Recent content by green-beans

Ohhh, okay, I got the answer! Thank you!

For (d) the angle measured by θ is the angle between the vertical and the line joining the centre of the semi circle with the centre of the particle (as it is indicated on the picture) :) For (f) I am still a little bit confused as I am getting something really weird. If I let (a,b) be the...

Thank you for your reply! For (d) it seems my diagram was not really representative (R wasn't entirely perpendicular) and so now I got that R = Rsin(theta)i + Rcos(theta)k For (f) I guess R will be in the direction of the normal at the point where the particle is. So, dz/dx = 2x and so the...

Homework Statement In each of the diagrams (please see attached file (I am sorry for the rotated, the original was in normal form but when uploading it, it was somehow rotated)) and the description of each case below, a particle is moving on a smooth surface, so that the reaction force R acting...

Homework Statement Consider the differential equation: mx'' + cx' + kx = F(t) Assume that F(t) = F_0 cos(ωt). Find the possible choices of m, c, k, F_0, ω so that resonance is possible. Homework EquationsThe Attempt at a Solution I know how to deal with such problem when there is no damping...

Oh right, this makes sense! Thank you so much - I now understand it :)

Yes, of course, it's when t=pi/nu. However, to obtain the answer from what I got it must be that (x(pi/nu)) = b which I don't quite understand why.

Thank you for your reply! At the highest point its velocity should become 0 so it will not go slack, is it right? If so I will need to solve v=0 as far as I understand. If yes, I then get that sin(nu)t *([μ2h/nu + h (nu)) = 0 but I cannot still see where the given expression for h comes from.

Should not ##x(0) = - h## since the mass starts below 0? Also I obtained that A = -h and B= -μh/nu but I don't quite understand how I can now find the greatest value of h. Also, since the string never becomes slack, does it mean that x(t) < 0? Thanks in advance!

Thank you for your reply! However, I keep on getting that A and B are zero if I use that x(0)=x'(0)=0 and this reduces the whole equation to 0 which is impossible.

Homework Statement A mass m is suspended by a light elastic string. When the mass remains at rest it is at a point 0, which is a distance a + b below the point from which the string is suspended from the ceiling, where a is the natural length of the string. The mass is pulled down a distance h...

Actually, I have just realized that the shaded area should be the area not under negative x-axis but under the parabola. So, it will be the entire area under x-axis and some of the area above it but below the parabola.

Thank you for your help!

Ohhh, I see! So, I get that I < {(α)^2 * (S)^2}/{β^2} which is a parabola starting at the origin. Therefore, the shaded region should be under S-axis (if I plot I as y-axis and S as x-axis).

Hmmmm... I am sorry but I still do not get what you are trying to say. As far as I understand, since the question asks to shade the region of the I-S space in which I increases with time, I need to obtain the function I(S) or S(I) which I can get by solving the differential equation. However, I...