Question

a buoy floating in the sea is bobbing in simple harmonic motion with amplitude 9 in and period 0.25 seconds. its displacement d from sea - level at time t = 0 seconds is 0 in, and initially it moves downward. (note that downward is the negative direction.) give the equation modeling the displacement d as a function of time t. d = 9 sin()

Explanation:

Step1: Recall the general form of simple - harmonic motion

The general form of a simple - harmonic motion equation is $d = A\sin(\omega t+\varphi)$, where $A$ is the amplitude, $\omega$ is the angular frequency, $t$ is the time, and $\varphi$ is the phase shift.

Step2: Determine the amplitude $A$

Given that the amplitude $A = 9$ inches.

Step3: Calculate the angular frequency $\omega$

The formula for the period $T$ and angular frequency $\omega$ is $T=\frac{2\pi}{\omega}$. Given $T = 0.25$ seconds. We can solve for $\omega$:
\[

$$\begin{align*} 0.25&=\frac{2\pi}{\omega}\\ \omega&=\frac{2\pi}{0.25}\\ \omega& = 8\pi \end{align*}$$

\]

Step4: Determine the phase - shift $\varphi$

Since $d(0)=0$ and the buoy initially moves downward. For $y = A\sin(\omega t+\varphi)$, when $t = 0$, $y=0$ and $y^\prime(0)<0$. The sine function $y=\sin x$ has $y = 0$ at $x = k\pi,k\in\mathbb{Z}$. The derivative of $y = A\sin(\omega t+\varphi)$ is $y^\prime=A\omega\cos(\omega t+\varphi)$. When $t = 0$, $y^\prime=A\omega\cos\varphi<0$. If we take $\varphi = 0$, $\cos\varphi=1>0$. If we take $\varphi=\pi$, $\cos\varphi=- 1<0$. So $\varphi=\pi$.

The equation of the simple - harmonic motion is $d = 9\sin(8\pi t+\pi)$.

Answer:

$d = 9\sin(8\pi t+\pi)$