In a harbour, sea level at low tide is 10m below the level of the sea at high tide. At low tide, the depth of the water in the harbour is 8m. On a particular day, low tide occurs at 1pm and the next high tide occurs at 1:30am. A ship can remain in the harbour safely when the depth of water is at least 12m. The sea level is modelled as rising and falling with simple harmonic motion.
Find the length of time, on this particular day for which it is safe for the ship to remain in the harbour.
I have found the amplitude = 5. and w=2pi/25
However, in the solutions, it uses total time as 12.5 + 2t.
I understand the 12.5 hours is how long high tide lasts for as it will be over 18m for this length of time, but how does the 2t come about?
I would have thought total time would be (12.5 + t) when the tide is 12m and above.
Also, using x=asin(wt), the value of x is taken as 1, am I correct in assuming that x was calculated as being 13-12, where 13 is the middle of the amplitude with 8 being lowest and 18 being highest?
Many Thanks in advance!