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Chapter 2: Problem 108

Wave Motion A buoy oscillates in simple harmonic motion \(y=A \cos \omega t\) as waves move past it. 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 \(t\) .

Short Answer

Expert verified
\(y = 3.5 \cos (0.2\pi t)\) feet; \(v = -3.5 \times 0.2\pi \sin (0.2\pi t)\) feet/second.

Step by step solution

01

Identify the amplitude

The amplitude A of a simple harmonic motion is the maximum displacement from the mean position. In this case, the problem states that the buoy moves a total of 3.5 feet from its low point to its high point. That means the amplitude of the motion \(A\) is \(3.5\) feet.
02

Identify the angular frequency

The problem states that the buoy returns to its high point every 10 seconds, which is the period \(T\) of the motion. The angular frequency \(\omega\) is given by \(\omega = \frac{2\pi}{T}\). Substituting \(T = 10\) seconds, we get \(\omega = \frac{2\pi}{10} = 0.2 \pi\) rad/s.
03

Write the equation of the motion

The equation that describes the motion of the buoy is \(y=A \cos \omega t\). Substituting the values for \(A\) and \(\omega\), we get \(y = 3.5 \cos (0.2\pi t)\) feet.
04

Determine the velocity of the buoy as a function of \(t\)

The velocity \(v\) of the buoy is given by the derivative of the displacement with respect to time. Differentiating \(y = 3.5 \cos (0.2\pi t)\) with respect to \(t\), we get \(v = -3.5 \times 0.2\pi \sin (0.2\pi t)\) feet/second.

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