I find your explanations clear and to the point, love you (no homo).
@ToppersMaster
9 жыл бұрын
Now, This was outstanding! I wanted to know why/how motion of a pendulum at larger angular displacement was not SHM. Thank you!
@jameshuang9568
10 жыл бұрын
thank you! that mgcos(theta) = T at turning point really explains a lot!! you will not believe our tutor just told us to memorize that!!!
@DocSchuster
10 жыл бұрын
Ouch. In a way, though, that is the difference between a teacher and a tutor. The tutor is primarily interested in your success in the class, not really whether you learn. In some cases, they might not see that those are often connected...
@wassysmith1261
9 жыл бұрын
I LOVE YOU!!!! YOU SAVE MY LIFE!
@DocSchuster
9 жыл бұрын
Wassy Smith And I'd do it again, since you're so GREAT!
@Pedritox0953
2 жыл бұрын
Nice explanation
@betchen7321
10 жыл бұрын
awesome.
@stsfoxfacel9171
10 жыл бұрын
thanks :)
@sirr3d488
9 жыл бұрын
Wrong. Sin(x) is ALWAYS = x - x^3/3 + x^5/5 ... ALWAYS. x can be 5 billion. the series will still converge to a number between -1 and 1. Apart from that awesome videos ! Very clear.
@DocSchuster
9 жыл бұрын
SirR3D That's an interesting point. You need EVERY term of that infinite series if x>1, though, right? I should emphasize that I need x
@sirr3d488
9 жыл бұрын
Yep, as x gets larger you need more and more terms of the Taylor series expansion for an accurate approximation. The reason is because that series is calculated around 0. The more you go away from 0 the more terms you need. I really like your videos, very clearly explained. I will start a series of my own courses about Systems and Signals and another one about electromagnetism and microwaves. I'm a 3rd year student in electronics&telecommunications , but I find it fun to make videos + you actually learn a lot about the topic when you try to explain something clearly. If I manage to explain half as good as you do I'd be so happy.
@DocSchuster
9 жыл бұрын
SirR3D You're very kind! You're right, too, about trying to teach others (videos or not). It's the very best for improving your understanding.
@ukgaming1084
4 жыл бұрын
I'm surprised Doc even replied to this... He made it more than clear enough that he was using the approximation for very small angles. Stop trying to enforce errors onto other people to try and demonstrate how smart you are
@carmelpule6954
4 жыл бұрын
Ah--Ha, you did not take into account the centrifugal force which would change the length of the pendulum when the blob is moving in a circle! If the length L was a spring, now that would be ver interesting!
@evanhagen7084
8 жыл бұрын
Doc Schuster, i know this is a very old video so hopefully you see this comment, but i am doing a lab in AP Physics about simple harmonic motion. For part of our lab report, there is some graph that we have to linearize, but i cant remember exactly what it is... it was something to do with 1/2Pi sqrt l/g or SOMETHING like that... she was saying that we needed to get it into the form y=mx+b.... then said something about how there was no addition in the equation so there is no b... she ended up saying we could have y^2=mx or y=sqrt of mx but i just dont remember exactly what she was talking about.... do you have any ideas??
@DocSchuster
8 жыл бұрын
+Evan Hagen Yes! Just set up your experimental data so it maps to the equation that you have. What do you suppose 1/2Pi sqrt l/g equals?
@MysticMD
9 жыл бұрын
Hellooo! Thanks very much for the vid! I have a question! Could you please explain why at turning point there is no acceleration? And why when the pendulum is at equilibrium position, there is acceleration? I just cant get my head around it! Thank you! :)
@DocSchuster
9 жыл бұрын
Zi Gyllenhaal Hi Zi! There IS acceleration at the turning points, because the net force points toward equilibrium. There's acceleration UPWARD (but not left or right) at equilibrium because the pendulum bob is tracing a circular path and centripetal acceleration is v*v/r.
@MysticMD
9 жыл бұрын
Thank you very much!
@sshirish9167
6 жыл бұрын
Isn't the restoring force for a mass on a spring equal to -kx^2?
@paulettec8861
5 жыл бұрын
The restoring force is equal to -kx. The potential energy of a spring, on the other hand is (1/2)kx^2
@tuck295q
10 жыл бұрын
There is one thing I don't understand about pendulum. Why is it that in a certain location of pendulum, the magnitude g (gravity) is somehow different. I thought that g is equal at all places to certain height on earth. Please explain.
@DocSchuster
10 жыл бұрын
Not true. At the bottom of its swing, the tension is not equal to mg. Now tell me why.
@tuck295q
10 жыл бұрын
Because the motion of the pendulum has centripetal force at the bottom is at maximum. On another hand, when pendulum is at max height, the motion pause in an instant so there is no need for centripetal force. sooo how does this affect g at different length of pendulum? I just encountered a problem where I have to use g=(4pi^2L)/T^2 and g doesn't always equal to 9.81 when L and T are different.
@DocSchuster
10 жыл бұрын
Your explanation is solid. Your confusion comes from the equation for the period of a pendulum (solved for g). There's probably some error in your T and L measurements. That's a common lab to do. You need to recognize that g depends on the mass near you and how far you are from it.
@francisfrancis4219
5 жыл бұрын
Your presentation was illogical: First you said “If L were longer it would be easier to displace along an arc length”. But later you said “L is a tendency not to get screwed up” - as though L is akin to the mass on a spring. You further said increasing L provides more resistance to being restored.
@anthonyfrantz884
7 жыл бұрын
I know it's an old video ,but h=h-hsin(θ)
@visualisationperfection9138
5 жыл бұрын
I find your explanations clear and to the point, love you (no homo).
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