Wave Motion
A buoy oscillates in simple harmonic motion as waves go past. At a given time it is noted that the buoy moves a total of 3.5 feet from its low point to its high point, and that it returns to its high point every 10 seconds. Write an equation that describes the motion of the buoy if it is at its high point at t=0.
(Note: Rationalize and think thoroughly throughout this solving process, don't stress your mind in a negative way. Just relax and look over the information carefully.)
since the amplitude is half the (max-min), and it oscillates about the middle, we can get that by using
y=7/4 cos(t) + 7/4
since cos(t) starts out at its max at t=0. But, that has period 2π. We're almost there. We want a period of 10, so since cos(kt) has period 2π/k,
y = 7/4 cos(2π/10 t) + 7/4
is our function.
To describe the motion of the buoy that oscillates in simple harmonic motion, we can use the equation of a sinusoidal function. Let's break down the given information to find the necessary parameters for the equation.
1. The total distance traveled by the buoy from its low point to its high point is 3.5 feet. This is called the amplitude (A) of the oscillation.
2. The buoy returns to its high point every 10 seconds. This is called the period (T) of the oscillation.
3. The problem states that the buoy is at its high point at t=0, which means the phase shift (ϕ) is 0.
Now we can use this information to write the equation for the motion of the buoy:
y(t) = A * sin(2π/T * t + ϕ)
In this equation:
- y(t) represents the vertical position of the buoy at time t.
- A is the amplitude, which is 3.5 feet in this case.
- T is the period, which is 10 seconds.
- t is the time variable.
- ϕ is the phase shift, which is 0 since the buoy is at its high point at t=0.
Substituting the given values, we get:
y(t) = 3.5 * sin(2π/10 * t + 0)
Simplifying further:
y(t) = 3.5 * sin(0.2πt)
Therefore, the equation that describes the motion of the buoy if it is at its high point at t=0 is y(t) = 3.5 * sin(0.2πt).