Lily asked in Science & MathematicsPhysics · 4 weeks ago

Advanced Functions Help!?

A large wheel is attached to a boat and spins as the boat moves. A rock becomes nudged in the wheel as it spins in the water. It is noticed that at t = 2 s, the rock is at the highest point 3 m above the water. At time t = 6 seconds, the rock is submerged in the water 5 m below the water(the lowest point).

a. Graph 5 points to represent one cycle of the above problem. Label these points clearly on your graph.

b. Determine a cosine model to represent the motion of the wheel.

c. Determine whether the rock was above the water’s surface at t = 0 or below the water’s surface and how far is the rock above or below the water at this time?

2 Answers

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  • NCS
    Lv 7
    4 weeks ago
    Favourite answer

    I see it this way:

    a. Your five points should be (y, t) such that

    (3, 2), (-5, 6), (3, 10) -- for crest, trough, crest (antinodes)

    and then (-1, 4) and (-1, 8) at the nodes

    where y = distance above surface in meters

    and t = time in seconds

    b. y = 4cos(2πt/8s - φ) - 1

    since the period is 8 seconds (4 s to go from crest to trough)

    At t = 2, y = 3 --

    3 = 4cos(2π*2/8 - φ) - 1

    1 = cos(π/2 - φ)

    0 = π/2 - φ

    φ = π/2

    so your model is

    y = 4cos(πt/4 - π/2) - 1

    c. y = 4cos(0 - π/2) - 1 = -1 m

    1 meter below the surface

    Attachment image
  • 4 weeks ago

     given,

    a large wheel is attached to a boat as the boat moves

    a rock becomes nudged in the wheel as it spins in the water

    at t = 2 , rock is at highest point 3 m above water

    at t = 6 s, rock is submerged in the water 5 m below water

    a. let angular speed of the wheel be w and let w be constant

    then

    angle made by the rock with vertical be theta

    then

    theta = 0 at t= 2 s and theta = pi at t = 6 s

    hence

    4*w = pi

    w = pi/4 rad/s

    hence

    at t = 2, theta = 0

    at t = 3, theta = w*1 = pi/4 rad

    at t = 4, theta = pi/2 rad

    at t = 5, theta = 3*pi/4 rad

    at t = 6, theta = pi rad

    coordinates of points are

    4(sin(0),cos(0)),4 (sin(pi/4),cos(pi/4), 4(sin(pi/2),cos(pi/2))), 4(sin(3*pi/4), cos(3*pi/4)), 4(sin(pi), cos(pi))

    b. let distance above the water be y

    then following cosine model is assumed

    y = A*cos(wt + phi)

    so, y(2) = A*cos(pi/2 + phi) = 3

    y(6) = A*cos(3*pi/2 + phi) = -5

    now

    A*cos(pi/2 + phi) = A(cos(pi/2)cos(phi) - sin(pi/2)sin(phi)) = -A*sin(phi)

    A*cos(3*pi/2 + phi) = A*cos(-pi/2 + phi) = A[cos(3*pi/2)cos(phi) - sin(3*pi/2)sin(phi)] = A*sin(phi)

    hence

    A = 8/2 = 4

    hence

    y = 4*cos(pi*t/4 + phi)

    for t = 2

    y = 4*cos(pi/2 + phi) = 3

    -4*sin(phi) = 3

    phi = -0.84806207 rad

    hence

    y = 4*cos(pi*t/4 -0.848062079)

      

    c. at t = 0

    y = 4*cos(-0.848062079) > 0

    hence the rock is above the water at t = 0 s

    y(0) = 2.6457513110090336146 m

    I owe you the graph .

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