A buoy floating in the ocean is bobbing in simple harmonic motion with amplitude 7ft and period 8 seconds. Its displacement d from sea level at time t=0 seconds is 0 ft, and initially it moves upward. (Note that upward is the positive direction.)
Give the equation modeling the displacement d as a function of time t.
Why did the buoy enroll in a physics class? Because it wanted to learn how to wave hello to all its sea buddies in a harmonious manner!
The equation modeling the buoy's displacement d as a function of time t is:
d(t) = 7cos(2πt/8)
Remember, t is measured in seconds. So, let the buoy's dance begin!
To find the equation modeling the displacement of the buoy as a function of time, we can use the equation for simple harmonic motion:
d(t) = A * sin(ωt + φ)
Where:
- d(t) is the displacement of the buoy from sea level at time t
- A is the amplitude of the motion
- ω is the angular frequency (2π divided by the period)
- φ is the phase constant or initial phase shift
Given:
- A = 7 ft (amplitude)
- Period = 8 seconds
First, we need to find the angular frequency ω. We know that the angular frequency is related to the period by the formula:
ω = 2π / T
Where T is the period. Substituting the given value, we have:
ω = 2π / 8 = π / 4
Now we can write the equation as:
d(t) = 7 * sin((π/4)t + φ)
Next, we need to determine the phase constant φ. It is given that initially, the buoy moves upward and the displacement at t = 0 is 0 ft. When the sine function is equal to 0, the displacement will be 0. Therefore, the phase constant φ is 0.
So, the equation modeling the displacement d as a function of time t is:
d(t) = 7 * sin((π/4)t)