M333 @@

the water level in a harbour can be assumed to rise and fall with simple harmonic motion. on a certain day, low tide will occur at 12 noon and the depth of the water will be 3 m. the subsequent high tide will occur at 6.20 pm and the water will then be 15 m deep. A ship which needs 5 m depth of water wishes to use the harbour. Find, to the nearest minute, the earliest time it can enter the harbour on this day and the time by which it must leave.

So hardd I dont get this ?? : (

I'm only covering this in physics now so my explanation may not be the best, but here we go:

Low tide occurs at 12:00, high tide at 18:20, the magnitude of both is 3m and 15m respectively.
The amplitude is therefore 6m, and the time difference is 6:20 = 380 mins
Consider a graph of the information above with 12:00 taken as t = 0. And the x-axis corresponding to a height of 9m (3m+amplitude = 3+6 = 9m). Since low tide occurs at t = 0 and we have a high tide at 380 mins we have a negative cosine graph as follows:



Take the equation $x = -Acos(\omega t)$. Remember that we know the period which is 2x380 mins = 760 mins. Use this to find omega. We want to find t, we know the amplitude and take x as -4 since we are trying to find the time at which the water level is 5m = 9m - 4m.

I hope I've helped. If you need clarification on anything let me know and I will try to explain further. Oh and I got the answer to be time = 13:41;44 (hh:mm;ss).