You need the propotion of the time.
Canīt one solve for when depth is 99cm and
, that is, the second time the depth is 99cm.
Then time spent "over" the 99cm mark is .
One end of a piece of elastic is attached to a point at the top of a door frame and the other end hangs freely. A small ball is attached to the free end of the elastic. When the ball is hanging freely it is pulled down a small distance and then release, so that the ball oscillates up and down on the elastic. The dept d cm of the ball from the top of the door frame after t seconds is given by:
d = 100 + 10 cos 500 t
(a) the greatest and least depths of the ball
Solution: Found 'em to be 110 and 90, respectively, when cos part is -1 or 1.
(b) the time at which the ball first reaches it highest position.
Solution: That would be at 90, which occurs when the cos equation first gets -1, i.e. when t = 180 / 500 = 0.36
(c) the time taken for a complete oscillation
Solution: This would be when t = 360 / 500 = 0.72
(d) the proportion of the time during a complete oscillation for wihch the depth of the ball is less than 99cms.
I need help with (d) part.