Question

A buoy floating in the ocean is bobbing in simple harmonic motion with period 5 seconds and amplitude 3ft . Its displacement d from sea level at time t=0 seconds is −3ft , and initially it moves upward. (Note that upward is the positive direction.) Give the equation modeling the displacement d as a function of time t .

Answer 1

d = 3sin(2π/5 t + 3π/2)

Explanation:

Equation for simple harmonic motion is:

d = A sin(2π/T t + B) + C

where A is the amplitude,

T is the period,

B is the horizontal shift (phase shift),

and C is the vertical shift.

Given that A = 3, T = 5, and C = 0:

d = 3 sin(2π/5 t + B)

At t = 0, the buoy is at d = -3:

-3 = 3 sin(2π/5 (0) + B)

-1 = sin(B)

3π/2 = B

d = 3sin(2π/5 t + 3π/2)

Notice you can also use cosine instead of sine and get a different phase shift.

d = 3 cos(2π/5 t + π)

You can even use phase shift properties to simplify:

d = -3 cos(2π/5 t)

Any of these answers are correct.