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A buoy floating in the ocean is bobbing in simple harmonic motion with period 7 seconds and amplitude 6ft. Its displacement d from sea level at time t=0 seconds is -6ft, 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.

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MathPhys

Answer:

d = 6 sin(2Ï€/7 t + 3Ï€/2)

Step-by-step explanation:

Equation for simple harmonic motion is:

d = A sin(2Ï€/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 = 6, T = 7, and C = 0:

d = 6 sin(2Ï€/7 t + B)

At t = 0, the buoy is at d = -6:

-6 = 6 sin(2Ï€/7 (0) + B)

-1 = sin(B)

3Ï€/2 = B

d = 6 sin(2Ï€/7 t + 3Ï€/2)

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

d = 6 cos(2π/7 t + π)

You can even use phase shift properties to simplify:

d = -6 cos(2Ï€/7 t)

Any of these answers are correct.

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