Recent content by bullet_ballet

Thanks for the reply. You helped me to conceptualize it, so your post is much appreciated. :)

Forgive me for the long post, but I'm in some desperate need of clarity on this matter. I just can't seem to grasp the whole shock wave concept, or at least the meaty part of it . I only have a couple of problems left to do to finish my HW I'm at an impasse until I dispel my confusion. I...

I don't think I was specific enough. I was asking why I could use x-ct to satisfy the PDE and show that f(x-ct) is therefore a solution. Having done this for a couple of different problems, it seemed like I could draw the conclusion that if the PDE held for some a(x,t), then the solution...

"Verify that, for any C¹ function f(x), u(x, t) = f(x - ct) is a solution of the PDE u_t + c u_x = 0, where c is a constant and u_t and u_x are partial derivatives." I managed to get the solution for this and a similar problem by showing that the new variable (x - ct in this case) satisfies...

My bad. I had the wrong signs in there. Since I set y=0 at the windowsill, all values should be negative. Therefore, if -y is the distance to the bottom of the window from the windowsill, it should be broken up into -u-1.9 where u is defined as before. Therefore, -u-1.9=-\frac{1}{2}gt^2. I...

Makes sense to me, but the back of the book doesn't seem to agree. It has x_0=\frac{f (\omega^2_0-\omega^2)}{r^2} and v_0=\frac{2\gamma\omega^2f}{r^2}.

Using the equations in my first post, I get 1.03m as the distance between the sill and the window.

(a) Since the skateboarder is on an incline, a = g sinθ. To find the time it takes to descend, find the average velocity v = (vf - vi)/2. We know that t = distance/velocity, or in this case, 11/3.6. Using a = v/t, you can find that θ = ArcSin v/gt. (b) If you draw a diagram of the problem...

If the flowerpot is just falling from rest and not being thrown or propelled downwards, then v0 will equal zero.

Let u be the distance between the windowsill and the top of the window. If you set the windowsill at y=0, then use the formula y=y_0-v_0t-\frac{1}{2}gt^2. We have y0=0 by our reference point and v0=0 since the flowerpot falls. Therefore: y=-\frac{1}{2}gt^2 -1.9-u=-\frac{1}{2}gt^2...

I need to find the initial conditions such than an underdamped harmonic oscillator will immediately begin steady-state motion under the time dependent force F = m f cosωt. For the underdamped case: x(t) = ae^{-\gamma t}cos(\Omega t+\alpha)+\frac{f}{r}cos(\omega t-\theta) and if it matter...

Thanks a lot.

I see that I made the mistake of dividing g by .01m and not .001m, but that only makes the answer worse at 5950 rpm. My mistake is definitely more fundamental, but I can't see it.

"An electric motor of mass 100 kg is supported by vertical springs which compress by 1 mm when the motor is installed. If the motor's armature is not properly balanced, for what revolutions/minute would a resonance occur?" I set my frame of reference at the end of the spring. Therefore, F =...

Thanks for the reply! I can't believe I made such a basic mistake. With that sign change, I'm able to manipulate the terms and obtain the final answer. Much appreciated.