Recent content by McCoy13

Yes. If you plot the pixel intensities against their angle, you can easily see that it's not just random scatter.

Maybe being more specific will help. I am analyzing image data. Basically I have images of bright rings on a dark background from several samples prepared with differing concentrations of a chemical reagent. I want to plot the characteristic intensity of the rings against the concentration...

I'm not sure, because your wording is ambiguous. I'll try to be as clear as possible (don't be offended if some of this is elementary). If you are making measurements, your apparatus has some limited precision, so your measurement has some uncertainty. For a quantity you're interested in, f...

The uncertainties are a result of the precision of my measurement. If I were measuring lengths of rods, it'd be how precise my ruler is. However, I could also have a very broad or very narrow distribution of rod length. These two quantities (the uncertainty and the spread) are unrelated.

I'm doing an analysis where I have a set of random variables with some known uncertainties (the uncertainties are different for each random variable). The random variable is roughly Gaussian distributed. I'd like to get a meaningful characteristic value and uncertainty for the whole set. I can...

I know this question has been asked plenty of times, but nonetheless I wanted to get some recommendations that might better jive with my situation than other the other threads I've read about this topic. I'm a first year grad student in physics that hasn't had a course in fluid dynamics, but...

I don't really understand what equation you've written down here. Why is this expression equal to 0? What is b? Is dx^2/dt^2 the second derivative of x with respect to t or is it (dx/dt)^2?

You don't need to plug in something for the dx/dt term. You may choose to plug in something for dx/dt. We are describing the motion of a particle on a track. Because it is on the track, if you know its x-coordinate then you know its y-coordinate. Similarly (because the track is not U-shaped) if...

Why don't you plug in the formula you've found for dy/dt into the Lagrangian? And you can replace y with f(x). If you really have your heart set on using y and replacing dx/dt, then what you need to do is invert f(x), so that you have x=f^{-1}(y), and then you can calculate dx/dt using the...

Why do you want to solve for dx/dt?

I think this is correct provided that y' means dy/dt. However, you are given y as a function of x, which suggests you find dy/dt in terms of dx/dt. There is no reason to invert y(x) to get x(y) except to have a simpler expression for your potential, but making your potential U(x)=mgf(x) would...

If y=f(x)=\frac{-x^3}{a^2} then \frac{dy}{dt}=\frac{df}{dx}\frac{dx}{dt} where \frac{df}{dx}=\frac{-3x^2}{a^2}. I do not think I can make this more clear without depriving you of a learning opportunity.

As an object is raised (an increase in y-coordinate) its gravitational potential goes up, so U=mgy. To get dy/dt, we simply apply the chain rule. y=f(x) \Rightarrow \frac{dy}{dt}=\frac{df}{dx}\frac{dx}{dt} The first derivative of f(x) is \frac{df}{dx}. Is this clear? I will explain it another...

You only have one degree of freedom in the problem. If you specify the x-coordinate position and velocity, the motion of the particle is fully determined. Therefore you should start with L=\frac{1}{2}m(\frac{dx}{dt}^2+\frac{dy}{dt}^2)-mgy recognize that y=f(x)\Rightarrow...

Homework Statement Calculate the deformation of a sphere of radius R and density \rho under the influence of its own gravity. Assume Hooke's law holds for the material. Homework Equations Not applicable; my question is simply one of understanding. The Attempt at a Solution I want...