M3 Simple Harmonic Motion help!

Ok, basically I am hopeless at Mechanics haha, and I need help. This question completely stumps me, I find the situation awfully hard to understand!

It's Q3 on CCEA M3 January 2005 in case anybody has a copy, because my laptop isn't connected to a scanner so I can't scan it in!

"The end O of a rod OB of length 0.5m is fixed.
The rod is rotated in a vertical plane with a constant angular speed about O. It completes one revolution every 30 seconds.
A horizontal beam of light is shone towards the rod and a shadow O'B' is cast on a wall.
The light beam and plane of rotation of the rod are perpendicular to the plane of the wall. O' and B' are the images of O and B.

As the rod is rotated anticlockwise the point B' moves in simple harmonic motion about O'.

(i) Write down the periodic time and the amplitude of the motion of B'.
[INDENT]I've done this part, T = 30 secs, amplitude = 0.5m[/INDENT]


(ii) When OB is horizontal find the speed of B'.

(iii) Find the time taken for the shadow to decrease from 0.4m in height until it is next at its maximum height."


Any help is really really appreciated. It's (ii) and therefore (iii) that I'm completely flummoxed by - I don't even have any method down on paper yet.

when B is horizontal it has turned pi/2 radians so B' returns to O , so x=0
and use v=wa

Ohh thankyou, thankyou! Have you got any tips on how to do (iii)?? What should I be considering etc?

if I have understood the question properly then,

find the time taken for x=acoswt where x=0.4 , and take away the time preiod (T=2pi/w)

Hate to say it, but that isn't giving me the right answer I'll post my working:

$s = a \cos wt$

Subbing in s, a and omega:

$0.4 = 0.5 \cos (\frac {\pi}{15}t)$

$0.8 = \cos (\frac {\pi}{15}t)$

$\frac {\pi}{15}t = 0.6435... \mathrm{or\ } \frac {\pi}{15}t = 5.6397$

$t = 3.0725... \mathrm{or\ } t = 26.927...$

From part (i), T = 30 secs, and when I take 30 away from either of those answers, I'm going to get a negative time. The answer that I'm supposed to be getting is 11.9 seconds?

no worries just as I thouhgt I misunderstood the bit "maximium height" as when B comes back to the top of the amplitude ..

but looking from the answer they simply wanted the time from 0.4 to the time when it is next at its amplitude which is simply
x=acoswt , this time x=-0.4

which should give the answer

It does, yep. Is there any way you could explain the wording of that question to me? If that makes sense. If I was in an exam, I wouldn't have had a clue that I had to put -0.4...