# Search results

1. ### Stability physics problem

Just come across a question and I'm at a point where i see no further. A uniform rod AB, of mass m and length 2a, is free to rotate in a vertical plane, about the end A. A light elastic string of modulus kmg and natural length a, has one and attached to B and the other end to a fixed point O...
2. ### Simple Harmonic Motion

A particle P of mass 0.2kg is attatched by an elastic spring of modulus 15N and natural length 1m to a point A of the smooth horizontal surface on which P rests. P recieves an impulse of magnitude 0.5Ns in the direction AP. Show that while the string is taught, the motion of P is simple harmonic...
3. ### Simple Harmonic Motion - Alevel M3

This is probably a really easy question. But alas the answer has eluded me thus far. Anyway, here is the question:- Points O, A and B lie in that order on a straight line. A particle P is moving on the line with S.H.M period of 4s, amplitude 0.5m and centre O. OA is 0.1m and OB is 0.3m. When...
4. ### Breakbeat techno metal thingy

Hey guys, Have a listen to what i do when im not doing maths and physics (music wise). www.myspace.com/thevolatilegentlemen I hope you enjoy. It might make for some interesting late night (Britain) conversation. Or at the very least mildly annoy you.
5. ### Mechanics - Elastic springs and strings

A particle P of mass m is attached to one end of a light elastic string of natural length L whose other end is attached to a point A on a ceiling. When P hangs in equilibrium AP has length \frac{5l}{3}. Show that if P is projected vertically downwards from A with speed \sqrt(\frac{3gl}{2}), P...
6. ### Mechanics - Elastic springs and strings

A particle P of mass m is attached to one end of a light elastic string of natural length L whose other end is attached to a point A on a ceiling. When P hangs in equilibrium AP has length \frac{5l}{3}. Show that if P is projected vertically downwards from A with speed \sqrt(\frac{3gl}{2}), P...
7. ### Messed up somewhere in my integration

\frac{dv}{dt}= -x^{-3} when t=0, the particle is at rest with x=1 Therefore by integrating i get v = \sqrt(x^-2 - 1) \frac{dx}{dt}= \sqrt(\frac{1 - x^2}{x^2}) dx\frac{x}{(/sqrt(1 - x^2))} = dt -\sqrt(1 - x^2) = t + C C=-1 Therefore:- t = 1 - \sqrt(1 - x^2) However i cant...
8. ### Yet another FODE question

Hey. I've been doing more mechanics recently - further kinematics in M3. I've come across another question i'm confused by. A particle moves on the positive x-axis. When its displacement from O is x meters, is acceleration is of magnitude x^-3 m/s^2 and directed towards O. Given that...
9. ### Mechanics 3 - Further kinematics

I'm studying M3 as one of the three modules for my further maths alevel, and have just started the first chapter. I'm fine with acceleration as a funtion of time, and integrating to find velocity and displacement, and most of acceleration as a function of displacement, but have come stuck on one...
10. ### Differential of 2^x

Just came across a question today with 2^x and realised i didnt know how to differentiate it. The entire function i had to differentiate was [math]y = 2^x + x -4[math] I tried taking logs but couldnt get anywhere near the true answer. What is the correct method for this? Thank's alot!
11. ### Second order differential equation

I just came across this one, was going really well until i came across this one. (d^2y/dx^2) + (dy/dx) = e^(-x) m^2 + m = 0 m = -1 and m = 0 Now i get the particular integral Try y = ke^(-x) (dy/dx) = -ke^(-k) (d^2y/dx^2) = ke^(-x) ke^(-x) - ke^(-x) = e^(-x) I get stuck here...
12. ### First order differential equation

Just need a hand with this one. (dy/dx)x + 2y = x^3.ln(x) (dy/dx) = (x^3.ln(x) - 2y)/x Integrating factor = x^2 (dy/dx)x^2 + 2xy = (x^3.ln(x))x^2 yx^2 = INT[(x^3.ln(x))x^2] I'm having trouble integrating the last part to complete it. Thanks alot and in advance for any help.
13. ### Using an identity to find the sum to n terms of a series

Just working through my FP1 book and have got stuck on a question. Use the identity (r+1)^3 - r^3 \equiv3r^2 + 3r + 1 to find \sum\limits_{r = 1}^n r(r+1) I've tried using the method of differences to get n^3 + 3n^2 + 3n, but cant see how to get it back into its original form, not sure how...