Asked by Sagar
A buoy oscillates in simple harmonic motion y = A cos(ωt) as waves move past it. The buoy moves a total of 2.7 feet (vertically) from its low point to its high point. It returns to its high point every 18 seconds.
(a) Write an equation describing the motion of the buoy if it is at its high point at t = 0.
b)Determine the velocity of the buoy as a function of t.
(a) Write an equation describing the motion of the buoy if it is at its high point at t = 0.
b)Determine the velocity of the buoy as a function of t.
Answers
Answered by
Steve
if it's a max at t=0, it is a cosine curve.
So, y(t) = (2.7/2) cos((2π/18) t)
the velocity is y', so
v(t) = -(1.35)(π/9) sin(π/9 t)
So, y(t) = (2.7/2) cos((2π/18) t)
the velocity is y', so
v(t) = -(1.35)(π/9) sin(π/9 t)
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