Difficult motion question - trial tomoz plz help (1 Viewer)

AGB

Member
Joined
Feb 7, 2003
Messages
860
Gender
Male
HSC
2004
Hey,

I have been trying to get this question out but I really have no idea...

The rise and fall in a sea level due to tides can be modelled by simple harmonic motion. On a certain day, a channel is 10 metres deep at 9am when it is low tide, and 16 metres deep at 4pm when it is high tide. If a ship needs 12 mteres of water to sail down a channel safely, at what times (correct to the nearest minute) between 9am and 9pm can the ship pass through?

If someone could help me out with this I would be eternally grateful :)

Thanks,
AGB
 

Heinz

The Musical Fruit
Joined
Oct 6, 2003
Messages
419
Location
Canberra
Gender
Male
HSC
2004
Low tide is 10 and high is 16m so the amplitude is 3 and the centre of motion is 13 (13 - 3 and 13 + 3). The equation becomes x = 13 - 3cosnt. T= 2pi/n or T/2 = pi/n. Half the period is equal to the time between low tide and high tide (7 hours) and so n = 2pi/7. So x = 13 - 3cos(2*pi*t/7). The ship needs 12 metres so then the equation becomes 12 = 13 - 3cos(2*pi*t/7). cos(2*pi*t/7) = 1/3
t = 7*cos<sup>-1</sup>(1/3)/2pi and 7*[2pi - cos<sup>-1</sup>(1/3)]/2pi which gives times of 1hour and 22 minutes and also 5 hours 38 minutes.

9 + 1hour and 22 minutes = 10:22am
9 + 5 hours 38 minutes = 2:38pm

hope thats right :p

Edit: pic shows y = 13 - 3cos(pi*t/7) intersecting with y = 12
 
Last edited:

AGB

Member
Joined
Feb 7, 2003
Messages
860
Gender
Male
HSC
2004
Hey Hien,

How is cosmo going? haha

Nah sorry the answer is 11:45am to 8:15pm...
 

wogboy

Terminator
Joined
Sep 2, 2002
Messages
654
Location
Sydney
Gender
Male
HSC
2002
pi/n. Half the period is equal to the time between low tide and high tide (7 hours) and so n = 2pi/7
Nope, n = 2pi/14 = pi/7 :)

Then it becomes:
x = 13 - 3cos(pi*t/7)
13 - 3cos(pi*t/7) >= 12
cos(pi*t/7) <= 1/3

t >= 7/pi * arccos(1/3) AND t <= 7/pi * (2pi - arccos(1/3))
t >= 2.742785864 AND t <= 11.25721414

The ship can sail between 11:44:34 and 20:15:26
 

Heinz

The Musical Fruit
Joined
Oct 6, 2003
Messages
419
Location
Canberra
Gender
Male
HSC
2004
Oops, my bad... n = pi/7

so t = 7cos<sup>-1</sup>(1/3)/pi = 2 hours 45 7[2pi - cos<sup>-1</sup>(1/3)]/pi = 11 hours 15. Add those two times to 9 and you should get it :)

Edit: what wogboy said. Have fun in your trials andrew :D
 

AGB

Member
Joined
Feb 7, 2003
Messages
860
Gender
Male
HSC
2004
i dont understand how you derived the original equation of x = 13 - 3cosnt ????

could you please explain this??

cheers
 

Estel

Tutor
Joined
Nov 12, 2003
Messages
1,261
Gender
Male
HSC
2005
We choose 13 as the centre of motion, being halfway between 10m low tide and 16m high tide. Since initially the particle is at an extreme point 3 units away, we choose 3cosnt for purposes of simplicity. And because it begins at the minimum, it's negative 3 cos.

So x = 13 - 3cosnt is your equation.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top