Question

A buoy oscillates in simple harmonic motion y= A cos cos 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.

Answer 1

Answer:

a) $y = 1.75 ft cos (\frac{\pi}{5}t)$

b) $v(t) = -1.0995 sin (\frac{\pi}{5}t)$

Step-by-step explanation:

Part a

Since the buoy oscillates in simple harmonic motion the equation to model this is given by:

$y = A cos (wt +\theta)$

For this case from the info given we know that:

$2A = 3.5 , A = \frac{3.5}{2}= 1.75 ft$

"It returns to its high point every 10 seconds." That means period =10 $T =10s$, and the angular frequency can be founded like this:

$w = \frac{2\pi}{10}= \frac{\pi}{5}$

Assuming that the value for the phase is $\theta =0$ our model equation is given by:

$y = 1.75 ft cos (\frac{\pi}{5}t)$

Part b

From definition we can obtain the velocity with the derivate of the position function and if w calculate the derivate we got this:

$\frac{dy}{dt}= v(t) = -1.75 ft (\frac{\pi}{5}) sin (\frac{\pi}{5}t)$

$v(t) = -1.0995 sin (\frac{\pi}{5}t)$