check my answer please :) SIMPLE HARMONIC MOTION -

Question: The rise and fall of the ride at the mouth of a river is assumed to be simple harmonic motion. The depth of water at low tide is 0.7m and the depth at high tide is 3.7m.
If low tide occurred at 8:55am and high tide at 3:05pm, find the earliest time at which the boat could enter the mouth of the river if it required the water to be at least 2m deep.
My solution: http://prntscr.com/7d1ex1

Question

check my answer please :) SIMPLE HARMONIC MOTION -

Question: The rise and fall of the ride at the mouth of a river is assumed to be simple harmonic motion. The depth of water at low tide is 0.7m and the depth at high tide is 3.7m. 
If low tide occurred at 8:55am and high tide at 3:05pm, find the earliest time at which the boat could enter the mouth of the river if it required the water to be at least 2m deep.
My solution: http://prntscr.com/7d1ex1

butterflydreamer · Answer

Low tide at 8:55am is 0.7m
High tide at 3:05 pm is 3.7m
Therefore, centre at 12:00pm will be 2.2m
Amplitude = 1.5 

Period = 2pi/n = ?

irishboy123 · Answer

i suggest you calculate the period as you have the amplitude

the period, T, is time for a complete oscillation.  you go from trough to peak during times 8:55am to 3:05 pm.  isn't that half an oscillation?

once you have T, you can then use $ \omega = \frac{2 \pi}{T}$ in your shm equation: $y - y_0 = A cos( \omega t + \phi)$, if that is how you are presenting it.

butterflydreamer · Answer

ohhh okayyy! 
Give me a minute or so to work it out :)

butterflydreamer · Answer

wait.. if 8:55am - 3:05 pm is HALF an oscillation.. that is 6hrs and 10 minutes...
So it takes 12 hrs and 20 minutes for a complete oscillation..
Therefore w = 740 (12hrs10 mins)
Sooo...
2pi/n = 740
n = pi/370
|dw:1433425233117:dw| 
|dw:1433425384627:dw|