The trajectory of a body with two simultaneous, mutually perpendicular oscillations is determined by x=asin(pomegat)""(1) y=bsin(qomegat+phi)""(2) where x and y are the projections of the body displacement on X and Y axes For simplicity, assume pomega=qomega=omega_(0). Then, the equation of trajectory will be y^(2)/b^(2)+x^(2)/a^(2)-(2xy)/(ab)cosphi=sin^(2)phi""(3) which is a general equation of ellipse. But if qomeganepomegaandpneq, then the graph of trajectory on the X-Y plane is either a closed curve, whose loop number is defined by the ratio n = p/q, or an open curve. Note that at the point where the curve reverses along the same trajectory, the velocities along the X-axis and Y-axis become equal to zero simultaneously. The body moving along the curve stops exactly at this moment at a certain point, say P, and then starts moving back, If p = 2, q = 1 and phi=0, the path of trajectory of motion of the body with the above two oscillations (1) and (2) is