Question

A ship needs 11m of water to pass down a channel safely. At low tide, the channel is 6.0m deep and at high tide, 14.0m deep. Low tide is at 5.30am and high tide at 11.30am. Assume that the tidal motion is SHM(simple harmonic motion).

a) State the amplitude and the period of the motion

b) Show that the earliest time that a ship can pass through the entrance is approximately 9am.

Answer 1

High Tide 14 m 11 m Safe Level -5-30 AM 11-30 AM 6 m Low Tide Low Tide

Figure shows variation of water level from low tide to hightide as a function of time.

As seen from figure, starting from low tide to high tide takes half of period of SHM .

Hence period of SHM = 2 $\times$ 6 hour = 12 hour

At low tide water level 6 m , water level at high tide 14 m .

Hence equilibrium position is 10 m and Amplitude = 4 m

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If we consider the starting point of SHM from low tide , we can write the equation of SHM as

h(t) = -4 cos ( 2$\pi$ t / 12 ) + 10 .....................(1)

where t is time in hours and t= 0 is considered at 5-30 AM

since safe level is 11m, we need to find long time duration when water level is above 11 m for the ship to cross the channel.

This can be seen from figure as the marked points A and B .

time for water level to reach 11 m at point A is obtained from eqn.(1) as

11 = - 4 cos ( 2$\pi$ t / 12 ) + 10

cos ( 2$\pi$ t / 12 ) = -0.25 ,   ( 2$\pi$ t / 12 ) = cos-1 (-0.25 ) = 1.823

Hence we get t = 3.48 hr = 3 hr 29 min

Hence warer level 11 m as marked by point A will reach at , ( 5 hr 30 min ) + ( 3 hr 29 min ) $\approx$ 9-00 AM