Questions: A buoy oscillates in simple harmonic motion as waves go past. The buoy moves a total of 2.5 feet from its low point to its high point (see figure), and it returns to its high point every 16 seconds. Write an equation that describes the motion of the buoy if its high point is at t=0, in terms of its height h.

A buoy oscillates in simple harmonic motion as waves go past. The buoy moves a total of 2.5 feet from its low point to its high point (see figure), and it returns to its high point every 16 seconds. Write an equation that describes the motion of the buoy if its high point is at t=0, in terms of its height h.

Transcript text: A buoy oscillates in simple harmonic motion as waves go past. The buoy moves a total of 2.5 feet from its low point to its high point (see figure), and it returns to its high point every 16 seconds. Write an equation that describes the motion of the buoy if its high point is at $t=0$, in terms of its height $h$.

Solution

Solution Steps

Step 1: Find the amplitude

The amplitude is half the distance between the high and low points, which is 2.5 ft / 2 = 1.25 ft.

Step 2: Find the period

The buoy returns to its high point every 16 seconds, which is the period.

Step 3: Find the angular frequency

The angular frequency (ω) is related to the period (T) by the formula ω = 2π/T. In this case, ω = 2π/16 = π/8.

Step 4: Determine the phase shift

Since the buoy is at its high point at t = 0, we use the cosine function. There is no horizontal shift, so the phase shift is 0.

Step 5: Write the equation

The equation for simple harmonic motion is of the form h(t) = Acos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase shift. Substituting the values we found, we get:

h(t) = 1.25cos(πt/8)