Recent content by Chopo

The string is of length 5L and the bead is positioned 1/5 th of the way along the string. 4/5 ths of the way in the other direction. Hence 4/5 L

The angle of a triangle of adjacent side 4L/5 and and opposite side A (amplitude)

Ok, So for part a, obtaining the equation of motion, i guess i should start by looking at the generic equation of motion: displacement = amplitude * cosine ( angular frequency of oscillation * time + phase shift ) x(t) = A cos( omega*t + phi) When x(t) = A, cos( om*t + phi) = 1 and om*t...

The problem is the following: a.) Obtain the equation of motion for the very small oscillations of a bead of mass m attached 1/5th of the way along a massless string of length 5l, which is under tension T. b.) Hence show that the angular frequency of oscillation is omega=sqrt(5T/4ml)...