Question

A buoy oscillates in simple harmonic motion y= A cos cos as waves move past it. The buoy moves a total of 3.5 feet (vertically) from its low point to its high point. It returns to its high point every 10 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.

Answer 1

The equation describing the motion of the buoy in simple harmonic motion is y = 1.75 cos(2π/10 t). The velocity of the buoy as a function of time is v = -1.75(2π/10) sin(2π/10 t).

Step-by-step explanation:

(a) Equation describing the motion of the buoy:

The equation for simple harmonic motion is given by y = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, t is time, and φ is the phase angle.

In this case, the buoy moves a total of 3.5 feet vertically, which corresponds to the amplitude A. It returns to its high point every 10 seconds, which corresponds to the period T. The angular frequency ω can be calculated using the formula ω = 2π/T.

Therefore, the equation describing the motion of the buoy is y = 1.75 cos(2π/10 t).

(b) Velocity of the buoy as a function of time:

The velocity of an object in simple harmonic motion can be obtained by taking the derivative of the position equation with respect to time.

For the equation y = A cos(ωt + φ), the velocity is given by v = -Aω sin(ωt + φ).

Therefore, the velocity of the buoy as a function of time is v = -1.75(2π/10) sin(2π/10 t).

Answer 2

a)
`y = 1.75 ft cos ((\pi)/(5)t)`

b)
`v(t) = -1.0995 sin ((\pi)/(5)t)`

Step-by-step explanation:

Part a

Since the buoy oscillates in simple harmonic motion the equation to model this is given by:

`y = A cos (wt +\theta)`

For this case from the info given we know that:

`2A = 3.5 , A = (3.5)/(2)= 1.75 ft`

"It returns to its high point every 10 seconds." That means period =10
`T =10s`, and the angular frequency can be founded like this:

`w = (2\pi)/(10)= (\pi)/(5)`

Assuming that the value for the phase is
`\theta =0` our model equation is given by:

`y = 1.75 ft cos ((\pi)/(5)t)`

Part b

From definition we can obtain the velocity with the derivate of the position function and if w calculate the derivate we got this:

`(dy)/(dt)= v(t) = -1.75 ft ((\pi)/(5)) sin ((\pi)/(5)t)`

`v(t) = -1.0995 sin ((\pi)/(5)t)`