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How do you derive the period of oscillation for a pendulum?
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- čas přidán 13. 07. 2022
- Just in case you can't remember the formula for the period of oscillation of a period (for small oscillations), here's how you find that.
This video just makes the entire thing more confusing lol
you're just dumb and not prepared yet to this
When you go to the next blank page, always keep the equation from last one visible. Love these videos btw!
I second that
Physics was easy but greek letters ruined it
True
I'll have to do this one. I don't remember how I use to do this from scratch, but it'll be a great refresher. :)
when you said θ(t) = Acosωt + Bsinωt, why did you decide to put omega inside the sine and cosine?
look up on how to solve a second order linear differential equation
I was so confused with this before. thank you
Why can we say w=2*pi*f?
angular frequency (ω) is defined as the amount of radians an oscillator undergoes per second, which would be 2πf because frequency is the amount of cycles and there are 2π radians/cycle
Thank you I finally understand this! I didn’t anticipate that it would involve a little Taylor series action!
Thank you for this video. At 3:25, isn't the acceleration always directed towards the centre in circular motion? Surely therefore there is no component in the direction of s to plug into F=ma?
The tension does indeed pull in the center direction, but not the gravitational force. This means the net force is not in the r-hat direction.
What you're referring to here is the centripetal force which facilitates circular motion. The centripetal force here is provided by the tension in the string. The gravitational acceleration is provided by the gravitational force.
General Relativity has a lot of explaining to do! :)
Could I ask why you have to let theta is small before the next step?
If theta is small, then you can let sin(theta) = theta. That makes the differential equation solvable by guessing a solution. It will look just like a simple harmonic oscillator.
hey professor, could you teach us how to solve extreme distance free fall problems?
What is an example of an "extreme distance free fall" problem?
like when should consider the gradient of gravity, like a tennis ball falling the same distance from the moon to earh, since the gravity will change in some rate, and obviously the acceleration will not be as the same as earth surface.
i think would be very nice see how it is done.
@@HigorMadeira97 You'd need to use differential equations and g (now a variable) would be GM / R where G is the universal gravitation constant, M is the mass of the earth and R is the distance between the point mass (assume tennis ball to be a point mass) and Earth's centre . Essentially, we are having g to vary with distance here.
I don't see how x(t) = Acos(wt).
I found a different way to derive, but im not sure if its correct.
force of gravity = centripetal force?
mg = mw^2 L
w^2 = g/L
w = (g/L)^1/2
is it a correct way to derive?
But the gravitational force is down and the acceleration is up. This doesn't work.
@@DotPhysics ahh alr
If you really want to understand this.. you need to know how to solve linear differential equations with constant coefficients and complex numbers. Requires more than just physics.
Why the θ(t) = Acosωt + Bsinωt ?
not theta(t), but f(t) - right? It's a function that satisfies the differential equation. If you take the derivative twice, you get the same function with a negative constant out front.
❤good
messy presentation
it was nice