$\text { A buoy oscillates in simple harmonic motion } y = A \cos \text { at as waves move past }$ it. The buoy moves a total of $3.5$ feet (vertically) between its low point and its high point. It returns to its high point every 14 seconds. Write an equation describing the motion of the buoy if it is at its high point at $t = 0$.
A) $y = 3.5 \cos \frac { \pi } { 7 } t$
B) $y = 1.75 \cos 7 \pi t$
C) $y = 1.75 \cos \frac { \pi } { 14 } t$
D) $y = 1.75 \cos \frac { \pi } { 7 } t$
E) $y = 3.5 \cos 14 \pi t$

Question

Quizplus · Accepted Answer

The general equation for simple harmonic motion is given by:

$$y = A\cos(\omega t + \phi)$$

where $A$ is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase angle.

We are given that the buoy oscillates through a total vertical distance of 3.5 feet. Since the motion is symmetric, the amplitude is half of this distance, or:

$$A = \frac{1}{2}(3.5) = 1.75$$

We are also given that the buoy returns to its high point every 14 seconds. This means that one complete cycle of the motion takes 28 seconds (since the buoy also reaches its low point once during this time). The angular frequency is therefore:

$$\omega = \frac{2\pi}{T} = \frac{2\pi}{28} = \frac{\pi}{14}$$

Finally, we are given that the buoy is at its high point when $t = 0$. This means that the phase angle is zero:

$$\phi = 0$$

Substituting these values into the general equation, we get:

$$y = 1.75\cos\left(\frac{\pi}{14}t\right)$$

Thus, the best choice is D.

A buoy oscillates in simple harmonic motion y=Acos⁡ at as waves move past \text { A buoy oscillates in simple harmonic motion } y = A \cos \text { at as waves move past } A buoy oscillates in simple harmonic motion y=Acos at as waves move past

$\text { A buoy oscillates in simple harmonic motion } y = A \cos \text { at as waves move past }$