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would appreciate some help working out the sine equation for the following question please:

Depth of water is 6m at low tide and 16m at high tide, which is 6 hours later. Assuming the motion of the water is a simple harmonic, draw a graph to show how the water varies with time over a period of 12 hours, commencing at low tide.

We have the general form

$$y = asin b(t-d)+c$$

where:

• 
  $vert avert$ is the amplitude.


• 
  $b$ is the frequency.


• 
  $(d, c)$ is the horizontal and vertical shift.

Ignoring units for now, at low tide, the depth is $6$ and at high tide, the depth is $16$. Therefore, the middle point/equilibrium point is at $frac{16+6}{2} = frac{22}{2} = 11$. From here, you can deduce that:

• 
  $a$ will be the height of the crest or the depth of the trough relative to the equilibrium point. Hence, $a = 16-11 = 5$ or $a = vert 6-11vert = vert -5vert = 5$.


• The graph is shifted vertically. Initially, the low-tide would be located at $-5$ (negative amplitude) but it’s now at $+6$ (low-tide), so $c = 11$.


• It takes $6$ hours to go from low-tide to high-tide. This means it would take $6$ hours to go back to low-tide and complete the cycle. Hence, there is one cycle in $12$ hours. We will treat $12$ as our $2pi$ in normal trigonometry, so we can obtain $b$ through a ratio:

$$frac{1}{12} = frac{b}{2pi} iff b = frac{pi}{6}$$

• The graph is shifted horizontally. In a regular sine graph would start from the point of equilibrium. Here, however, you start from the low-tide. The shift must be by $frac{T}{4} = 3$, so $c = 3$.

Adding all this together, you get

$$y = 5sinfrac{pi}{6}(x-3)+11$$

which becomes

$$y = 5sinleft(frac{pi}{6}x-frac{pi}{2}right)+11$$

If you’ve studied cosine graphs as well, you would notice $c = 3$ is really just $frac{T}{4}$ (a quarter of the period), and the shift $left(x-frac{T}{4}right)$ would produce a negative cosine graph, so you could also use

$$y = -5cosleft(frac{pi}{6}xright)+11$$

Here are the plots of the graphs.