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
$$y = a\sin b(t-d)+c$$
where:
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:
$$\frac{1}{12} = \frac{b}{2\pi} \iff b = \frac{\pi}{6}$$
Adding all this together, you get
$$y = 5\sin\frac{\pi}{6}(x-3)+11$$
which becomes
$$y = 5\sin\left(\frac{\pi}{6}x-\frac{\pi}{2}\right)+11$$
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
$$y = -5\cos\left(\frac{\pi}{6}x\right)+11$$
Here are the plots of the graphs.