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# Equation describing the motion of a buoy

Please explain how to solve to the following problem:

A buoy oscillates in simple harmonic motion y = A cos omega(t) The buoy moves a total of 3.5 feet (vertically) from its low point to its high point. It returns to its high point every 10 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

#### Solution Preview

We have the equation of oscillation as,

y = A cos [omega t]

where, A is called the amplitude and omega is the angular frequency.
omega = 2 pi * frequency = 2 pi * (1/T) = 2 pi * (1/10) = 2*3.14*(1/10) = 0.628
or omega = pi/5

T = 10 sec is the period of oscillation as the buoy returns to the high point every ...

#### Solution Summary

The solution includes and explains all steps required (more than 20 detailed steps). Solving problems of this type should be a trivial matter after you have gone through this solution.

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