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# Simple Harmonic Motion watch

1. 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, But couldnt get the Answer
w is angular frequency
2. Your equation should include a constant offset for the (given) mean water height. If that doesn’t fix it please post your working.
3. Can u post ur working out? Its the easiest way to see where u made a mistake so we can work to fix it.
4. Is t not supposed to be inside the bracket next to w?
5. (Original post by Shaanv)
Can u post ur working out? Its the easiest way to see where u made a mistake so we can work to fix it.
Sorry for late reply, i hadnt seen a notification so i thought no one replied to the post.

my working:

Amplitude =1.6
Time Period=12hours*3600(to convert to seconds)
time elapsed=4 hours

x=Acos(2pi/T)t
x=1.6cos(2pi/(12*3600)* (4*3600)
6. (Original post by H3n)
Is t not supposed to be inside the bracket next to w?
I dont think so, but ill try that
7. (Original post by old_engineer)
Your equation should include a constant offset for the (given) mean water height. If that doesn’t fix it please post your working.
im not sure what you mean by constant offset
8. (Original post by joyoustele)
I dont think so, but ill try that
It is supposed to be inside the bracket

Edit: This definitely belongs in Physics, not Maths! I’ve moved it across.
9. Omg lol OCR A question? I did that one yesterday 😂😂😂
10. (Original post by Protostar)
It is supposed to be inside the bracket

Edit: This definitely belongs in Physics, not Maths! I’ve moved it across.
I tried it, it didnt work
11. (Original post by joyoustele)
I tried it, it didnt work
Your calculator is in radians right? I haven’t actually tried the question myself but surely it must work...
12. (Original post by joyoustele)
I tried it, it didnt work
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
13. (Original post by Protostar)
Your calculator is in radians right? I haven’t actually tried the question myself but surely it must work...
It is in radians. Have you seen my working? is that correct?
14. (Original post by Shaanv)
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
Oh wow, It worked Thanks so much.
15. (Original post by joyoustele)
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?
16. (Original post by joyoustele)
Sorry for late reply, i hadnt seen a notification so i thought no one replied to the post.

my working:

Amplitude =1.6
Time Period=12hours*3600(to convert to seconds)
time elapsed=4 hours

x=Acos(2pi/T)t
x=1.6cos(2pi/(12*3600)* (4*3600)
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?
17. (Original post by Kyrpt)
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
18. (Original post by joyoustele)
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
ahhh thankyou

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