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The problem involves a buoy bobbing in simple harmonic motion with amplitude and a period of 5 seconds, with its displacement from sea level, \( d(t) \), modeled as a function of time \( t \). Initially, the buoy moves downward, indicating a negative direction of displacement.

### To model the displacement, we use the general equation for simple harmonic motion:
\[
d(t) = A \cos(\omega t + \phi)
\]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase shift.

Given:
- Period \( T = 5 \) seconds, so \( \omega = \frac{2\pi}{T} = \frac{2\pi}{5} \).
- Initially, the buoy moves downward, so the cosine function will likely have a phase shift of \( \phi = \pi \) to start in the negative direction.

Thus, the equation becomes:
\[
d(t) = A \cos\left(\frac{2\pi}{5} t + \pi\right).
\]

If the amplitude \( A \) is known, it can be substituted into the equation.

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Would you like further clarification on how to apply this model? Here are five follow-up questions for deeper exploration:
1. How do we calculate amplitude in simple harmonic motion problems?
2. What is the significance of phase shift in harmonic motion?
3. How does the angular frequency relate to the period of motion?
4. Can this equation be used for vertical oscillations in other scenarios?
5. How does the initial condition (moving downward) affect the cosine function choice?

**Tip:** Remember, for any simple harmonic motion, the period and angular frequency are inversely related: \( T = \frac{2\pi}{\omega} \).

A buoy floating in the ocean is bobbing in simple harmonic motion with amplitude and period of 5 seconds. Its displacement from sea level at time t seconds is d(t), and initially it moves downward. Give the equation modeling the displacement as a function of time.

Explore how to model the displacement of a buoy bobbing in simple harmonic motion with a period of 5 seconds and an initial downward movement. Learn the key trigonometric concepts and formulas needed to solve the problem.

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