A buoy floating in the sea is bobbing in simple harmonic motion with period 1 second and amplitude 11 in. 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.)

Answer: incomplete question

Complete question:

A buoy floating in the sea is bobbing in simple harmonic motion with period 1 second and amplitude 11 in. 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.). Give the equation modelling displacement (d) as a function of time (t)

Answer:

d = - 11sin2πt

Step-by-step explanation:

Given,

Amplitude = 11 in

Period = 1 seconds.

Since, d at t = 0 in

Let displacement be d = asin (wt+∅)

So, a = amplitude = - 11 in (it moves downward)

w = angular velocity

= 2π/period

= 2π/1

= 2πrad/sec

Therefore, at t = 0

So, since sin (∅) = 0

∅ = 0

So, inputing values of a,∅, and w we get

d = - 11sin (2πt + 0)

d = - 11sin2πt