SHM question (1 Viewer)

[Dumbarse]

The deck of a ship was 2.4m below the level of a wharf at low tide and 0.6m above wharf level at high tide. Low tide was at 8.30a.m and high tide at 2.35 p.m.
Find when the deck was level with the wharf, if the motion of the tide was simple harmonic??

[i know how to do it, i just dont have an answer to check with so i wanna know what u get so i can see if i'm right, thanx]

 

[p00_p00]

did it say where it started???

If not ill assume it started at low tide

 

[wogboy]

I got 12:47 pm or to be exact in hh/mm/ss form, 12:47:15.84 pm

Is that correct?

 

[BlackJack]

I got 12.57pm (and 16.92 seconds)...
edit: wrong! not this!

 

[p00_p00]

Originally posted by Dumbarse
The deck of a ship was 2.4m below the level of a wharf at low tide and 0.6m above wharf level at high tide. Low tide was at 8.30a.m and high tide at 2.35 p.m.
Find when the deck was level with the wharf, if the motion of the tide was simple harmonic??

[i know how to do it, i just dont have an answer to check with so i wanna know what u get so i can see if i'm right, thanx]

man u sure thats the right time????


Cause if its between 0800 and 1435 hours, then it means for one complete oscillation from low tide, back 2 low tide takes 12 1/6 hrs. But for my equation I got x=1.5 Cos (nt + & ) -.9
and i got &=0 (assuming i started from low tide) and so when
t=6 1/12, x=.9 I got n= 24/73 ........ but the period for one complete oscillation from low 2 low is 12 1/6 hrs, and my period (that i worked out) is 6 1/12 .......... so is that rite......... so far?

 

[McLake]

OK:

T = 73/6

n = 12/37 pi

t = 0, x = -1.5, a = 1.5

x = a cos (nt + B)
-1.5 = 1.5 cos(0 + B)
-1 = cos B
B = pi

so x = 1.5 cos ([12/37 pi]t + pi)

where x = -0.9
-0.9 = 1.5 cos ([12/37 pi]t + pi)
cos ([12/37 pi]t + pi) = -0.6
[12/37 pi]t + pi = arccos[-0.6]
[12/37 pi]t + pi = pi - 0.927 or pi + 0.927
[12/37 pi]t = +/- 0.927
t = -0.91

so time = 9:24 am

Where's the mistake?

 

[BlackJack]

Wha... Lake?
low tide @ 8.30am
high tide @ 2.35pm
Therefore, median (let x = 0m, 0.9m = level of wharf) @ 11.32am; 30 secs
edit: note heights:
| 1.5m | 0.9m | 0.6m |
(low td) (med) (wharf) (high)
Therefore time passed b/twn med & high tide is 3hrs 2mins 30secs

*insert diag here of sin graph, level of wharf 0.9 metres, high tide @ 1.5 m at t=3hrs 2.5mins. draw line down from where they first meet.*

Since high tide is 0.6m above...
| 0.9m | 0.6m |
(med) (wharf) (high)
Which corresponds to a sin ratio of 9/15

Therefore time passed from when water at median to water at wharf is {[arcsin(9/15)] / [pi/2]} * 3hrs 2mins 30secs.

Add time to 11.32am.30sec
New answer: 12.47pm 16 secs.
EDIT: I suck at simple arithmetics... 2/3 instead of 9/15

 

[spice girl]

I got 17minutes and 16secs after 12:30

it's 12:47:16 pm

Here's why:

Low tide is -2.4m, high is +0.6m, thus amplitude = 1.5m, centre of oscillation = -0.9m
Low tide: 8:30a.m, high: 2.35pm
Period T = 2 * (6hours 5 minutes = 6 1/12) = 12 1/6 hours = 73/6 hours

We begin with low tide, so it looks like a -cosx graph:

disp = -0.9 -(1.5)cos(nt)
where n = 2pi / T = 12pi / 73

disp = 0
1.5cos(12pi * t / 73) = -0.9
cos(12pi*t / 73) = -0.6
12pi*t / 73 = 2.2143 (calc)
t = 4.2877 hours after 8:30a.m
which is 12:47:16 pm

 

[McLake]

Opps!

Wrote n as 12/37 not 12/73

Then I calulated arccos wrong.

SO, it should be ...

T = 73/6

n = 12/73 pi

t = 0, x = -1.5, a = 1.5

x = a cos (nt + B)
-1.5 = 1.5 cos(0 + B)
-1 = cos B
B = pi

so x = 1.5 cos ([12/73 pi]t + pi)

where x = -0.9
-0.9 = 1.5 cos ([12/73 pi]t + pi)
cos ([12/73 pi]t + pi) = -0.6
[12/73 pi]t + pi = arccos[-0.6]
[12/73 pi]t + pi = pi +/- 2.214
[12/73 pi]t = +/- 2.214

t = 4.288 ....

so time is 8:30 + 4hr 17min
which is 12:47pm

THANK GOD THAT'S FIXED ...

Not a good day ...

 

[Dumbarse]

yehha i got that as my answer!, sweet

 

[p00_p00]

Originally posted by spice girl
I got 17minutes and 16secs after 12:30

it's 12:47:16 pm

Here's why:

Low tide is -2.4m, high is +0.6m, thus amplitude = 1.5m, centre of oscillation = -0.9m
Low tide: 8:30a.m, high: 2.35pm
Period T = 2 * (6hours 5 minutes = 6 1/12) = 12 1/6 hours = 73/6 hours

We begin with low tide, so it looks like a -cosx graph:

disp = -0.9 -(1.5)cos(nt)
where n = 2pi / T = 12pi / 73

disp = 0
1.5cos(12pi * t / 73) = -0.9
cos(12pi*t / 73) = -0.6
12pi*t / 73 = 2.2143 (calc)
t = 4.2877 hours after 8:30a.m
which is 12:47:16 pm

Why is it disp = -0.9 -(1.5)cos(nt)
I got x=(1.5)Cos(nt) -.9

 

[BlackJack]

It started at low tide, going up.

Comparing the trig graphs, it matches the -cosx one.

Therefore it's -1.5cosnt -.9

 

[p00_p00]

cheers mate!