Question

A swimming dock oscillates up and down as waves go by. At time (in seconds) t = 0 the dock is 15 feet above the bottom of the lake. It takes ten seconds to move one cycle (from its highest point at t = 0 back to its next high point at t = 10) It moves down 3.6 feet from high point to low point. In other words it is 11.4 feet above the bottom of the lake at t = 5 seconds.

Answer 1

Answer:

The motion, y of the dock with time , t can be represented as a cosine function as presented in the following equation;

$y = 1.8 \times cos(\frac{2\pi }{10} \cdot t) + 13.2$

Step-by-step explanation:

Here we approximate the motion of the dock as a cosine function as follows

y = a·cos(b(x - c)) + d

a = Amplitude

The period = 2π/b

d = Vertical shift

c = Horizontal shift

Here we have the period given as the time to complete one cycle = 10

The vertical shift, d, is the height above the bottom of the lake + amplitude = 11.4 ft + 3.6 ft/2 =13.2 ft

The amplitude is the distance from the highest point to the neutral point = 3.6/2 = 1.8 ft

Therefore, a = 1.8 ft

Again, we have that at t = 0, we have the dock is at its highest point, therefore, the dock agrees well with the cosine function that has the highest value at 0, therefore c = 0

The cosine function representing the motion of the dock is thus;

y = 1.8·cos(2π/10·t) + 13.2.