A buoy oscillates in simple harmonic motion according to the motion of waves at sea. An observer notes that the buoy moves a total of 6 feet from its lowest point to its highest point. The buoy returns to its highest point every 15 seconds.
Amplitude: 6 feet
Period: ???
Frequency: ???
Possible equation:???
Period: 2pi/15, or one oscillation every 15 minutes.
Freq: 1/15
Equation: let d be the displacement then
d = 6cos(2pi/15t)
To find the period and frequency of the buoy's motion, we can use the given information. Let's start with the period.
The period (T) is the time it takes for the buoy to complete one full oscillation or cycle. In this case, the buoy returns to its highest point every 15 seconds. So the period (T) is 15 seconds.
The frequency (f) is the number of cycles per unit of time. It is the reciprocal of the period. Therefore, to find the frequency, we can use the formula: f = 1/T.
Substituting the value of T we found earlier:
f = 1/15 s⁻¹
Now let's move on to the amplitude.
The amplitude (A) is the maximum displacement from the equilibrium position (in this case, the lowest point to the highest point). The given amplitude is 6 feet.
Finally, we can use the equation for simple harmonic motion to represent the motion of the buoy:
x(t) = A * cos(2πft)
Where:
x(t) is the displacement of the buoy from its equilibrium position at time t.
A is the amplitude.
f is the frequency.
So for this specific case, the equation would be:
x(t) = 6 * cos(2π * (1/15) * t)
To summarize:
Amplitude: 6 feet
Period: 15 seconds
Frequency: 1/15 s⁻¹
Equation of motion: x(t) = 6 * cos(2π * (1/15) * t)