Question

Stan observes a raft floating on the water bobbing up and down through a total amplitude of 5 feet. Beginning at the top of the wave, if the raft completes a full cycle every 5 seconds, what is the height of the raft relative to its lowest point after 32 seconds?

Answer 1

The raft is performing a form of oscillatory motion; to model it mathematically, we need to assume an expression for its variation with time. A simple choice, applicable in many circumstances, is to take either the sine or cosine waveform as the basis. In this case, where the height of the raft is at a maximum at the beginning of the observations, at t = 0, the cosine is more appropriate, as it avoids the need to include an initial compensating factor or phase shift.

So if we assume that

H(t) = H0.cos(ω.t) + C

where H0 is a measure of the amplitude of the motion, and ω reflects its frequency. Since the cosine function varies from -1 to +1 in each complete cycle, and the total amplitude of the motion of the raft is 5 feet, we should set H0 = 2.5 feet. That is, the excursion of the cosine wave is measured about its mean position, not an extreme one. The constant C adjusts the calculated height H so that this varies from 0 up to 5 feet, rather than from -2.5 to + 2.5 feet. This requires that C = H0, so the equation becomes

H(t) = H0.[cos(ω.t) + 1]

Finally, we need to introduce the angular frequency ω in terms of the measured period of 5 seconds. A cosine wave will perform a complete cycle as ω.t varies from 0 to 2.π radians, in this case ω.5 = 2.π or ω = 2.π/5. So our final equation is

H(t) = H0.[cos(2.π.t/5)] + 1] with H0 = 2.5 ft.

Let's check it: at t = 0, H(0) = 2.5 x [cos(0) + 1] = 5 ft.

Minimum height at t = 2.5 s, Hmin = 2.5 x [cos(2.π.2.5)/5 + 1] = 0 ft

Height after 32 seconds H(32) = 2.5 x [cos(2.π.32/5) + 1] = 0.477 ft

or about 6 inches above its lowest position.

Answer 2

-5cos[2pi/5(32)]=4.04 ft.

Answer 3

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Answer 4

thank you so much

Answer 5

The raft starts at the top of a wave, so every 5 seconds it will be at the top of a wave. After 32 seconds it will have competed 6 full cycles and be 2 seconds into the next cycle. So it will be (2/5) * 2 * pi radians into the seventh cycle. Measuring from the starting point on what is a Cosine wave, it will be Cos (4pi/5) * 2.5 feet from the mid point of the cycle = -2.0225 So it 2.5 - 2.0225 = 0.4775 feet above to lowest point.

Answer 6

The raft is performing a form of oscillatory motion; to model it mathematically, we need to assume an expression for its variation with time. A simple choice, applicable in many circumstances, is to take either the sine or cosine waveform as the basis. In this case, where the height of the raft is at a maximum at the beginning of the observations, at t = 0, the cosine is more appropriate, as it avoids the need to include an initial compensating factor or phase shift. So if we assume that H(t) = H0.cos(ω.t) + C where H0 is a measure of the amplitude of the motion, and ω reflects its frequency. Since the cosine function varies from -1 to +1 in each complete cycle, and the total amplitude of the motion of the raft is 5 feet, we should set H0 = 2.5 feet. That is, the excursion of the cosine wave is measured about its mean position, not an extreme one. The constant C adjusts the calculated height H so that this varies from 0 up to 5 feet, rather than from -2.5 to + 2.5 feet. This requires that C = H0, so the equation becomes H(t) = H0.[cos(ω.t) + 1] Finally, we need to introduce the angular frequency ω in terms of the measured period of 5 seconds. A cosine wave will perform a complete cycle as ω.t varies from 0 to 2.π radians, in this case ω.5 = 2.π or ω = 2.π/5. So our final equation is H(t) = H0.[cos(2.π.t/5)] + 1] with H0 = 2.5 ft. Let's check it: at t = 0, H(0) = 2.5 x [cos(0) + 1] = 5 ft. Minimum height at t = 2.5 s, Hmin = 2.5 x [cos(2.π.2.5)/5 + 1] = 0 ft Height after 32 seconds H(32) = 2.5 x [cos(2.π.32/5) + 1] = 0.477 ft or about 6 inches above its lowest position.