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Mechanical Oscillations

Problem 4.42

A particle of mass mm moves in the plane xyx y due to the force varying with velocity as F=a(y˙i−x˙j),\mathbf{F}=a(\dot{y} \mathbf{i}-\dot{x} \mathbf{j}), where aa is a positive constant, i and j are the unit vectors of the xx and yy axes. At the initial moment t=0t=0 the particle was located at the point x=y=0x=y=0 and possessed a velocity v0\mathbf{v}_{0} directed along the unit vector j.\mathbf{j} . Find the law of motion x(t),y(t)x(t), y(t) of the particle, and also the equation of its trajectory.

Reveal Answer
4.42. Let us write the motion equation in projections on the xx and yy axes: x¨=ωy˙,y¨=−ωx,\ddot{x}=\omega \dot{y}, \ddot{y}=-\omega x, where ω=a/m\omega=a / m Integrating these equations, with the initial conditions taken into account, we get x=(v0/ω)(1−cos⁡ωt),y=(v0/ω)sin⁡ωt.x=\left(v_{0} / \omega\right)(1-\cos \omega t), y=\left(v_{0} / \omega\right) \sin \omega t . Hence (x−v0/ω)2+y2=(v0/ω)2\left(x-v_{0} / \omega\right)^{2}+y^{2}=\left(v_{0} / \omega\right)^{2}. This is the equation of a circle of radius v0/ωv_{0} / \omega with the centre at the point x0=v0/ω,y0=0x_{0}=v_{0} / \omega, y_{0}=0
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