Recent content by Abourque72

Indeed: ##T^2 = {T_{Pend}}^2 + {T_{Spring}}^2.##

Ah, yes. I get $$\ddot{x} = \frac{-2kx}{m+ \frac{2kr}{g}}$$ which is the (hopefully) correct SHO which I can get the period from. Interesting to see that "correcting" mass term of ##\frac{2kr}{g}##; I wonder what kind of physical interpretation it entails.

I don't see a way to cancel out theta without cancelling out the ##\ddot{x}##, unless I am misunderstanding your suggestion.

I'm assuming you can treat the mass m of the pendulum as a mass at point p, and since Hooke's Law makes the forces on the point in the same direction, its just a simple harmonic oscillator with spring constant 2k and mass m: $$\ddot{x} = \frac{-2kx}{m}$$ Hence, the period of motion is $$\pi...

Thanks for the response. I looked back and sure enough I was missing that cosine. Anyways, your suggestion makes me think that the term ##{\dot{\theta}}^2 \theta## should certainly be negligible since, if ##\theta## is small, ##\dot{\theta}## ought to be pretty small too. So this term is a...

I would like to preface this by saying that I solved the homework problem, but my professor gave me an added challenge of finding the period of the motion described in this problem. 1. Homework Statement The pendulum bob of mass m shown in the figure below is suspended by an in-extensible...

Hi, My name is Andre. I am an eager and curious learner, and I enjoy doing mathematics and physics. I look forward to being a part of this community.