Equations of Damped Harmonic Motion

Question

The path of a fast-moving particle traces a circle with equation

(x + 7)^2 + (y - 5)^2 = 36

It starts at point (-1,5), moves clockwise, and passes the point ( -7,11) for the first time after traveling 6 microseconds. Where is the particle after traveling 15 microseconds? (Hint: Let x= asinb(t-c)+d and y= esinf(t-g)+h)

Answer 1

This can be solved the easy way or the hard way.

The hard way is to follow the hint.

Here is the easy way.

First draw a picture.

The circle has center at the point (-7, 5) and has radius 6.

The start point of (-1, 5) is the rightmost point of the circle.

The point (-7, 11) reached after 6 ms is the topmost point of the circle.

Since the particle moves clockwise, it covered the angle of 3π/2 in 6 ms.

Therefore after travelling 15 ms the particle covers the angle

(3π/2)(15/6) = 15π/4 = 4π - π/4

That means after 15 ms the particle is halfway between its start point and end point, which is

(-7+6/√2, 5+6/√2) = (-2.75736, 9.24264)