On a given port , the tides cause the ocean surface to rise and fall a distance d (from highest level to the lowest level ) in simple harmonic motion with a period of 11.5 hours. How long ( in hours ) it takes for the water to fall a distance d / 4 its highest level ?

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#1**Anonymous**Answered at 2013-05-12 18:48:43

Assuming that at time t = 0 corresponds to a ridge, the equation:

h = ( d / 2 ) cos ( 2πt / 11.5 )

gives the height of the tide above / below average at time t in hours. The peak is at d / 2, so you want to solve

h = d / 2 - d / 4 = d / 4

( d / 2 ) cos ( 2πt / 11.5 ) = d / 4

cos ( 2πt / 11.5 ) = 1/2

The smallest positive value of t also gives the smallest positive argument of the cosine possible , so that the inverse cosine gives the answer directly :

2πt / 11.5 = cos ⁻ ¹ ( 1/2 ) = π / 3

t = 11.5 / 6 .... after multiplying both sides by 11.5 / ( 2π )

t = 23 /12 hour or 1 hour and 55 minutes

h = ( d / 2 ) cos ( 2πt / 11.5 )

gives the height of the tide above / below average at time t in hours. The peak is at d / 2, so you want to solve

h = d / 2 - d / 4 = d / 4

( d / 2 ) cos ( 2πt / 11.5 ) = d / 4

cos ( 2πt / 11.5 ) = 1/2

The smallest positive value of t also gives the smallest positive argument of the cosine possible , so that the inverse cosine gives the answer directly :

2πt / 11.5 = cos ⁻ ¹ ( 1/2 ) = π / 3

t = 11.5 / 6 .... after multiplying both sides by 11.5 / ( 2π )

t = 23 /12 hour or 1 hour and 55 minutes

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