# Recent content by Malby

1. ### Natural Frequency of a Spring System

You're right. As \phi goes to 90 degrees, the natural frequency should decrease to 0, as there would be no component of force acting in the x-direction then. It appears as though multiplying by cosine would make this correct, but I don't know where that comes in in the working: In the...
2. ### Natural Frequency of a Spring System

I worked through it again and found the natural frequency to be: sqrt(4k/m)/(2 * Pi). I basically let the angles on both sides change by phi +/- dphi, and the same with the length L -/+ dL, and with x. If this is the correct answer then I suppose it worked out ok.
3. ### Natural Frequency of a Spring System

A mass m is confined to move in a channel in the x-direction and is connected to four identical springs with spring constant k, which are oriented at angles \phi = 45° as shown, if the system is in static equilibrium. a) Ignoring friction, determine the natural frequency of vibration of the...
4. ### Integral using hyperbolic substitution

Aha! I was wondering how it got to that step. Thanks very much for your help!
5. ### Integral using hyperbolic substitution

Oops, that was meant to be nice neat LaTeX... Well, I tried doing that but ended up with int(cosh^4(u))du... not sure if there is a nice little trick for solving that or if I've ended up with a harder problem than I started with. Also, why didn't that tex code just work?
6. ### Integral using hyperbolic substitution

[tex]\begin{eqnarray*} \int\left(1+x^{2}\right)^{\frac{3}{2}}dx & = & \int\left(1+\sinh^{2}\left(u\right)\right)^{\frac{3}{2}}\cosh\left(u\right)\, du\\ & = & \int\sqrt{\left(\cosh^{2}\left(u\right)\right)^{3}}\cosh\left(u\right)\, du\\ & = &...
7. ### Integral using hyperbolic substitution

Homework Statement \int\left(1+x^{2}\right)^{\frac{3}{2}}dx Homework Equations The hyperbolic functions. The Attempt at a Solution We've been going over hyperbolic substitutions in class so I assume I'm meant to use one of those, but I'm just not sure how to choose which one. Any help...
8. ### Linear second order non-homogeneous ODE question

Aha! Of course. It's all so simple once you know what to do... :rofl: Thanks very much!
9. ### Linear second order non-homogeneous ODE question

I'm not sure I understand this. Are you able to explain it a little more? Cheers
10. ### Linear second order non-homogeneous ODE question

Determine the general solution to the ODE: y'' + 2y' = 1 + xe-2x I know the solution will be of the form y = yh + yp. The homogeneous solution is y = c1 + c2e-2x. For the particular solution, I have been using the method of undetermined coefficients. c3e-2x won't work as it is not...
11. ### Any ideas on what material this is?

Hi everyone, I'm having trouble identifying the material used in the oil seal of a hydraulic gear pump. There is a picture of the microstructure of the oil seal with it's hardness. Any ideas? Cheers
12. ### Electric Flux Through A Box

So you've been given the angle theta which is 13 degrees. You'll need to find the cross sectional area which will be given by the distance x multiplied by the height of the box which I believe is given by the question. The angle that you need to work out the length of x is also in the...
13. ### Please help: Normal force for box on an inclined plane = mg/cos(theta)?

Should there not also be a force due to friction pointing back up the slope? When you put that into your free body diagram have a look at all the individual x and y components of each force, F_n, f_s and F_g.
14. ### MATLAB Trouble with Euler's method in MATLAB

I'm trying to solve an ODE using MATLAB and Euler's method but I've having some trouble understanding what's going on with the code. This is something relatively simple but I'm new to MATLAB so I'm not really sure what's going on. The question: Write a Matlab M-file that uses Euler’s method to...
15. ### Optics and a thin wedge

In this case isn't m the fringe number and not the fringe number per unit length? I stumbled upon another answer that said m/x was 2*theta/Lambda which gives you a value. I'm not sure about that though because surely you need to know the distance the "viewing screen" (in this case the glass) is...