Recent content by giodude

Oh, think I figured it out. This might be where the inexactitude you mentioned is coming in: The ##x## in the previous message is the amplitude. Now we can plug it into ##mgh = \frac{1}{2}kA^{2}## which we derive from the conservation of energy equation: ##h = \frac{kA^{2}}{2mg} = 0.2536m##...

We also know ##T = 216 N## from the original post, given the ultimate strength of the wire. So, ##T = kx - mg## ##216 N = kx - (0.5 kg)(9.8 \frac{m}{s^{2}})## ##kx = 220.9 N## ##x = \frac{220.9 N}{k} = 0.0113 m## Dimensionally this resolves properly. However, the solution in the textbook is...

Whats the inexactitude? Also how does this reconcile with ##v_{f} = \sqrt{2gh}## if ##v = 0## at time ##t##?

Oh, I see. If ##T = kx - mg = ma## is at its maximum then acceleration needs to be maximum. As a result, we want ##cos(\omega t + \alpha) = 1## and ##sin(\omega t + \alpha) = 0##. This makes sense because this sets ##x = Acos(\omega t + \alpha)## to its max ##x = A##.

We also know that in SHM ##x = A cos(\omega t + \alpha)##, where ##\omega^{2} = \frac{k}{m}##. So, ##v = \frac{dx}{dt} = - \omega A sin(\omega t + \alpha)## ##a = \frac{d^{2}x}{dt^{2}} = - \omega^{2} A cos(\omega t + \alpha)## KE is maximum where ##a = 0## at ##cos(\omega t + \alpha) = 0## and...

I think I'm starting to get somewhere. We know, due to the starting state of lifting and dropping that the initial speed, ##v_{0} = 0##. Additionally, we know that ##KE = mgh## at the point of peak tension. Therefore, ## \Delta KE = \frac{1}{2} \times m \times v_{f}^{2} - \frac{1}{2} \times m...

It'd occur just before the sudden jerk? So, at PE = 0 and KE = mgh?

The change in GPE will be mgh from start to finish. Which means KE will have changed from 0 to mgh at the moment of maximum tension.

Simple harmonic motion?

Hi, I solved part (a) and will provide my solution below. However, I've been working on part (b) for quite a bit and reviewed the provided, relevant text a few times now but haven't been able to find what I'm missing: Solution (a): Using ##A = \frac{\pi \times d^{2}}{4}##, ##k =...

Hi! In reading about Superconductivity and its current state of only being achieved in super cooled or heated materials. This sparked a question the following question: What is the result of the trade off between energy saved by avoiding dissipation through the natural resistance of a material...

I took images of them on my phone and then airdropped them to my laptop, I'll fix them up when I get back to my laptop and edit the post. Thank you.

Hello, I'm currently working through Purcell and Morin, Electricity and Magnetism textbook and came across a problem in which the goal is to verify the inverse square law. I'm worked through and completed the problem. However, I'm confused how this verifies the inverse square law, I'm posting...

Ah, I think my sign error is a result of improperly setting up my cross product calculation with respect to the final velocity. Once fixed it looks like the sign is fixed because what was previously ##L_{P, f}^{sys} = I_{cm} + md^{2} + \frac{1}{2}dmv_{f}## becomes ##L_{P, f}^{sys} = I_{cm} +...

Fixed it using conservation of angular momentum: System: ##L_{P,i}^{sys} = L_{P,f}^{sys}## ##k_{i} = k_{f}## Solve for initial and final angular momentum about ##P##: ##L_{P,i}^{sys} = -\frac{d}{2}mv_{i}## ##L_{P,f}^{sys} = I_{cm} + md^{2} + \frac{d}{2}mv_{f} = \frac{1}{3}md^{2}\omega_{f} +...