A buoy oscillates in simple harmonic motion y = A cos(ωt) as waves move past it. The buoy moves a total of 2.7 feet (vertically) from its low point to its high point. It returns to its high point every 18 seconds.
(a) Write an equation describing the motion of the buoy if it is at its high point at t = 0.
b)Determine the velocity of the buoy as a function of t.
if it's a max at t=0, it is a cosine curve.
So, y(t) = (2.7/2) cos((2π/18) t)
the velocity is y', so
v(t) = -(1.35)(π/9) sin(π/9 t)
a) Well, well, well, look at Mr. Buoy here, getting all high and mighty! So, the equation describing the motion of this snazzy buoy is y = A cos(ωt), where A represents the amplitude and ω represents the angular frequency. Since the buoy is at its high point when t = 0, we can plug that in to find the equation:
y = A cos(ωt)
y = A cos(ω * 0)
y = A cos(0)
y = A * 1
y = A
So, the equation describing the motion of the buoy when it's at its high point is simply y = A.
b) Now, let's dive into the velocity of this buoy. The velocity of an object undergoing simple harmonic motion can be defined as the derivative of its displacement with respect to time. In other words, we need to find dy/dt.
We know that y = A cos(ωt), so let's differentiate this bad boy with respect to time:
dy/dt = -Aω sin(ωt)
And there you have it! The velocity of the buoy as a function of time is dy/dt = -Aω sin(ωt). Now, you can go impress all your friends with your knowledge of buoy motion and calculus. Enjoy the waves! 🌊
(a) To write an equation describing the motion of the buoy if it is at its high point at t = 0, we can use the equation for simple harmonic motion:
y = A cos(ωt)
Given that the buoy moves a total of 2.7 feet from its low point to its high point, the amplitude, A, is half of that distance:
A = 2.7/2 = 1.35 feet
Since the buoy is at its high point at t = 0, the phase angle, φ, is 0:
φ = 0
Therefore, the equation describing the motion of the buoy is:
y = 1.35 cos(ωt)
(b) To determine the velocity of the buoy as a function of t, we need to find the derivative of the position equation with respect to time (t):
v = dy/dt = d/dt (1.35 cos(ωt))
Using the chain rule, we differentiate the cosine function:
v = -1.35ω sin(ωt)
So, the velocity of the buoy as a function of t is:
v = -1.35ω sin(ωt)
To write an equation describing the motion of the buoy, we need to find the values for amplitude and angular frequency.
(a) The given equation for simple harmonic motion is y = A cos(ωt). In this case, the buoy moves 2.7 feet from its lowest to highest point, so the amplitude A is 2.7 feet.
The period T of the motion is the time it takes for the buoy to complete one full cycle. From the information given, we know the buoy returns to its high point every 18 seconds. The period is related to the angular frequency ω by the formula T = 2π/ω.
So, let's calculate the angular frequency ω first:
T = 18 seconds
2π/ω = 18 seconds
ω = 2π/18 = π/9 radians per second
Now we can write the equation describing the motion of the buoy if it is at its high point at t = 0:
y = A cos(ωt)
y = 2.7 cos(πt/9)
(b) To determine the velocity of the buoy as a function of t, we need to take the derivative of the equation for displacement with respect to time. The derivative of cos(ωt) with respect to t is -ω sin(ωt).
So the velocity v of the buoy can be found by taking the derivative of y with respect to t:
v = dy/dt = d/dt (2.7 cos(πt/9))
v = -2.7(π/9) sin(πt/9)
Therefore, the velocity of the buoy as a function of t is v = -2.7(π/9) sin(πt/9).