At an ocean depth of 20 meters, a buoy bobs up and then down 2 meters from the ocean's depth. Four seconds pass from the time the buoy is at its highest point to when it is at its lowest point. Assume at x = 0, the buoy is at normal ocean depth.
The first point must be on the mid-line and the second point must be a maximum or minimum value on the graph closest to the first point.
To solve this problem, we need to understand the equation of motion for the buoy and how it relates to the given information. The equation we will be using is the sinusoidal equation for simple harmonic motion, which describes the up and down motion of the buoy.
Let's break down the given information:
- The buoy bobs up and then down 2 meters from the ocean's depth: This means the amplitude of the motion is 2 meters.
- Four seconds pass from the time the buoy is at its highest point to when it is at its lowest point: This indicates the time period of the motion is 4 seconds.
- At x = 0, the buoy is at normal ocean depth: This tells us that the initial phase of the motion is 0.
Now we can form the equation for the motion of the buoy. The general equation for simple harmonic motion is given by:
y(t) = A * sin(2π/T * (t - t₀))
where:
- y(t) is the displacement of the object at time t
- A is the amplitude of the motion
- T is the time period of the motion
- t₀ is the initial phase or time offset
Substituting the given values into the equation, we have:
y(t) = 2 * sin(2π/4 * (t - 0))
Simplifying further:
y(t) = 2 * sin(π/2 * t)
This is the equation that describes the up and down motion of the buoy.
To find the first point on the mid-line, we set y(t) = 0:
0 = 2 * sin(π/2 * t)
To solve for t, we need to find the values of t that satisfy this equation. Since sin(π/2 * t) = 0 when π/2 * t = nπ (where n is an integer), we have:
π/2 * t = nπ
Solving for t, we get:
t = 2n
So the possible values for t are 0, 2, 4, 6, and so on. However, since we are interested in the first point on the mid-line and the second point closest to the first point, we choose t = 0 and t = 2 as the first and second points, respectively.
Plugging these values into the equation, we get:
y(0) = 2 * sin(π/2 * 0) = 2 * sin(0) = 0
y(2) = 2 * sin(π/2 * 2) = 2 * sin(π) = 0
So the first point on the mid-line is when t = 0, and the second point (a maximum or minimum) closest to the first point is when t = 2. Therefore, the two points are (0, 0) and (2, 0).
I did this a few minutes ago.
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