Fluid Dynamics: describing difference in behaviour of a particle released from the origin into flow

Question

I've been given a flow field $$\mathbf{u} = (\cos(\omega t), x\sin(\omega t), 0)$$ and have found the streamlines and path lines etc. However I am stuck on trying to figure out what the difference in behaviour of a particle released at the origin would be if this was instead $$\mathbf{u} = (\cos(\omega t), x\sin(\sigma t), 0)$$ with $\sigma \neq \omega$ I've tried trying to find the path lines of this but it got complicated and I'm after an intuitive description of what actually would happen to the particle. Any help appreciated

Answer 1

So you have to solve the system $$\dot{x} = \cos(\omega t), \quad \dot{y} = x\sin(\sigma t),\quad \dot{z} = 0.$$Taking the easy things out of the way first, we get $$x = \frac{1}{\omega}\sin(\omega t),\quad z= 0,$$as the particle is released from the origin. So we plug that expression for $x$ and get $$\dot{y} = \frac{1}{\omega}\sin(\omega t)\cos(\sigma t).$$But we can join them using a product-to-sum formula, and get $$\dot{y} = \frac{1}{2\omega}(\cos((\omega-\sigma)t)-\cos((\omega+\sigma)t)).$$Then it follows that $$y = \frac{1}{2\omega(\omega-\sigma)}\sin((\omega-\sigma)t) - \frac{1}{2\omega(\omega+\sigma)}\sin((\omega+\sigma)t).$$Then you can proceed with the analysis using these formulas. I'm not sure if there's some immediate intuition to be drawn from this solution (at least I can't see it right now myself).