A particle moves around the circle x^2 + y^2 = r^2 with a constant angular velocity of omega knot radians per unit time. Show that the projection of the particle on the x axis satisfies the equation x double dot + omega knot x = 0.

Can anyone help me out with this? Not sure on how to begin this problem.

Question

A particle moves around the circle x^2 + y^2 = r^2 with a constant angular velocity of omega knot radians per unit time. Show that the projection of the particle on the x axis satisfies the equation x double dot + omega knot x = 0.

Can anyone help me out with this? Not sure on how to begin this problem.

anonymous · Answer

"knot?" You mean, "naught?"

anonymous · Answer

Think of the circle as a unit circle (like from trigonometry), and the projection onto the x-axis is the cosine of the angle as you go around the circle.

anonymous · Answer

You have a second-order DE, $x''+\omega_0x=0$
This DE is typical of simple harmonic motion (i.e. things that can be modeled with sines and cosines).

anonymous · Answer

yeah I meant naught, so I see what you mean, so then would I have to solve the DE and then get a general result? If not, then how would I start this?

anonymous · Answer

Yes, solve the DE and compare it to the x coordinate of where the particle is at any moment, t. Note also (if you haven't already), that $\omega_0 $ has units of $\Delta \theta/\Delta t$

anonymous · Answer

In other words, x is a function of the angle, Θ in radians.