force acting on a spring is f = -kx and \ddot{}x = \omegao2x
i know how to arrive to that equation f = - kx but also f = ma therefore ma = -kx re arrange and you get \ddot{}x = \omegao2x
woah why did this get moved? lol this is 2nd year physics :p and i saw a fellow forum user ask about a 1st year physics question in the advance section. no fair ;[
Homework Statement
A horizontal massless spring of spring constant k is attached to a immovable wall at one end and a mass of 0.45kg at the other end. The spring which was not originally under tension, is now extended by 0.18 m by pulling the mass horizontally. The mass was then released by...
found the problem sorry, i was reading the question wrongly. you can use this thread as an example of why you should read questions properly, i thought the square root was covering the whole wave function but it was only covering one of the constants. thank you for all your help.
you have the wave function, you have what the answer should be and you have the identity integral needed to solve this. I gave you the answer i kept receiving, my friend who is a theoretical physicist also received the same answer, if you do in fact obtain the right answer can you show a step by...
Hi, 2nd year physics student here
doing a past paper on quantum mechanics everything is going nice and dandy then this happens..
question: prove that the normalisation constant A is given by A = \frac{1}{2^1^/^2} (\frac{a}{\pi})^1/4
ok seems fairly straight forward but i keep getting...