A buoy floating in the sea is bobbing in simple harmonic motion with amplitude 13 in and period 0.25 seconds. Its displacement d from sea level at time t=0 seconds is 0in, and initially it moves downward. (Note that downward is the negative direction.)Required:Give the equation modeling the displacement d as a function of time t.

The equation is

Step-by-step explanation:

From the question we are told that

      The  amplitude is  

       The period is  

Generally the displacement function for a simple harmonic motion is mathematically represented as

Here   is the angular frequency which is mathematically represented as

substituting values

Given that at t =  0  the displacement is equal to 0 it means that there is no phase shift and also  we are told that it is initially moving downward which implies that its Amplitude is  

So the equation modeling the displacement d as a function of time t is mathematically represented as

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92

your answer is

a.

5

step-by-step explanation: