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A point moves in the plane `xy` according to the law `x=a sin omega t, y=b cos omegat , ` where `a,b` and `omega` are positive constants. Find `:` `(a)` the trajectory equation `y(x)` of the point and the direction of its motion along this trajectory , `(b)` the acceleration `w` of the point as a function of its radius vector `r` relative to the orgin of coordinates. |
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Answer» From the Eqn `: x= a sin omega t` `sin^(2)omegat=x^(2)//a^(2)` or `cos^(2) omegat =1-(x^(2))/(a^(2)) ....(1)` And from the equation `: y=b cos omegat ` `cos^(2)omegat =y^(2)//b^(2) ...(2)` From Eqns `(1)` and `(2)`, we get `:` `1-(x^(2))/(a^(2))=(y^(2))/(b^(2))` or `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` Which is the standard equation of the ellipse shown in the figure. we observe that, at `t=0, x=0` and `y=b` and at `t=(pi)/(2 omega), x=+a` and `y=0` Thus we observe that at `t=0`, the point is at point (figure) and at the following moments, the co- ordinat `y` diminishes and `x` beomes positive. Consequenctly the motion is clock-wise `(b)` As `x=a sin omegat ` and `y=b cos omegat` so we amy write `vec(r)= a sin omega t vec(i+)b cos omegat vec(j)` Thus `ddot(r)=vec(w)=-omega^(2)vec(r)` |
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