Recent content by hyperon

You don't have to do that actually, you can simply equate the sum of the forces to zero along the x and y axes. F_1=k_1\left(\left(\begin{array}{c}2\\4\end{array}\right)-\left(\begin{array}{c}x\\y\end{array}\right)\right)=\left(\begin{array}{c}2k_1-k_1x\\4k_1-k_1y\end{array}\right)...

This is the DE you need. The boundary condition is just continuity with the earlier function you found.

My apologies. After pondering about this problem for a while, I realized that this is wrong. The linear acceleration A of the cylinder is indeed just g - T/M. and the problem of inconsistency arose from another equation, namely one of the 2 for the rotation. I will present the solution which I...

The cylinder is acted on by 2 forces: the tension T and its weight Mg. Its acceleration in the frame of the rope is g - T/M. Since the rope is accelerating in the opposite direction at a, the linear acceleration of the cylinder's centre of mass in the frame of the observer is A = g - T/M - a...

The superscript greek letters seems to be just a problem of using latex next to normal text. First Problem: I assume that A refers to the acceleration of the centre of mass of the cylinder downwards. In that case, I can get equation 2: a = g - T/m. I get a different equation for 3 and 4...

The extra 100m in path A means that by the time light from A has reached the detector, B has undergone a pi/2 phase change to be in phase with A, so the total phase difference is zero. B leads A in path difference because it is closer to the detector, and this makes up for the initial phase lead...

Hmm I have always thought that the Euler-Lagrange equations can be derived using F=\dot{p}=-\nabla U, albeit a little crudely. The way I remember seeing it goes something like this, with the same assumption of a curlless field, For one dimension, \dot{p} = -\frac{\partial U}{\partial x}...

Perhaps that's exactly what the question is asking for, it wants you to take U(π/2) = 0 where the dipole is parallel with the x-axis. Then U will be most negative when θ = 0. The limits would then be from wherever U=0 to the current position, from θ = π/2 to θ, if the integral used is U = -...

I am not exactly convinced that finding which path information necessarily collapses the photon's wavefunction. If a single detector is placed in front of only one slit, this does collapse the wavefunction if the photon passes through that slit, since the photon would have interacted with the...

Mathematica is better than Maple because it was designed to be an all purpose math software. It is capable of a greater range of computations because it has more functions than Maple. You can find a comparison between them http://amath.colorado.edu/computing/mmm/. One feature of Maple which I...

You can use the figure command to control the size of the plot window. First, you find the size of your screen, as a vector in the form of [left, bottom, width, height]. Then, you set the size of the figure window for the property 'Position'. The property value for position is in the same form...