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

#### AGB

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

Last edited:

#### AGB

##### Member
Hey Hien,

How is cosmo going? haha

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

#### wogboy

##### Terminator
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
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 #### AGB

##### Member
i dont understand how you derived the original equation of x = 13 - 3cosnt ????

cheers

#### Estel

##### Tutor
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.

#### AGB

##### Member
thanks... i get it now 