isn't SAA derived from Taylor polynomials where sin of theta is just approximately theta when we assume small and displacement and SHM? Just wanted some clarification because physical pendulums seem to be more involved than this.

[@kryptynite76]

Wouldn't it be sin(x) = opp/hyp instead of tan(x)? And also, when you solved for the period near the end, Wouldn't T = 2pi/w? I just feel confused about how you concluded that in the end

[@roywastaken]

superb.

[@jordabox]

I think you forgot to change the video title

[@tıbhendese]

6:41 please dont do that sound ! its annoying. thanks