# Thread: Wave Motion and Velocity

1. ## Wave Motion and Velocity

A buoy oscillates in simple harmonic motion $\displaystyle y = A cos \omega t$ 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.

2. Originally Posted by seuzy13
A buoy oscillates in simple harmonic motion $\displaystyle y = A cos \omega t$ 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.

A is the amplitude of motion ... the maximum distance from equilibrium.

in this case $\displaystyle A = \frac{3.5}{2} = \frac{7}{4}$

$\displaystyle \omega = \frac{2\pi}{T} = \frac{2\pi}{10} = \frac{\pi}{5}$

$\displaystyle y = \frac{7}{4}\cos\left(\frac{\pi}{5} \cdot t\right)$

the velocity function is the derivative of the position function ...

$\displaystyle v = -\frac{7\pi}{20}\sin\left(\frac{\pi}{5} \cdot t\right)$

3. All makes since except $\displaystyle A = \frac {3.5}{2}$. Why divide by two?

Thanks. You've helped a lot.

4. Originally Posted by seuzy13
All makes since except $\displaystyle A = \frac {3.5}{2}$. Why divide by two?

Thanks. You've helped a lot.
because equilibrium is half-way between the high and low positions

5. Okay. I understand now. Thank you so much!