2007 HSC Q7 (a) Simple Harmonic Motion Question (1 Viewer)

[blackops23]

http://boardofstudies.nsw.edu.au/hsc_exams/hsc2004exams/pdf_doc/maths_ext1_04.pdf


I got 4:28 am -- can anyone check my answer?

My solution:

equation for simple harmonic motion @ harbour entrance (not the wharf) is y = 3cos(4pi(t)/25) + b --> where b is the centre of motion.

As the difference between high tide and low tide is 6m, lets let the height of low tide be y = b-3. Therefore the requisite height of the tide necessary for the ship is y = b - 3 + 2 which is y= b-1

so substituting that into the equation:

b - 1 = 3cos(4pi(t)/25) + b

-1/3 = cos (4pi(t)/25)

t = 3.80 = 3hr 48 min

Now as the tides at harbour entrance occur one hour earlier than the wharf, therefore t = 0 is at 1am --> therefore ship must be at the harbour entrance at 1am + 3hr + 48 min = 4:48 am

It takes 20 min to get there, so ship must leave wharf at 4:28 am.

 

[Assets]

thats piss easy. Im going to wait for one of the 4units to fuck up on this question XD Thoth, care to get on your high horse here you fukin idiot

 

[AAEldar]

Your solution is correct, as is your answer.

 

[blackops23]

Your solution is correct, as is your answer.

fuuuckk that 2mark question took me bloody 11 minutes, mainly because i wrongly put in a "12" instead of a "25" so I got an answer that didn't accord with my graph and it took me 8minutes to realise what i had done wrong-- fuuuuu so stupid, brb gonna fail my hsc

Also, I meant 2004 HSC question lol