This is supposed to be a physics question, But maybe the Mathematicians could help out.
The height of the water on a beach can be approximated as simple harmonic motion with a period of 12hours. If the mean water height is 3.5m, the amplitude of the tide is 1.6m, and `high water' occurs at 7am one day, what would you predict the height of the water to be at 11am?
I used the equation, x=Acos(w)t But couldnt get the Answer w is angular frequency
U can keep everything in hours as thats what all the relevant measurements are going to be in. The equilibrium position is at 3.5m.
So if u imagine a graph of Acos(wt) in this situation it would not suffice as the equilibrium point is 0 not 3.5, so ur equation should be along the lines of Acos(wt) + 3.5
U can keep everything in hours as thats what all the relevant measurements are going to be in. The equilibrium position is at 3.5m.
So if u imagine a graph of Acos(wt) in this situation it would not suffice as the equilibrium point is 0 not 3.5, so ur equation should be along the lines of Acos(wt) + 3.5
It is in radians. Have you seen my working? is that correct?
I think that you need to take the 3.5m into account - this is effectively the equilibrium position and so the amplitude will be 1.6m above the 3.5m. So you’d have to add this on somewhere?
I know this sounds dumb but wondering how you got t=4 (i can see it is from 11-7) but dont know why it isnt just 11?
BTW im no expert, but the simple harmonic motion graph is displacement (Y axis) against time (x axis). Because I started when the tide had maximum amplitude at 7am (cos graph) t=0 at 7 am. And therefore at 11am, t has to = 4hours
BTW im no expert, but the simple harmonic motion graph is displacement (Y axis) against time (x axis). Because I started when the tide had maximum amplitude at 7am (cos graph) t=0 at 7 am. And therefore at 11am, t has to = 4hours