Shm Q (1 Viewer)

[Hotdog1]

This is from Arnold and Arnold:

The depth of water in a harbour is 7.2 m at low water and 13.6 m at high water. On MOnday, the low water is at 2:05pm and high water at 8:20 pm. The captain of a ship drawing 12.3 m of water want to leave harbour on Monday afternoon. Between what times can he leave on Monday?

How do you make it so that you can find the range in 1 go rather than repeated trials, which is what I did( I got 4 different velues and only 2 fitted the times required).

 

[McLake]

These questions are hard.

Always use RADIANS

Answer:

we know its 7.2 @ 2:05 and 13.6 @ 8:20 so we wnat some time in between where hight is >=12.3.

Period = T = 6hr 15 mins * 2
T = 12.5 hrs
but T = 2(pi)/n
so 2(pi)/n = 12.5
n = 2(pi)/12.5 = 4(pi)/25

NOW

[A = alpha]

x = a*cos(nt + A)
when t = 0, x = -6.4 (if center is origin), a = 6.4
so -6.4 = 6.4 cos (0 + A)
-1 = cosA
A = (pi)

so x = 6.4 cos (4(pi)/25*t + (pi))

NOW
want depth of 12.3 = 1.9 (with origin 10.4m)

so 1.9 = 6.4 cos (4(pi)/25*t + (pi))
19/64 = cos (4(pi)/25*t + (pi))
so (4(pi)/25*t + (pi)) = (pi) - arccos(19/64) or (pi) + arccos(19/64)

so 4(pi)/25*t = +/- arccos(19/64)
t = +/- 2.5253 ....
t = +/- 2hrs 32 mins

so 2:05pm + 2hrs 52mins = 4:57pm
and next low tide (2:35 am)
so 2:35am - 2hrs 52mis = 11:43pm

 

[freaking_out]

hey do u have that cambridge 3u text book??...theres a question like this on the examples.

 

[Hotdog1]

You champion

 

[freaking_out]

Originally posted by Hotdog
You champion

why, thank u .....or were u refering to Mclake?:mad1:

 

[MyLuv]

I dont understand why when t =0 ,x=-6.4

 

[McLake]

Originally posted by MyLuv
I dont understand why when t =0 ,x=-6.4

The origin is set to the average of the high/low tides = 10.4m. At the low tide the water is 6.4m below this origin ...

 

[redslert]

i dont get why u double the times??

6h 15m to 12h 30m???

 

[redslert]

Originally posted by McLake
so 1.9 = 6.4 cos (4(pi)/25*t + (pi))
19/64 = cos (4(pi)/25*t + (pi))
so (4(pi)/25*t + (pi)) = (pi) - arccos(19/64) or (pi) + arccos(19/64)

why is it (pi) +- cos^-1(19/64)
isn't it just cos^-1(19/64)??????

hmmmm but if you are using the cos general solution isn't it:
= n2pi (+-)(cos^-1(x))

so 2:05pm + 2hrs 52mins = 4:57pm
and next low tide (2:35 am)
so 2:35am - 2hrs 52mis = 11:43pm

hmmm where did u get the times for this one??
where did 2hr 52min come from??? and the next low tide?????
basically can u explain slowly, the times, step by step

thank you!!!

and what is the final answer?? between 4.57pm and 11.43pm???
ARRRRR i dont get it!!!!

 

[McLake]

Originally posted by redslert
i dont get why u double the times??

6h 15m to 12h 30m???

Full Period (Up, Down, and Up again)

 

[McLake]

Originally posted by redslert
why is it (pi) +- cos^-1(19/64)
isn't it just cos^-1(19/64)??????

hmmmm but if you are using the cos general solution isn't it:
= n2pi (+-)(cos^-1(x))


hmmm where did u get the times for this one??
where did 2hr 52min come from??? and the next low tide?????
basically can u explain slowly, the times, step by step

thank you!!!

and what is the final answer?? between 4.57pm and 11.43pm???
ARRRRR i dont get it!!!!

2hr 52mis should be 2hr 32mins from the line above (Oops!)

Final Answer is between 4:57 and 11:43.

Not sure why pi +/- cos (), its jsut the method that you use


Ask if you need more help

 

[Constip8edSkunk]

high tide is at 13.6, low tide is at 7.2, taking 10.4 as the origin, wouldnt the amplitude be 3.2?

 

[redslert]

Originally posted by Constip8edSkunk
high tide is at 13.6, low tide is at 7.2, taking 10.4 as the origin, wouldnt the amplitude be 3.2?

hmmmmm i agree

if u take amplitude = a = 6.4
then low tide would be 4m

 

[McLake]

Yes.

Any more mistakes you would care to point out ...