Group 1: Sea Port

(a) State the: (i) period (ii) amplitude (iii)phase shift
= 12.4 A = 1.8 C= +4.00


(b) What is the maximum depth of the water? When does it occur?

1.8+3.1= 4.9m = maximum depth


4.9 = 1.8sin[2π (t-4.00) ] +3.1
/12.4

1.8 =1.8sin 12.4 [2π (t-4.00) ]
/12.4
1.8 =1.8sin [24.8π(t-400)] let θ = [24.8π(t-400)]

1.8 = 1.8sinθ
sinθ = 1
θ = 0.8414709848

0.8414709848 = [24.8π (t-4.00)]
/77.91149781 /77.91149781

0.0108003441 + 4.00 = (t–4.00)+4.00

t = 4.01 At 4:01 the water will reach its maximum depth of 4.9 m.


(c) Determine the depth of the water at 5:00 am and at 12:00 noon.

(d) Determine one time where the water is 2.25 meters deep.

So far this is all I understand so if any more of my group members have any corrections for me or anwsers to c or d. Comment and I'll get too them if they are sent early enough. Sergio.