Simple Harmonic Motion

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

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Reply 1

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.

Reply 2

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.

Reply 3

H3n

Is t not supposed to be inside the bracket next to w?

Reply 4

joyoustele

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)

(edited 6 years ago)

Reply 5

joyoustele

Original post by H3n

Is t not supposed to be inside the bracket next to w?

I dont think so, but ill try that

Reply 6

joyoustele

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

Reply 7

niamhdoesmaths

Original post by joyoustele

I dont think so, but ill try that

It is supposed to be inside the bracket :smile:

Edit: This definitely belongs in Physics, not Maths! I’ve moved it across.

(edited 6 years ago)

Reply 8

Texxers

Omg lol OCR A question? I did that one yesterday 😂😂😂

Reply 9

joyoustele

Original post by Protostar

It is supposed to be inside the bracket :smile:

Edit: This definitely belongs in Physics, not Maths! I’ve moved it across.

I tried it, it didnt work

Reply 10

niamhdoesmaths

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...

Reply 11

Shaanv

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

Reply 12

joyoustele

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?

Reply 13

joyoustele

Original post by Shaanv

Oh wow, It worked Thanks so much.

Reply 14

niamhdoesmaths

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?

Reply 15

Kyrpt

Original post by joyoustele

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?

Reply 16

joyoustele

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