At a certain harbour, the tides cause the ocean surface to rise and fall a distance d(from highest level to lowest level) in simple harmonic motion, with a period of 12.5h. How long does it take for the water to fall a distance 0.250d from its highest level?
h = (d/2) cos (2 pi t/ 12.5)
h = .25 d
so
.5 = cos theta
theta = 60 degrees = pi/3
so
2 pi t/12.5 = pi/3
t = 12.5/6
= 2.08 hours
To find the time it takes for the water to fall a distance 0.250d from its highest level, we can use the equation for simple harmonic motion:
d(t) = A * sin(ωt + φ),
where d(t) is the displacement at time t, A is the amplitude (in this case, d), ω is the angular frequency (2π divided by the period, T), and φ is the phase constant.
First, let's find the angular frequency:
ω = 2π / T = 2π / 12.5 = π/6.25
Next, we'll find the time it takes for the water to fall a distance 0.250d. Since the displacement is given by d(t), we can set it equal to 0.250d and solve for t:
0.250d = d * sin(π/6.25 * t + φ)
To simplify the equation, let's assume the phase constant, φ, is zero (as it doesn't affect the time it takes for the water to fall a certain distance):
0.250 = sin(π/6.25 * t)
Now, we need to find the inverse sine (also called arcsine) of both sides to isolate t:
sin^(-1)(0.250) = π/6.25 * t
Using a calculator, we find that the arcsine of 0.250 is approximately 14.477 degrees.
Converting this to radians (since the angular frequency is in radians):
π radians = 180 degrees, so 1 degree = π/180 radians.
Therefore, 14.477 degrees * (π/180 radians/degree) ≈ π/12.4 radians.
Now, we can solve for t:
π/12.4 = π/6.25 * t
To isolate t, we can divide both sides by π/6.25:
t = (π/12.4)/(π/6.25) = (π/12.4) * (6.25/π) = 6.25/12.4 ≈ 0.5040 hours.
So, it takes approximately 0.5040 hours for the water to fall a distance 0.250d from its highest level.
To find how long it takes for the water to fall a distance of 0.250d from its highest level, we need to use the equation for simple harmonic motion:
y = A * sin(2π * f * t + φ)
where:
- y is the displacement from the equilibrium position
- A is the amplitude, which in this case is d (the full distance from highest to lowest level)
- f is the frequency, which is the reciprocal of the period (f = 1 / T, where T is the period)
- t is the time
- φ is the phase angle, which we can ignore in this case since we are starting at the highest level
Given that the period T is 12.5 hours, we can calculate the frequency f:
f = 1 / T = 1 / 12.5 = 0.08 (per hour)
Now we can rearrange the formula to solve for time t when the displacement y is 0.250d:
0.250d = d * sin(2π * 0.08 * t)
To simplify the equation, we can divide both sides by d:
0.250 = sin(2π * 0.08 * t)
To solve for t, we need to take the inverse sine (sin⁻¹) of both sides:
sin⁻¹(0.250) = 2π * 0.08 * t
Now divide both sides by 2π * 0.08:
t = sin⁻¹(0.250) / (2π * 0.08)
Using a calculator, we find:
t ≈ 0.4245407 / (0.502654825) ≈ 0.843 hours
Therefore, it takes approximately 0.843 hours (or 50.6 minutes) for the water to fall a distance of 0.250d from its highest level.