In a harbour the cycle of tides can be modelled as SHM with a period of 12 hours 30 minutes. On a certain day high water is 10 m above low water. i) Sketch a graph of the height x m of the water above (or below) the mean level against t, the time in hours since the water was at mean level and

Question

In a harbour the cycle of tides can be modelled as SHM with a period of 12 hours 30 minutes. On a certain day high water is 10 m above low water.
i) Sketch a graph of the height x m of the water above (or below) the mean level against t, the time in hours since the water was at mean level and rising.
ii) Find a suitable expression to model x in terms of t.
iii) Determine for how long the water is more than 6 m above the low water mark.
iv) Find the rate at which the tide is rising or falling when the water is 6 m above the low water mark.
v) Find the maximum rate at which the tide rises.

Accepted Answer

i) figure 14.7

ii) First find ω and a .

The period of the tide T = 12.5 hours.

since [latex]T = \frac{2π}{ω}[/latex]

[latex]ω = \frac{2π}{T}[/latex]

ω = 0.503

From the graph in part i), the amplitude of the oscillation is 5 m, so a = 5.

Since the water has risen x m in t hours after it was at the mean level, a suitable equation is

x = a sin ωt

x = 5 sin (0.503t).

iii) Remember that you are working with time in hours, so any values of t will be in hours. The tide is 6 m above low water when x ≥ 1.

Now x = 1 when 5 sin(0.503t) = 1

⇒ sin(0.503t) = 0.2 = sin 0.201

The first two values of t satisfying this equation are [latex]t_{1}[/latex] and [latex]t_{2}[/latex] where

0.503 [latex]t_{1}[/latex] = 0.201 ⇒ [latex]t_{1}[/latex] = 0.400

and 0.503 [latex]t_{2}[/latex] = π - 0.201 ⇒ [latex]t_{2}[/latex] = 5.846

The water is 6 m above its lowest level for ([latex]t_{2} - t_{1}[/latex]) hours i.e. about [latex]5\frac{1}{2}[/latex] hours.

iv) The water is 6 m above its lowest level when x = 1. The velocity of the oscillation at this point is given by:

v² = ω²(a² - x²)

⇒ v = ±0.503 [latex]\sqrt{25 - 1}[/latex]

= ± 2.46.

The rate of rise or fall is 2.46 m per hour, or about 4 cm per minute.

v) Since v² = ω²(a² - x²), the maximum value of v is aω when x = 0.
The water is rising fastest at the centre of its motion and the rate at which it is rising is :

aω = 0.503 × 5

= 2.515 m per hour

= 4.2 cm per minute.

Note
A tidal range of 10 m is large, but is realistic for several places around the British coastline.
The rate at which the water is rising, 4.2 cm per minute, does not sound high, but if you think about how fast the water would approach your deck chair up a typical beach (with a gradient of say 3°), you will see that it is quite dramatic.

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